Elementary Algebra

🧭 Overview

🧠 One-sentence thesis

Linear inequalities are solved using properties similar to equations, but multiplying or dividing by a negative number reverses the inequality sign, and solutions are represented on number lines and in interval notation.

📌 Key points (3–5)

• Graphing inequalities: Use parentheses for strict inequalities (< or >) and brackets for inclusive inequalities (≤ or ≥); shade the solution region on a number line.
• Interval notation: Express solution sets using parentheses and brackets with infinity symbols; e.g., (3, ∞) means all numbers greater than 3.
• Addition and subtraction properties: Adding or subtracting the same number from both sides preserves the inequality direction.
• Multiplication and division properties: Multiplying or dividing by a positive number preserves direction; by a negative number reverses direction.
• Common confusion: The inequality sign reverses only when multiplying or dividing by a negative number, not when adding or subtracting negative numbers.

📊 Graphing and notation

📊 Number line representation

An inequality's solution is graphed by shading all values that satisfy it.

• Open parenthesis ( or ): The endpoint is not included (strict inequality).
• Closed bracket [ or ]: The endpoint is included (inclusive inequality).
• Direction: Shade to the right for "greater than," to the left for "less than."

Example: For x > 3, place an open parenthesis at 3 and shade everything to the right.

📐 Interval notation

Interval notation uses parentheses and brackets to describe solution sets compactly.

Inequality	Interval notation	Meaning
x > 3	(3, ∞)	All numbers greater than 3
x ≤ 1	(−∞, 1]	All numbers less than or equal to 1
x ≥ −3	[−3, ∞)	All numbers greater than or equal to −3

• The symbol ∞ (infinity) means "no upper bound"; −∞ means "no lower bound."
• Always use parentheses with ∞ or −∞ because infinity is not a specific number.

Don't confuse: The notation symbols match the graph symbols—parentheses for open endpoints, brackets for closed.

➕ Addition and subtraction properties

➕ How these properties work

For any numbers a, b, and c: if a < b, then a − c < b − c and a + c < b + c. The same holds for >.

• Adding or subtracting the same value from both sides does not change the inequality direction.
• This is true whether the number added or subtracted is positive or negative.

Example: Solve n − 1/2 ≤ 5/8.

• Add 1/2 to both sides: n ≤ 5/8 + 1/2 = 5/8 + 4/8 = 9/8.
• Graph: closed bracket at 9/8, shade left.
• Interval notation: (−∞, 9/8].

➖ Isolating the variable

• Use the same techniques as solving equations: add or subtract to move constants to one side.
• The inequality sign stays the same throughout.

Don't confuse: Subtracting a negative number (e.g., subtracting −5) is the same as adding a positive; the inequality direction still does not change.

✖️ Multiplication and division properties

✖️ Positive multiplier or divisor

For any numbers a, b, and c where c > 0: if a < b, then a/c < b/c and ac < bc.

• Multiplying or dividing both sides by a positive number keeps the inequality direction the same.

Example: Solve 7y < 42.

• Divide both sides by 7 (positive): y < 6.
• Graph: open parenthesis at 6, shade left.
• Interval notation: (−∞, 6).

✖️ Negative multiplier or divisor

For any numbers a, b, and c where c < 0: if a < b, then a/c > b/c and ac > bc.

• Multiplying or dividing both sides by a negative number reverses the inequality sign.

Example: Solve −10a ≥ 50.

• Divide both sides by −10 (negative): a ≤ −5.
• The ≥ becomes ≤.
• Graph: closed bracket at −5, shade left.
• Interval notation: (−∞, −5].

Why the reversal? Consider −2 < 5. Multiply both sides by −1: 2 > −5. The relationship flips.

Don't confuse: Only multiplication/division by a negative reverses the sign; adding or subtracting a negative does not.

🔧 Solving multi-step inequalities

🔧 General strategy

1. Simplify each side (distribute, combine like terms).
2. Collect variable terms on one side using addition/subtraction.
3. Collect constant terms on the other side.
4. Multiply or divide to isolate the variable (watch the sign if dividing by a negative).
5. Graph and write in interval notation.

Example: Solve 4m ≤ 9m + 17.

• Subtract 9m from both sides: −5m ≤ 17.
• Divide by −5 (negative): m ≥ −17/5.
• Interval notation: [−17/5, ∞).

🔧 Identities and contradictions

• Identity: The inequality is true for all real numbers (e.g., −10 < 36 after variables cancel).
  • Solution: all real numbers, (−∞, ∞).
• Contradiction: The inequality is false for all values (e.g., 0 < −18 after variables cancel).
  • Solution: no solution, written as "no solution" or ∅.

Example: Solve 8x − 2(5 − x) < 4(x + 9) + 6x.

• Simplify: 10x − 10 < 10x + 36.
• Subtract 10x: −10 < 36 (always true).
• Solution: all real numbers, (−∞, ∞).

Don't confuse: An identity means every number works; a contradiction means no number works.

🗣️ Translating word problems

🗣️ Common inequality phrases

Phrase	Symbol
is greater than, is more than, exceeds	>
is at least, is no less than, minimum	≥
is less than, is smaller than, is lower than	<
is at most, is no more than, maximum	≤

Example: "Twelve times c is no more than 96."

• Translate: 12c ≤ 96.
• Solve: c ≤ 8.
• Interval notation: (−∞, 8].

🗣️ Solving application problems

1. Read and identify what you're looking for.
2. Choose a variable.
3. Translate the English sentence into an inequality.
4. Solve the inequality.
5. Write the solution in interval notation and graph.
6. Answer with a complete sentence.

Example: "Thirty less than x is at least 45."

• Translate: x − 30 ≥ 45.
• Solve: x ≥ 75.
• Interval notation: [75, ∞).

Don't confuse: "At least 21" means 21 or more (≥ 21), not less than 21.

🧭 Overview

🧠 One-sentence thesis

This section teaches how to solve formulas for a specific variable, apply the distance-rate-time formula to real-world problems, and solve and graph linear inequalities using properties that mirror those for equations—except that multiplying or dividing by a negative number reverses the inequality sign.

📌 Key points (3–5)

• Distance-rate-time formula: d = rt connects distance, rate (speed), and time; you can solve for any one variable if you know the other two.
• Solving formulas for a variable: isolate the desired variable using the same algebraic steps as solving equations (add, subtract, multiply, divide).
• Graphing inequalities: use open parentheses for strict inequalities (< or >) and brackets for inclusive inequalities (≤ or ≥); interval notation matches these symbols.
• Common confusion—inequality direction: when you multiply or divide both sides by a negative number, the inequality sign reverses; with a positive number, it stays the same.
• Translating phrases: "at least" means ≥, "at most" means ≤, "more than" means >, "less than" means <.

🚗 Distance, Rate, and Time Applications

🚗 The d = rt formula

The distance formula: d = rt, where d is distance, r is rate (speed), and t is time.

• This formula connects three quantities; if you know any two, you can find the third.
• Example: If someone drives at 60 miles per hour for 3 hours, distance = 60 × 3 = 180 miles.

🔧 Solving d = rt for different variables

• To solve for t: divide both sides by r, giving t = d/r.
• To solve for r: divide both sides by t, giving r = d/t.
• The excerpt shows practice problems asking to solve both with specific numbers and "in general" (as a formula).
• Example: If distance is 240 miles and rate is 60 mph, then t = 240/60 = 4 hours.

🧮 Other formulas in the exercises

The excerpt includes practice with:

• Area of a triangle: A = (1/2)bh (solve for base b or height h).
• Simple interest: I = Prt (solve for principal P, rate r, or time t).
• Perimeter of a rectangle: P = 2L + 2W (solve for length L or width W).
• Volume of a box: V = LWH (solve for any dimension).

Don't confuse: solving "when x = 3" (substitute first, then solve) vs. solving "in general" (isolate the variable algebraically without substituting).

📏 Graphing and Notating Inequalities

📏 Number line conventions

• Open parenthesis ( or ): the endpoint is not included (for < or >).
• Bracket [ or ]: the endpoint is included (for ≤ or ≥).
• Shade to the right for "greater than," to the left for "less than."
• Example: x > 3 is graphed with an open circle at 3 and shading to the right.

📐 Interval notation

Interval notation uses parentheses and brackets to describe solution sets, with ∞ (infinity) or −∞ (negative infinity) for unbounded ends.

Inequality	Interval Notation	Meaning
x > 3	(3, ∞)	All numbers greater than 3
x ≤ 1	(−∞, 1]	All numbers less than or equal to 1
x ≥ −3	[−3, ∞)	All numbers greater than or equal to −3

• The parenthesis/bracket in interval notation matches the symbol on the number line.
• Infinity always gets a parenthesis, never a bracket (it is not a number you can reach).

➕➖ Solving Inequalities: Addition and Subtraction

➕ Addition and Subtraction Properties of Inequality

For any numbers a, b, and c: if a < b, then a + c < b + c and a − c < b − c (and similarly for >).

• These properties work exactly like the equality properties: the inequality sign stays the same when you add or subtract.
• Example: To solve x + 5 > 9, subtract 5 from both sides: x > 4.

🔍 Step-by-step process

1. Isolate the variable by adding or subtracting the same quantity from both sides.
2. Simplify.
3. Graph the solution on the number line.
4. Write the solution in interval notation.

Example from the excerpt: Solve n − 1/2 ≤ 5/8.

• Add 1/2 to both sides.
• Simplify to get n ≤ (some value).
• Graph with a bracket at the endpoint, shading left.

✖️➗ Solving Inequalities: Multiplication and Division

✖️ The critical rule: negative multipliers reverse the inequality

When you divide or multiply an inequality by a positive number, the inequality stays the same. When you divide or multiply by a negative number, the inequality reverses.

Operation	Effect on Inequality
Multiply/divide by positive	Sign stays the same
Multiply/divide by negative	Sign reverses

• Example: Solve 7y < 42. Divide both sides by 7 (positive), so y < 6 (sign stays).
• Example: Solve −10a ≥ 50. Divide both sides by −10 (negative), so a ≤ −5 (sign reverses).

⚠️ Common mistake

Don't forget to reverse the inequality when multiplying or dividing by a negative number—this is the most common error.

🔄 Variable on the right side

• If you end up with something like −20 < (4/5)u, you can rewrite it as u > −25 (flip the entire inequality).
• Think: "If Xavier is taller than Alex, then Alex is shorter than Xavier."

🧩 Multi-step Inequalities and Special Cases

🧩 Solving complex inequalities

Follow the same steps as solving equations:

1. Simplify each side (distribute, combine like terms).
2. Collect variable terms on one side, constants on the other.
3. Isolate the variable.
4. Remember: reverse the inequality if you multiply or divide by a negative.

Example from the excerpt: Solve 8p + 3(p − 12) > 7p − 28.

• Distribute: 8p + 3p − 36 > 7p − 28.
• Combine: 11p − 36 > 7p − 28.
• Subtract 7p: 4p − 36 > −28.
• Add 36: 4p > 8.
• Divide by 4: p > 2.

🔁 Identities and contradictions

• Identity: the variable cancels and you get a true statement (like −10 < 36) → solution is all real numbers, (−∞, ∞).
• Contradiction: the variable cancels and you get a false statement → no solution.
• Don't confuse: these are the same concepts as with equations, just applied to inequalities.

🗣️ Translating Words to Inequalities

🗣️ Key phrases and their symbols

Phrase	Symbol	Example
is greater than, is more than, exceeds	>	x is greater than 5: x > 5
is at least, is no less than, minimum	≥	x is at least 21: x ≥ 21
is less than, is smaller than, is fewer than	<	x is less than 10: x < 10
is at most, is no more than, maximum	≤	x is at most 8: x ≤ 8

🧠 Understanding "at least" and "at most"

• "At least 21 years old" means 21 or more: age ≥ 21.
• "At most 8 items" means 8 or fewer: items ≤ 8.
• These phrases include the boundary value (use ≥ or ≤, not strict inequalities).

📝 Translation examples from the excerpt

• "Twelve times c is no more than 96" → 12c ≤ 96 → c ≤ 8.
• "Thirty less than x is at least 45" → x − 30 ≥ 45 → x ≥ 75.
• "Twenty times y is at most 100" → 20y ≤ 100 → y ≤ 5.

🧭 Overview

🧠 One-sentence thesis

This chapter provides a systematic strategy for solving linear equations and inequalities by applying properties of equality and inequality, then extends these techniques to word problems and formula manipulation.

📌 Key points (3–5)

• Core solving tools: Addition, Subtraction, Multiplication, and Division Properties of Equality let you perform the same operation on both sides of an equation to isolate the variable.
• General strategy: Simplify each side, collect variable terms on one side and constants on the other, then make the coefficient equal to 1.
• Equation types: A conditional equation is true for specific values; an identity is true for all values; a contradiction has no solution.
• Common confusion: When solving inequalities, multiplying or dividing by a negative number reverses the inequality sign (e.g., if you divide both sides of an inequality by −2, flip < to >).
• Word-problem approach: Read carefully, identify what you're looking for, name it with a variable, translate into an equation, solve, check, and answer in a complete sentence.

🔧 Properties of Equality

➕ Addition and Subtraction Properties

Addition Property of Equality: For any numbers a, b, and c, if a = b, then a + c = b + c.

Subtraction Property of Equality: For any numbers a, b, and c, if a = b, then a − c = b − c.

• These properties say you can add or subtract the same number from both sides without changing the equality.
• Why it works: The two sides stay balanced, like a scale.
• Example: To solve x + 7 = 19, subtract 7 from both sides → x = 12.

✖️ Multiplication and Division Properties

Multiplication Property of Equality: For any numbers a, b, and c, if a = b, then a·c = b·c.

Division Property of Equality: For any numbers a, b, and c (c ≠ 0), if a = b, then a/c = b/c.

• You can multiply or divide both sides by the same non-zero number.
• Example: To solve 8x = 72, divide both sides by 8 → x = 9.
• Don't confuse: Division by zero is undefined; you cannot divide both sides by 0.

🧩 General Strategy for Solving Equations

📝 Step-by-step process

1. Simplify each side: Use the Distributive Property to remove parentheses; combine like terms.
2. Collect variable terms: Use Addition or Subtraction Property to get all variable terms on one side.
3. Collect constants: Move all constant terms to the other side.
4. Make the coefficient 1: Use Multiplication or Division Property so the variable stands alone.
5. Check: Substitute your solution back into the original equation to verify.

🔀 Variables and constants on both sides

• Choose a "variable side" and a "constant side" to organize your work.
• Example: For 7y = 6y − 13, subtract 6y from both sides → y = −13.
• Why this matters: Keeping terms organized prevents sign errors.

🧪 Classifying equations

Type	Definition	Solution
Conditional	True only for specific values	One or more specific numbers
Identity	True for all values	All real numbers
Contradiction	False for all values	No solution

• Example of identity: 2(x + 3) = 2x + 6 simplifies to 6 = 6 (always true).
• Example of contradiction: x + 1 = x + 2 simplifies to 1 = 2 (never true).

🧮 Equations with Fractions and Decimals

🍰 Fraction coefficients

• Strategy: Find the least common denominator (LCD) of all fractions, then multiply both sides by the LCD to clear the fractions.
• Example: For (2/5)n − (1/10) = (7/10), the LCD is 10. Multiply every term by 10 to eliminate denominators.
• Why it helps: Working with whole-number coefficients is simpler and reduces errors.

💵 Decimal coefficients

• You can solve directly with decimals or multiply both sides by a power of 10 to convert to whole numbers.
• Example: For 0.8x − 0.3 = 0.7x + 0.2, you can multiply by 10 to get 8x − 3 = 7x + 2.

📐 Solving Formulas for a Specific Variable

🔄 Rearranging formulas

To solve a formula for a specific variable: isolate that variable on one side with a coefficient of 1, moving all other variables and constants to the other side.

• Use the same properties of equality as for numerical equations.
• Example: The distance formula is d = r·t. To solve for t, divide both sides by r → t = d/r.

🚗 Distance, Rate, and Time

Formula: d = r·t, where d = distance, r = rate (speed), t = time.

• If you know any two quantities, you can find the third.
• Example: If Natalie drove 7.5 hours at 60 miles per hour, distance = 60 × 7.5 = 450 miles.
• Don't confuse: Rate is speed (miles per hour), not total distance.

📊 Solving Linear Inequalities

⚖️ Properties of Inequality

• Addition and Subtraction: For any a, b, c, if a < b then a + c < b + c (same for subtraction).
• Multiplication and Division by positive: If a < b and c > 0, then a·c < b·c and a/c < b/c (inequality stays the same).
• Multiplication and Division by negative: If a < b and c < 0, then a·c > b·c and a/c > b/c (inequality reverses).

🔁 When to flip the inequality

• Key rule: When you multiply or divide both sides by a negative number, reverse the inequality sign.
• Example: Solve −3x > 12. Divide both sides by −3 → x < −4 (the > flips to <).
• Common confusion: Forgetting to flip the sign when dividing by a negative leads to wrong answers.

📏 Graphing and interval notation

• Graph the solution on a number line: use an open circle for < or >, a closed circle for ≤ or ≥.
• Interval notation: (a, b) means all numbers between a and b, not including endpoints; [a, b] includes endpoints.
• Example: x ≥ −2 is graphed with a closed circle at −2 and shading to the right; in interval notation: [−2, ∞).

🧠 Word-Problem Strategy

📖 Seven-step approach

1. Read the problem carefully; look up unfamiliar words.
2. Identify what you are looking for.
3. Name it with a variable (choose a letter that makes sense).
4. Translate the problem into an equation (restate in one sentence if helpful; look for clue words).
5. Solve the equation using good algebra techniques.
6. Check that your answer makes sense in the original problem.
7. Answer the question with a complete sentence.

🔑 Translation tips

• Clue words: "sum" → addition; "difference" → subtraction; "product" → multiplication; "quotient" → division; "is" or "equals" → =.
• Example: "The sum of twice a number and seven is 15" translates to 2n + 7 = 15.
• Don't confuse: "Difference of a and b" means a − b (order matters in subtraction).

🎯 Positive mindset

• The excerpt emphasizes replacing negative thoughts ("I can't do word problems") with positive ones ("I can learn this step by step").
• Why it matters: Confidence and a calm approach improve problem-solving success.
• Think of skills you've mastered over time; word problems are no different—practice builds competence.

🧭 Overview

🧠 One-sentence thesis

Number word problems can be solved systematically by translating English sentences into algebraic equations, defining unknowns carefully in terms of a single variable, and following a consistent seven-step problem-solving strategy.

📌 Key points (3–5)

• The seven-step strategy: Read, identify what you're looking for, name variables, translate to an equation, solve, check, and answer the question.
• Single-variable approach: Even when finding multiple numbers, define all unknowns in terms of one variable (e.g., if one number is "five more than another," use n and n + 5).
• Consecutive integer patterns: Consecutive integers differ by 1 (n, n+1, n+2); consecutive even or odd integers both differ by 2 (n, n+2, n+4).
• Common confusion: Adding 2 to get from one odd integer to the next may seem strange, but the pattern is the same for both even and odd—always add 2.
• Why it matters: This structured approach turns confusing word problems into manageable algebraic equations that can be solved with familiar techniques.

🔢 The seven-step problem-solving strategy

📖 Step 1: Read the problem

• Read carefully to understand what the problem is asking.
• This step ensures you grasp the situation before attempting to solve.

🎯 Step 2: Identify what you are looking for

• Clearly state what quantity or quantities you need to find.
• Example: "the number," "two consecutive integers," "how much the husband earns."

🏷️ Step 3: Name the variable(s)

• Choose a variable to represent the unknown.
• If multiple unknowns exist, express them all in terms of one variable using the relationships given in the problem.
• Example: Let n = first number; if "one number is five more than another," then n + 5 = second number.

🔄 Step 4: Translate into an equation

• Restate the problem as one sentence containing all important information.
• Translate that sentence into an algebraic equation.
• Substitute any variable expressions you defined in Step 3.

⚙️ Step 5: Solve the equation

• Use algebraic techniques: combine like terms, add/subtract from both sides, divide/multiply both sides.
• Find the value of your variable.
• If needed, calculate the other unknowns using the expressions from Step 3.

✅ Step 6: Check your answer

• Verify that your solution satisfies all conditions stated in the original problem.
• Substitute back into the problem context, not just the equation.

💬 Step 7: Answer the question

• Write a complete sentence that directly answers what was asked.
• Example: "The number is 19" or "The three consecutive integers are −13, −14, and −15."

🔗 Defining multiple unknowns with one variable

🧩 Why use one variable for multiple numbers

• The excerpt emphasizes avoiding multiple variables because "so far we have only solved equations with one variable."
• Instead, read the problem carefully to discover how all numbers relate to each other.
• Define the second (or third) number in terms of the first.

🔗 Common relationship patterns

Relationship phrase	First number	Second number
"One number is five more than another"	n	n + 5
"One number is four less than the other"	n	n − 4
"One number is ten more than twice another"	x	2x + 10
"$16,000 less than twice what her husband earns"	h (husband)	2h − 16,000 (wife)

• Example: If the sum of two numbers is 21 and one is five more than the other, let n = first number and n + 5 = second number, so n + (n + 5) = 21.

🧮 Solving for all unknowns

• After solving for your variable, substitute back into the expressions to find the other numbers.
• Example: If n = 8, then the second number is n + 5 = 8 + 5 = 13.

🔢 Consecutive integer problems

🔢 Consecutive integers

Consecutive integers: integers that immediately follow each other.

• Examples: 1, 2, 3, 4 or −10, −9, −8, −7 or 150, 151, 152, 153.
• Each number is one more than the preceding number.
• Pattern: If the first integer is n, then:
  • First integer: n
  • Second consecutive integer: n + 1
  • Third consecutive integer: n + 2
  • And so on.

🔢 Consecutive even integers

Consecutive even integers: even integers that immediately follow one another.

• Examples: 18, 20, 22 or 64, 66, 68 or −12, −10, −8.
• Each integer is 2 more than the preceding one.
• Pattern: If the first even integer is n, then:
  • First even integer: n
  • Second consecutive even integer: n + 2
  • Third consecutive even integer: n + 4

🔢 Consecutive odd integers

Consecutive odd integers: odd integers that immediately follow one another.

• Examples: 77, 79, 81.
• Each integer is also 2 more than the preceding one.
• Pattern: If the first odd integer is n, then:
  • First odd integer: n
  • Second consecutive odd integer: n + 2
  • Third consecutive odd integer: n + 4

❓ Don't confuse: Why add 2 to get from one odd to the next?

• The excerpt addresses this directly: "Does it seem strange to add 2 (an even number) to get from one odd integer to the next?"
• When you add 2 to any odd number, you get the next odd number (e.g., 3 + 2 = 5, 11 + 2 = 13, 47 + 2 = 49).
• Key point: "Whether the problem asks for consecutive even numbers or odd numbers, you don't have to do anything different. The pattern is still the same—to get from one odd or one even integer to the next, add 2."

📝 Worked example patterns

📝 Basic "find the number" problems

• Example from excerpt: "The difference of a number and eight is 17. Find the number."
• Let n = the number.
• Translate: n − 8 = 17.
• Solve: n = 25 (not shown in excerpt, but follows the pattern).

📝 Two-number sum problems

• Example: "One number is five more than another. The sum of the numbers is 21."
• Let n = first number, n + 5 = second number.
• Equation: n + (n + 5) = 21.
• Solve: 2n + 5 = 21 → 2n = 16 → n = 8.
• Second number: 8 + 5 = 13.
• Check: Is 13 five more than 8? Yes. Is 8 + 13 = 21? Yes.

📝 Consecutive integer sum problems

• Example: "Find three consecutive integers whose sum is −42."
• Let n = first integer, n + 1 = second, n + 2 = third.
• Equation: n + (n + 1) + (n + 2) = −42.
• Solve: 3n + 3 = −42 → 3n = −45 → n = −15.
• The integers are −15, −14, −13.
• Check: (−15) + (−14) + (−13) = −42. ✓

📝 Consecutive even integer problems

• Example: "Find three consecutive even integers whose sum is 84."
• Let n = first even integer, n + 2 = second, n + 4 = third.
• Equation: n + (n + 2) + (n + 4) = 84.
• Solve: 3n + 6 = 84 → 3n = 78 → n = 26.
• The integers are 26, 28, 30.
• Check: 26 + 28 + 30 = 84. ✓

📝 Real-world application problems

• Example: "A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?"
• Let h = amount husband earns.
• Then 2h − 16,000 = amount wife earns.
• Equation: h + (2h − 16,000) = 110,000.
• Solve: 3h − 16,000 = 110,000 → 3h = 126,000 → h = 42,000.
• Husband earns $42,000; wife earns 2(42,000) − 16,000 = $68,000.
• Check: $42,000 + $68,000 = $110,000. ✓

💡 Tips for success

💡 Write out all steps initially

• The excerpt notes: "Did you notice that we left out some of the steps as we solved this equation? If you're not yet ready to leave out these steps, write down as many as you need."
• Don't skip steps until you're confident.

💡 Always check in the original problem context

• Don't just verify that your equation is solved correctly.
• Verify that the numbers satisfy the relationships described in the word problem.
• Example: Check both "Is 13 five more than 8?" and "Is the sum 21?"

💡 Answer the actual question asked

• Step 7 requires a complete sentence.
• Make sure you're answering what was asked, not just stating what n equals.

🧭 Overview

🧠 One-sentence thesis

Percent equations translate directly from English sentences into algebra, enabling systematic solutions to everyday problems involving tips, interest, discounts, and mark-ups by identifying the unknown quantity and applying the appropriate percent relationship.

📌 Key points (3–5)

• Translation method: Convert percent word problems into algebraic equations using "is" = equals and "of" = multiply, then solve for the unknown variable.
• Three basic percent patterns: finding a percent of a number, finding what percent one number is of another, and finding the whole when given a percent and part.
• Percent increase/decrease: Calculate the change amount first, then express it as a percent of the original amount (not the new amount).
• Common confusion: In discount problems, the sale price equals original price minus discount; in mark-up problems, the list price equals original cost plus mark-up—don't reverse the operations.
• Simple interest formula: I = Prt connects principal, rate (as decimal), time (in years), and interest earned.

🔤 Translating percent sentences into equations

🔤 The basic translation rules

Translation key: "is" means equals (=); "of" means multiply (×); "what" or unknown quantities become variables.

• Choose a variable that reminds you of what you're finding (n for number, p for percent, t for tip, etc.).
• Always convert percents to decimals before calculating: 35% becomes 0.35.
• The structure of the English sentence directly maps to the equation structure.

📝 Three fundamental patterns

Pattern	Example sentence	Equation form	What you solve for
Part unknown	What number is 35% of 90?	n = 0.35 × 90	The part (result)
Whole unknown	6.5% of what number is $1.17?	0.065 × n = 1.17	The whole (base)
Percent unknown	144 is what percent of 96?	144 = p × 96	The percent (rate)

Example: "What number is 45% of 80?" translates to n = 0.45 × 80, giving n = 36.

Don't confuse: When finding percent, your final answer must be converted to percent form (multiply the decimal by 100 and add the % symbol).

🧮 Real-world percent applications

💵 Tips and basic percentages

The problem-solving strategy follows seven steps: Read → Identify what you're looking for → Name it with a variable → Translate to equation → Solve → Check if it makes sense → Answer in a complete sentence.

Example: A restaurant bill is $68.50 and you want to leave an 18% tip.

• Let t = amount of tip
• The tip is 18% of the total bill
• t = 0.18 × 68.50
• t = 12.33
• The tip should be $12.33

🥣 Finding the whole from a part

When you know a part and what percent it represents, solve for the total recommended amount.

Example: One serving provides 85 mg of potassium, which is 2% of the daily amount.

• Let a = total daily amount
• 85 is 2% of the total amount
• 85 = 0.02 × a
• a = 85 ÷ 0.02 = 4,250 mg

📊 Finding what percent

When comparing two quantities, find what percent one is of the other.

Example: A brownie has 480 total calories and 240 calories from fat. What percent is from fat?

• Let p = percent from fat
• 240 is what percent of 480?
• 240 = p × 480
• p = 240 ÷ 480 = 0.5 = 50%

Check tip: Does the answer make sense with your everyday experience? 240 is half of 480, so 50% is reasonable.

📈 Percent increase and decrease

📈 Calculating percent increase

Percent increase formula: First find the increase amount (new amount − original amount), then find what percent the increase is of the original amount.

Example: Community college fees rise from $26 to $36 per unit.

• Amount of increase = 36 − 26 = 10
• What percent is 10 of the original 26?
• 10 = p × 26
• p = 10 ÷ 26 ≈ 0.385 = 38.5%

Key point: Always calculate the percent based on the original amount, not the new amount.

📉 Calculating percent decrease

Percent decrease formula: First find the decrease amount (original amount − new amount), then find what percent the decrease is of the original amount.

Example: Gas price drops from $3.71 to $3.64 per gallon.

• Amount of decrease = 3.71 − 3.64 = 0.07
• What percent is 0.07 of the original 3.71?
• 0.07 = p × 3.71
• p = 0.07 ÷ 3.71 ≈ 0.019 = 1.9%

Don't confuse: The base for the percent calculation is always the original (earlier) amount, whether calculating increase or decrease.

💰 Simple interest applications

💰 The simple interest formula

Simple interest: I = Prt, where I = interest earned, P = principal (amount deposited or borrowed), r = annual interest rate (as a decimal), t = time in years.

• Organize information in a chart with I, P, r, and t to see what's given and what's unknown.
• Substitute known values and solve for the unknown variable.
• Interest rates must be converted to decimal form (4% becomes 0.04).

💵 Finding interest earned

Example: Nathaly deposits $12,500 at 4% interest for 5 years. How much interest does she earn?

• I = ?, P = $12,500, r = 4% = 0.04, t = 5 years
• I = Prt = (12,500)(0.04)(5)
• I = $2,500

📊 Finding the interest rate

Example: A loan of $3,000 is repaid with $660 interest after 4 years. What was the rate?

• I = $660, P = $3,000, r = ?, t = 4 years
• 660 = (3,000)(r)(4)
• 660 = 12,000r
• r = 660 ÷ 12,000 = 0.055 = 5.5%

💳 Finding the principal

Example: A car loan at 7.5% interest for 5 years costs $6,596.25 in interest. How much was borrowed?

• I = $6,596.25, P = ?, r = 0.075, t = 5
• 6,596.25 = P(0.075)(5)
• 6,596.25 = 0.375P
• P = 6,596.25 ÷ 0.375 = $17,590

🏷️ Discount and mark-up applications

🏷️ Discount problems

Discount formulas:

• Amount of discount = discount rate × original price
• Sale price = original price − amount of discount

Example: A dress originally priced at $140 is discounted 35%.

• Amount of discount = 0.35 × 140 = $49
• Sale price = 140 − 49 = $91

Sanity check: The sale price should always be less than the original price.

🔍 Finding the discount rate

When you know the original price and sale price, work backwards.

Example: A swimsuit originally $31 is on sale for $13.95.

• Amount of discount = 31 − 13.95 = $17.05
• What percent is 17.05 of 31?
• 17.05 = r × 31
• r = 17.05 ÷ 31 ≈ 0.55 = 55%

💲 Mark-up problems

Mark-up formulas:

• Amount of mark-up = mark-up rate × original cost
• List price = original cost + amount of mark-up

Example: A gallery buys a photograph for $250 and marks it up 40%.

• Amount of mark-up = 0.40 × 250 = $100
• List price = 250 + 100 = $350

Don't confuse: Mark-up adds to the original cost (list price is higher); discount subtracts from the original price (sale price is lower). The operations go in opposite directions.

🧭 Overview

🧠 One-sentence thesis

Coin word problems can be systematically solved by organizing information into a table that multiplies the number of each coin type by its value to find total value, then translating the relationships into an equation.

📌 Key points (3–5)

• Core model: For any coin type, number × value = total value; sum all coin types to get the total.
• Organization strategy: Use a table with columns for type, number, value, and total value to structure the problem.
• Variable assignment: Define one unknown quantity with a variable, then express other quantities in terms of that variable based on the problem's relationships.
• Common confusion: Don't confuse the number of coins (how many) with the value of coins (how much they're worth)—you must multiply them together.
• Seven-step process: Read, identify, name, translate, solve, check, and answer ensures complete solutions.

🪙 The fundamental coin value model

🪙 How coin value works

Total Value of Coins: For the same type of coin, the total value equals number · value = total value, where number is the count of coins, value is the worth of each coin, and total value is the combined worth of all those coins.

• This is not about counting coins alone—it's about calculating their monetary worth.
• Each coin type has a fixed value: quarters = $0.25, dimes = $0.10, nickels = $0.05, pennies = $0.01.
• Example: 17 dimes means 17 × $0.10 = $1.70 total value for all dimes.

📊 Why tables matter

• A table organizes all coin types in one place with clear columns.
• The structure prevents mixing up numbers with values.
• You can see at a glance: what you have (number), what each is worth (value), and the contribution to the total (total value).

🔢 Setting up coin problems

🔢 The table structure

Every coin problem uses this four-column format:

Type	Number	Value	Total Value
(coin name)	(count or variable)	(fixed $ amount)	(number × value)

• Fill in known information first: coin types, individual values, and the grand total.
• Leave blanks for what you're solving for.

🔤 Assigning variables

• Choose one coin type as your base variable (often the one mentioned second or defined "in terms of" another).
• Express other quantities using that variable plus the relationship described.
• Example: "nine more nickels than dimes" → let d = dimes, then nickels = d + 9.
• Example: "twice as many pennies as quarters" → let q = quarters, then pennies = 2q.

Don't confuse: "more than" means addition; "times as many" means multiplication.

🛠️ The seven-step solution method

🛠️ Steps 1–3: Setup

1. Read: Identify coin types, create the table, fill in values and total.
2. Identify: State clearly what you're looking for (number of each coin type).
3. Name: Assign a variable to one quantity, express others in terms of it, complete the table by multiplying number × value.

🛠️ Steps 4–5: Solve

4. Translate: Write an equation by adding all the "total value" entries; they must equal the given total.
5. Solve: Use algebra (distribute, combine like terms, isolate the variable) to find the variable's value, then calculate the other quantities.

🛠️ Steps 6–7: Verify and answer

6. Check: Substitute your answers back—does (number of coin A × value A) + (number of coin B × value B) = stated total?
7. Answer: Write a complete sentence stating how many of each coin type.

📝 Worked example patterns

📝 Pattern 1: One quantity defined by another

• Example: Adalberto has $2.25 in dimes and nickels; nine more nickels than dimes.
• Let d = dimes, then nickels = d + 9.
• Equation: 0.10d + 0.05(d + 9) = 2.25.
• Solution: d = 12 dimes, so 21 nickels.

📝 Pattern 2: Multiplicative relationship

• Example: Maria has $2.43 in quarters and pennies; twice as many pennies as quarters.
• Let q = quarters, then pennies = 2q.
• Equation: 0.25q + 0.01(2q) = 2.43.
• Solution: q = 9 quarters, so 18 pennies.

📝 Pattern 3: Complex expression

• Example: Danny has $2.14 in pennies and nickels; nickels are two more than ten times the pennies.
• Let p = pennies, then nickels = 10p + 2.
• Equation: 0.01p + 0.05(10p + 2) = 2.14.
• Solution: p = 4 pennies, so 42 nickels.

Don't confuse: "two more than ten times p" means 10p + 2, not 2 + 10 or 10(p + 2).

🎯 Extension to other mixture problems

🎯 Tickets and stamps

• The excerpt notes that ticket and stamp problems work exactly like coin problems.
• Each ticket type or stamp denomination has a fixed price (value).
• Use the same table structure and seven-step method.
• The model remains: number × value = total value for each type, then sum all types.

🧭 Overview

🧠 One-sentence thesis

These word problems use a systematic table-based approach to solve for unknown quantities when mixing items of different values—whether coins, tickets, stamps, or ingredients—by setting up equations that equate total value or total quantity.

📌 Key points (3–5)

• Core method: organize information in a table with columns for type, number, value per unit, and total value; the last row (total) gives the equation to solve.
• Two main scenarios: (1) when one quantity is defined in terms of the other (e.g., "nickels are two more than ten times pennies"), or (2) when the total number is known and you subtract to find the second quantity (e.g., if 810 total tickets and c children's tickets, then adult tickets = 810 − c).
• Common confusion: don't mix up "number of items" with "total value"—you must multiply number × value to get total value before adding.
• Why it matters: the same structure applies to coins, tickets, stamps, investments (simple interest), and ingredient mixtures, making it a versatile problem-solving framework.
• Key step: the equation always comes from adding the total values (or interests) of each type to equal the overall total value.

🪙 Coin problems

🪙 The table structure

Every coin problem uses a table with these columns:

Type of coin	Number	Value ($)	Total value ($)
Pennies	p	0.01	0.01p
Nickels	expression in p	0.05	0.05 × (expression)
Total	—	—	given total $

• The "Total value" column is number × value.
• The bottom row gives the equation: sum of individual total values = overall total value.

🔢 Scenario 1: One quantity defined in terms of the other

When the number of one coin type is described as a function of the other (e.g., "nickels are two more than ten times the number of pennies"), define the base variable first, then express the second quantity algebraically.

Example (from the excerpt): Danny has $2.14 in pennies and nickels; the number of nickels is two more than ten times the number of pennies.

• Let p = number of pennies.
• Then number of nickels = 10p + 2.
• Equation: 0.01p + 0.05(10p + 2) = 2.14.
• Solving gives p = 4 pennies, so nickels = 10(4) + 2 = 42.
• Check: 4(0.01) + 42(0.05) = 0.04 + 2.10 = 2.14 ✓

Don't confuse: The phrase "two more than ten times" means 10p + 2, not 2 + p × 10 (though algebraically the same, order matters for translation).

🔢 Scenario 2: Total number of coins is known

When the problem states the total number of items (e.g., "Galen sold 810 tickets"), use subtraction to express the second quantity: if total is T and one type is x, the other is T − x.

Example (from the excerpt): Galen sold 810 tickets (children's at $3, adults at $5) for $2,820 total.

• Let c = number of children's tickets.
• Then adult tickets = 810 − c.
• Equation: 3c + 5(810 − c) = 2,820.
• Solving: 3c + 4,050 − 5c = 2,820 → −2c = −1,230 → c = 615.
• Adult tickets = 810 − 615 = 195.
• Check: 615(3) + 195(5) = 1,845 + 975 = 2,820 ✓

Why this works: The total number constraint means the two quantities must add up, so expressing one as "total minus the other" automatically satisfies that constraint.

🎟️ Ticket and stamp problems

🎟️ Same structure, different items

Ticket and stamp problems are structurally identical to coin problems: each type has a value, and you multiply number × value to get total value.

Example (from the excerpt): A concert sold $1,506 in tickets; student tickets $6 each, adult tickets $9 each; adults sold were five less than three times students.

• Let s = student tickets.
• Adult tickets = 3s − 5.
• Equation: 6s + 9(3s − 5) = 1,506.
• Solving: 6s + 27s − 45 = 1,506 → 33s = 1,551 → s = 47.
• Adults = 3(47) − 5 = 136.
• Check: 47(6) + 136(9) = 282 + 1,224 = 1,506 ✓

🎟️ When total number is given

The excerpt emphasizes a key insight:

• If Bianca sold 100 tickets total and 20 were child tickets, adult tickets = 100 − 20 = 80.
• If she sold x child tickets, adult tickets = 100 − x.

Example (from the excerpt): 115 museum tickets sold for $1,163; adult $12, student $5.

• Let a = adult tickets.
• Student tickets = 115 − a.
• Equation: 12a + 5(115 − a) = 1,163.
• Solve to find a, then compute 115 − a for students.

Don't confuse: "Total number of tickets" vs. "total dollar value"—these are two separate constraints; use the number constraint to express one variable in terms of the other, then use the value constraint to write the equation.

📮 Stamp problems with decimals

Stamp problems often involve decimal values (e.g., $0.41, $0.02); the method is unchanged, but be careful with decimal arithmetic.

Example (from the excerpt): Monica paid $8.36 for stamps; 41-cent stamps were four more than twice the number of 2-cent stamps.

• Let x = number of 2-cent stamps.
• 41-cent stamps = 2x + 4.
• Equation: 0.41(2x + 4) + 0.02x = 8.36.
• Expand: 0.82x + 1.64 + 0.02x = 8.36 → 0.84x = 6.72 → x = 8.
• 41-cent stamps = 2(8) + 4 = 20.
• Check: 8(0.02) + 20(0.41) = 0.16 + 8.20 = 8.36 ✓

🧪 Mixture problems (ingredients and investments)

🧪 Mixing ingredients by weight and cost

Mixture problems extend the same table method to scenarios where you combine ingredients (or products) of different unit costs to achieve a target cost per unit for the mixture.

Example (from the excerpt): Henning makes 10 pounds of trail mix; raisins $2/lb, nuts $6/lb; he wants the mix to cost $5.20/lb.

• Let x = pounds of raisins.
• Pounds of nuts = 10 − x.
• Table:

Item	Number (lb)	Value ($/lb)	Total value ($)
Raisins	x	2	2x
Nuts	10 − x	6	6(10 − x)
Mix	10	5.20	10(5.20) = 52

• Equation: 2x + 6(10 − x) = 52.
• Solving: 2x + 60 − 6x = 52 → −4x = −8 → x = 2.
• Nuts = 10 − 2 = 8 pounds.
• Check: 2(2) + 8(6) = 4 + 48 = 52 ✓

Why the last row matters: The mixture's total value is (total weight) × (desired cost per unit); this gives the right-hand side of the equation.

💰 Investment problems with simple interest

Investment problems use the simple interest formula I = Prt; when t = 1 year, I = Pr, so "interest earned" replaces "total value" in the table.

Example (from the excerpt): Stacey invests $20,000 in two accounts: one at 3% per year, one at 5% per year; she wants overall interest of 4.5% per year.

• Let x = amount at 3%.
• Amount at 5% = 20,000 − x.
• Table:

Account	Principal ($)	Rate	Interest ($)
3%	x	0.03	0.03x
5%	20,000 − x	0.05	0.05(20,000 − x)
Total	20,000	0.045	0.045(20,000) = 900

• Equation: 0.03x + 0.05(20,000 − x) = 900.
• Solving: 0.03x + 1,000 − 0.05x = 900 → −0.02x = −100 → x = 5,000.
• Amount at 5% = 20,000 − 5,000 = 15,000.
• Check: 0.03(5,000) + 0.05(15,000) = 150 + 750 = 900 ✓

Don't confuse: The "rate" column is the interest rate (as a decimal), not the interest earned; you must multiply principal × rate to get interest.

🧪 General mixture principle

The excerpt states:

"Grocers and bartenders use the mixture model to set a fair price for a product made from mixing two or more ingredients. Financial planners use the mixture model when they invest money in a variety of accounts and want to find the overall interest rate. Landscape designers use the mixture model when they have an assortment of plants and a fixed budget, and event coordinators do the same when choosing appetizers and entrees for a banquet."

Key insight: Whether mixing raisins and nuts, or splitting money between two interest rates, the structure is the same:

• (value of part 1) + (value of part 2) = (value of whole).
• The "value" might be dollars, interest earned, or cost × quantity.

🛠️ The seven-step method

🛠️ Systematic approach

The excerpt uses a consistent seven-step process for all problems:

1. Read the problem; identify the types of items involved; create a table.
2. Identify what you are looking for (e.g., number of each coin type).
3. Name the variables; express one quantity in terms of the other (or use total − x).
4. Translate the word problem into an equation using the table's total-value row.
5. Solve the equation algebraically.
6. Check the answer by substituting back into the original conditions.
7. Answer the question in a complete sentence.

Why this matters: Following the same steps every time reduces errors and makes the method transferable across different contexts (coins, tickets, mixtures, investments).

🛠️ The table as the backbone

Every problem in the excerpt builds a table with:

• Rows: each type of item, plus a "Total" row.
• Columns: Type | Number | Value (per unit) | Total value.
• The Total value column is always Number × Value.
• The Total row gives the equation: sum of individual total values = overall total value.

Example structure:

Type	Number	Value	Total value
Item A	x	v₁	v₁x
Item B	expression or total − x	v₂	v₂(expression)
Total	given or sum	—	given total $

Don't skip the table: Even if you can solve the problem mentally, the table ensures you don't miss a step and makes checking easier.

🧭 Overview

🧠 One-sentence thesis

Geometry applications use formulas for triangles, rectangles, and the Pythagorean Theorem to solve real-world problems by translating word problems into equations and solving systematically.

📌 Key points (3–5)

• Triangle properties: angles sum to 180°, perimeter is the sum of sides, and area equals one-half base times height.
• Pythagorean Theorem: in right triangles, the sum of the squares of the two legs equals the square of the hypotenuse (a² + b² = c²).
• Rectangle properties: perimeter equals twice length plus twice width (P = 2L + 2W), and area equals length times width (A = L·W).
• Problem-solving strategy: read, draw and label a figure, identify what you're looking for, choose a variable, translate to an equation, solve, check, and answer.
• Common confusion: don't confuse perimeter (distance around) with area (space inside); perimeter uses addition, area uses multiplication.

🔺 Triangle fundamentals

🔺 Basic triangle properties

Triangle: a shape with three sides and three interior angles.

• Triangles have three vertices (plural of vertex), labeled with uppercase letters (A, B, C).
• Sides are labeled with lowercase letters (a, b, c) matching the opposite vertex.
• The excerpt names triangles by their vertices, e.g., △ABC.

📐 Angle sum property

The sum of the measures of the angles of a triangle is 180°.

• Formula: m∠A + m∠B + m∠C = 180°
• If you know two angles, you can find the third by subtracting from 180°.
• Example: If two angles measure 55° and 82°, the third angle = 180° - 55° - 82° = 43°.

📏 Perimeter of triangles

Perimeter: the sum of the lengths of the sides of the triangle.

• Formula: P = a + b + c
• If you know the perimeter and two sides, subtract those sides from the perimeter to find the third side.
• Example: Perimeter is 24 feet, two sides are 4 feet and 9 feet, so the third side = 24 - 4 - 9 = 11 feet.

📐 Area of triangles

Area: A = (1/2)bh, where b is the base and h is the height.

• The height is a line perpendicular (90° angle) to the base that reaches the opposite vertex.
• If you know area and base, solve for height: multiply area by 2, then divide by base.
• Example: Area is 90 square meters, base is 15 meters, so height = (90 × 2) ÷ 15 = 12 meters.

▶️ Right triangles

▶️ What makes a right triangle special

Right triangle: a triangle with one 90° angle, often marked with a small square at the vertex.

• The 90° angle is one of the three angles, so the other two angles must sum to 90°.
• Example: If one angle is 28° and another is 90°, the third angle = 180° - 90° - 28° = 62°.

▶️ Defining angles in terms of each other

• Sometimes one angle is described relative to another (e.g., "20 degrees more than the smallest angle").
• Let a variable represent the smallest angle, then write expressions for the other angles.
• Example: Let a = smallest angle, then (a + 20) = second angle, and 90 = third angle. Since a + (a + 20) + 90 = 180, solve to get a = 35°, so angles are 35°, 55°, and 90°.

📐 The Pythagorean Theorem

📐 What the theorem states

Pythagorean Theorem: In any right triangle, a² + b² = c², where a and b are the lengths of the legs and c is the length of the hypotenuse.

• The hypotenuse is the side opposite the 90° angle (the longest side).
• The legs are the two sides that form the 90° angle.
• This relationship only works for right triangles.

🔢 Finding the hypotenuse

• If you know both legs, square each, add them, then take the square root.
• Example: Legs are 3 and 4, so c² = 3² + 4² = 9 + 16 = 25, therefore c = √25 = 5.

🔢 Finding a leg

• If you know the hypotenuse and one leg, square both, subtract the leg² from hypotenuse², then take the square root.
• Example: Hypotenuse is 13, one leg is 5, so b² = 13² - 5² = 169 - 25 = 144, therefore b = √144 = 12.

🔢 Square roots reminder

If m = n², then √m = n (for n ≥ 0).

• Example: √25 = 5 because 5² = 25.
• When the result is not a whole number, approximate to the nearest tenth.
• Example: √50 ≈ 7.1.

🏗️ Real-world application

• Example: A 10-inch brace forms a right triangle with equal legs. To find leg length, solve x² + x² = 10², so 2x² = 100, x² = 50, x ≈ 7.1 inches.
• Don't confuse: the hypotenuse is always the longest side; if you get a leg longer than the hypotenuse, recheck your work.

🟦 Rectangle properties

🟦 Basic rectangle facts

Rectangle: a shape with four sides and four right (90°) angles; opposite sides are equal in length.

• One side is called the length (L), and the adjacent side is the width (W).
• Opposite sides have the same measurement.

📏 Perimeter of rectangles

Perimeter: P = 2L + 2W (sum of twice the length and twice the width).

• This is the distance around the rectangle.
• Example: Length is 32 meters, width is 20 meters, so P = 2(32) + 2(20) = 64 + 40 = 104 meters.
• If you know perimeter and one dimension, substitute and solve for the other.

📐 Area of rectangles

Area: A = L·W (product of length and width).

• This is the space inside the rectangle.
• Example: Area is 168 square feet, length is 14 feet, so W = 168 ÷ 14 = 12 feet.
• Don't confuse: area uses multiplication (square units), perimeter uses addition (linear units).

🔢 Defining one dimension in terms of another

• Sometimes width or length is described relative to the other (e.g., "two feet less than the length").
• Let a variable represent one dimension, then write an expression for the other.
• Example: Let L = length, and width = L - 2. If perimeter is 52 feet, then 2L + 2(L - 2) = 52. Solve: 2L + 2L - 4 = 52, so 4L = 56, L = 14 feet, and width = 12 feet.

🔢 More complex relationships

• Example: "Length is four more than twice the width."
  • Let W = width, then L = 2W + 4.
  • If perimeter is 32 cm, substitute: 2(2W + 4) + 2W = 32.
  • Solve: 4W + 8 + 2W = 32, so 6W = 24, W = 4 cm, L = 12 cm.

🛠️ Problem-solving strategy

🛠️ Seven-step process

The excerpt provides a systematic approach:

Step	Action
1. Read	Understand the problem; draw and label a figure
2. Identify	What are you looking for?
3. Label	Choose a variable to represent the unknown
4. Translate	Write the appropriate formula and substitute known values
5. Solve	Use algebra to solve the equation
6. Check	Substitute the answer back into the equation and verify it makes sense
7. Answer	State the answer in a complete sentence

🖼️ Why drawing helps

• A labeled diagram clarifies what is given and what is unknown.
• It helps you see relationships between quantities.
• Example: Drawing a triangle with two known angles makes it easier to see that the third angle is what remains to reach 180°.

✅ Checking your work

• Substitute your answer back into the original formula.
• Verify the units make sense (feet, square meters, degrees, etc.).
• Ask: does this answer fit the real-world context?
• Example: If you calculate a triangle's third side as 11 feet, check that 4 + 9 + 11 = 24 feet (the given perimeter).

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Note: The excerpt also includes numerous practice exercises and examples involving uniform motion applications (distance, rate, time), but those are part of a different section (3.5) and are not the focus of section 3.4 on geometry applications.

🧭 Overview

🧠 One-sentence thesis

Linear inequalities model real-world constraints—such as budgets, capacity limits, and profit goals—and solving them reveals the maximum or minimum values that satisfy those constraints.

📌 Key points (3–5)

• What inequalities represent: constraints like "at most," "at least," "no more than," or "minimum" in everyday situations (budgets, weight limits, sales targets).
• How to solve: translate the word problem into an inequality, solve algebraically, then interpret the solution in context.
• Rounding in context: when the solution must be a whole number (e.g., number of people or items), round up or down based on the constraint direction.
• Common confusion: "at least" means ≥, "no more than" means ≤; mixing these up reverses the constraint.
• Why it matters: inequalities help answer questions like "How many can I afford?" or "What sales do I need to meet my goal?"

🔢 Core concepts

🔢 What linear inequalities model

Linear inequality: a mathematical statement using <, >, ≤, or ≥ to express a constraint on a variable.

• Real-life situations often involve limits or thresholds, not exact values.
• Example: "Emma's monthly income must be at least three times the rent" translates to income ≥ 3 × rent.
• The inequality captures the boundary condition; the solution is a range of acceptable values.

🔍 Translating words into symbols

Key phrases and their inequality symbols:

Phrase	Symbol	Example context
At least, minimum, no less than	≥	"Income must be at least 3 times rent" → income ≥ 3r
At most, maximum, no more than	≤	"Budget is at most $500" → cost ≤ 500
More than, greater than	>	"Sales must exceed $925" → sales > 925
Less than, fewer than	<	"Expenses less than income" → expenses < income

• Don't confuse: "at least 10" means 10 or more (≥ 10), not "more than 10" (> 10).

🛠️ Solution strategy

🛠️ Seven-step method

The excerpt uses a consistent seven-step approach (identical to equation word problems, but with inequalities):

1. Read the problem carefully.
2. Identify what you are looking for.
3. Name it with a variable.
4. Translate the constraint into an inequality (write a sentence first, then convert).
5. Solve the inequality using algebra.
6. Check that the answer makes sense in the original problem.
7. Answer with a complete sentence.

🔄 When to round

• If the variable represents a countable quantity (people, boxes, tablets), the algebraic solution may not be a whole number.
• Round down when the inequality is ≤ or < (you cannot exceed the limit).
  • Example: "Dawn can buy at most 15.74 tablets" → round down to 15 tablets.
• Round up when the inequality is ≥ or > (you must meet or exceed the threshold).
  • Example: "Elliot must work at least 84.7 jobs" → round up to 85 jobs.
• Always check the rounded value in the original inequality to confirm it satisfies the constraint.

📦 Common application types

📦 Capacity and budget limits

Setup: A maximum weight, cost, or quantity constraint.

• Example (from excerpt): Alan loads boxes weighing 45 pounds each; the pallet supports no more than 900 pounds. How many boxes?
  • Let n = number of boxes.
  • 45n ≤ 900 → n ≤ 20.
  • Alan can load at most 20 boxes.

💰 Income and profit goals

Setup: Variable pay (base + commission) must meet or exceed a target.

• Example (from excerpt): Pete earns $500 + 12% of sales; he wants to exceed $925. What sales are needed?

  • Let s = total sales.
  • 500 + 0.12s > 925 → 0.12s > 425 → s > 3,541.67.
  • Sales must be more than $3,541.67.

• Example (profit): Elliot charges $60 per job, expenses are $1,100/month, wants profit ≥ $4,000.

  • Let j = number of jobs.
  • 60j − 1,100 ≥ 4,000 → 60j ≥ 5,100 → j ≥ 85.
  • He must do at least 85 jobs.

🚗 Rate-based expenses

Setup: A fixed fee plus a per-unit charge, with a budget cap.

• Example (from excerpt): Car rental costs $75/week + $0.25/mile; budget is $200. How many miles?
  • Let m = miles.
  • 75 + 0.25m ≤ 200 → 0.25m ≤ 125 → m ≤ 500.
  • They can drive up to 500 miles.

🎉 Multi-expense trip planning

Setup: Several costs (airfare, hotel, food) plus earnings; total expenses ≤ savings + earnings.

• Example (from excerpt): Brenda's trip costs $350 (airfare) + $375 (food) + $60 × 3 nights (hotel); she has $500 saved and earns $15/hour babysitting. How many hours?
  • Let h = hours.
  • 350 + 375 + 180 ≤ 500 + 15h → 905 ≤ 500 + 15h → 405 ≤ 15h → h ≥ 27.
  • She must babysit at least 27 hours.

⚠️ Common pitfalls

⚠️ Reversing inequality direction

• The inequality symbol does not flip when you add, subtract, multiply, or divide by a positive number.
• It does flip if you multiply or divide by a negative number (not shown in these examples, but important to remember).
• Don't confuse: "at least" (≥) vs. "more than" (>); "at most" (≤) vs. "less than" (<).

⚠️ Forgetting to round appropriately

• If the problem asks "how many people" and you get 15.74, you cannot have 0.74 of a person.
• Check the inequality direction: if it's ≤, round down; if it's ≥, round up.

⚠️ Misinterpreting the solution

• The solution to an inequality is a range, not a single number.
• Example: n ≤ 15 means "15 or fewer," so 0, 1, 2, …, 15 are all valid (if they make sense in context).
• Always state the answer in a complete sentence that answers the original question.

🧭 Overview

🧠 One-sentence thesis

This section teaches a general strategy for solving linear equations by systematically simplifying both sides, collecting variable terms on one side and constants on the other, and then isolating the variable.

📌 Key points (3–5)

• General strategy: Simplify each side first (distribute, combine like terms), then collect variables on one side and constants on the other, finally make the coefficient equal to 1.
• Clearing fractions and decimals: Multiply both sides by the LCD (for fractions) or an appropriate power of 10 (for decimals) to eliminate them early.
• Three types of equations: Conditional (one solution), identity (all real numbers are solutions), contradiction (no solution).
• Common confusion: Don't confuse an identity (always true, like 6 = 6) with a contradiction (always false, like 0 = -1).
• Formula manipulation: You can solve a formula for any variable by isolating that variable using the same algebraic steps.

🔧 General Strategy for Solving Linear Equations

🔧 The five-step process

The excerpt presents a systematic approach:

1. Simplify each side as much as possible (use Distributive Property, combine like terms)
2. Collect variable terms on one side (use Addition or Subtraction Property of Equality)
3. Collect constant terms on the other side (use Addition or Subtraction Property of Equality)
4. Make the coefficient equal to 1 (use Multiplication or Division Property of Equality)
5. Check by substituting back into the original equation

🎯 Choosing which side for variables

• Compare coefficients of the variable on each side
• Make the side with the larger coefficient the "variable side"
• This keeps arithmetic simpler and avoids negative coefficients when possible

Example: In 8n - 4 = -2n + 6, since 8 > -2, keep variables on the left side.

⚠️ Don't confuse: Steps vs. shortcuts

The strategy is a sequence—you must simplify first before moving terms around. Skipping simplification leads to errors.

🧹 Clearing Fractions and Decimals

🧹 Why clear fractions

Clearing fractions: Multiplying both sides of an equation by the LCD of all fractions to eliminate them.

• Fractions make arithmetic harder and increase error risk
• After clearing, the equation becomes a simpler form you already know how to solve

🔢 Strategy for fraction coefficients

1. Find the LCD of all fractions in the equation
2. Multiply both sides by that LCD
3. Distribute and simplify—fractions disappear
4. Solve using the general strategy

Example: For (1/6)y - 1/3 = 5/6, the LCD is 6. Multiply both sides by 6 to get y - 2 = 5.

💰 Strategy for decimal coefficients

• Think of decimals as fractions: 0.06 = 6/100, 0.3 = 3/10, 1.5 = 15/10
• The LCD is usually 10, 100, or 1000
• Multiply both sides by the appropriate power of 10

Example: For 0.06x + 0.02 = 0.25x - 1.5, multiply both sides by 100 to clear decimals.

📐 Distribute first when needed

If you see something like (1/2)(y - 5), distribute before clearing fractions—it may simplify the work.

🔍 Classifying Equations

🔍 Three types of outcomes

Type	What happens when solving	Result	Solution
Conditional equation	Variables cancel, leaving a solvable equation	True for specific value(s)	One or more values
Identity	Variables cancel, leaving a true statement (e.g., 6 = 6)	True for any value	All real numbers
Contradiction	Variables cancel, leaving a false statement (e.g., 0 = -1)	False for all values	No solution

🎭 Recognizing identities

When you solve and all variables disappear, leaving something like 6 = 6 or -3 = -3, the original equation is an identity.

Example: Solving 2y + 6 = 2(y + 3) leads to 2y + 6 = 2y + 6, then 6 = 6 after subtracting 2y.

🚫 Recognizing contradictions

When you solve and all variables disappear, leaving something false like 0 = -1 or 27 ≠ -22, the equation is a contradiction.

Example: Solving 5z = 5z - 1 leads to 0 = -1, which is never true.

⚖️ Don't confuse: Identity vs. conditional

• An identity is true for every value of the variable
• A conditional equation is true for only certain values
• Check: if simplifying leads to a variable-free true statement, it's an identity; if it leads to a normal equation like x = 3, it's conditional

📏 Solving Formulas for a Specific Variable

📏 What it means to solve for a variable

Solve a formula for a specific variable: Isolate that variable on one side of the equals sign with a coefficient of 1, with all other variables and constants on the other side.

• You're rearranging the formula so one variable is expressed in terms of the others
• Use the same algebraic steps as solving numerical equations

🚗 The distance formula

Distance formula: d = rt, where d = distance, r = rate, t = time.

• If you know any two quantities, you can find the third
• To solve for t: divide both sides by r to get t = d/r
• To solve for r: divide both sides by t to get r = d/t

Example: If distance is 520 miles and rate is 65 mph, then t = 520/65 = 8 hours.

🔄 Solving in general vs. with numbers

The excerpt shows solving formulas side-by-side:

• Left column: substitute specific numbers and solve
• Right column: use the same steps with variables only

This demonstrates that the algebraic process is identical whether you have numbers or variables.

🧮 Common formulas to manipulate

Area of a triangle: A = (1/2)bh

• To solve for h: multiply both sides by 2, then divide by b → h = 2A/b

Simple interest: I = Prt

• To solve for P: divide both sides by rt → P = I/(rt)

Linear equation in two variables: 3x + 2y = 18

• To solve for y: subtract 3x from both sides, then divide by 2 → y = (18 - 3x)/2

⚠️ Don't confuse: Solving for a variable vs. evaluating

• Solving for a variable: rearranging to isolate it (result is a formula)
• Evaluating: substituting numbers to find a numerical answer
• You can do either, depending on what the problem asks

🛠️ Properties of Equality (Reference)

🛠️ Four fundamental properties

All equation-solving relies on these:

Property	Statement	What it means
Subtraction	If a = b, then a - c = b - c	Subtract the same amount from both sides
Addition	If a = b, then a + c = b + c	Add the same amount to both sides
Division	If a = b, then a/c = b/c (c ≠ 0)	Divide both sides by the same nonzero number
Multiplication	If a = b, then ac = bc	Multiply both sides by the same number

🔐 Why these work

When you perform the same operation on both sides of an equation, you maintain equality—the balance stays balanced.

Example: If 5x = -27, dividing both sides by 5 gives x = -27/5, and the equation remains true.

🧭 Overview

🧠 One-sentence thesis

Linear inequalities are solved using the same properties as equations, except that multiplying or dividing both sides by a negative number reverses the inequality sign.

📌 Key points (3–5)

• What a solution to an inequality means: any value that makes the inequality true (unlike equations, which usually have one solution, inequalities often have infinitely many solutions).
• Graphing and interval notation: inequalities are represented on number lines using parentheses (not included) or brackets (included), and in interval notation using symbols like (a, ∞) or (−∞, b].
• Addition and subtraction properties: adding or subtracting the same number from both sides keeps the inequality sign the same direction.
• Critical rule for multiplication/division: when multiplying or dividing both sides by a positive number, the inequality stays the same; when multiplying or dividing by a negative number, the inequality reverses.
• Common confusion: forgetting to reverse the inequality when multiplying or dividing by a negative—this is the key difference from solving equations.

📊 Understanding solutions and graphs

📊 What makes a number a solution

A solution of an inequality is a value of the variable that makes a true statement when substituted into the inequality.

• Unlike equations (which typically have one or a few solutions), inequalities usually have infinitely many solutions.
• Example: for x > 3, the solutions include 4, 5, 20, 3.001, and any number greater than 3.

📈 Graphing on the number line

• Open parenthesis ( or ): the endpoint is not included (used for < or >).
• Bracket [ or ]: the endpoint is included (used for ≤ or ≥).
• Shade the region containing all solutions.
• Example: x ≥ 3 is graphed with a bracket at 3 and shading to the right.

🔢 Interval notation

Interval notation uses parentheses and brackets to match the number line representation:

Inequality	Interval notation	Meaning
x > 3	(3, ∞)	All numbers greater than 3
x ≤ 1	(−∞, 1]	All numbers less than or equal to 1
x ≥ −3	[−3, ∞)	All numbers greater than or equal to −3

• The symbol ∞ (infinity) is read as "infinity" and is not an actual number; always use a parenthesis with ∞ or −∞.
• Don't confuse: the bracket/parenthesis in interval notation matches the symbol on the number line endpoint.

➕ Addition and Subtraction Properties

➕ How these properties work

Subtraction Property of Inequality: For any numbers a, b, and c, if a < b then a − c < b − c (and similarly for >).

Addition Property of Inequality: For any numbers a, b, and c, if a < b then a + c < b + c (and similarly for >).

• The inequality sign stays the same when you add or subtract the same amount from both sides.
• This mirrors the properties for equations.

🔧 Solving with addition/subtraction

Example: Solve n − 1/2 ≤ 5/8

• Add 1/2 to both sides to isolate n.
• Simplify: n ≤ 5/8 + 1/2 = 5/8 + 4/8 = 9/8.
• The inequality sign does not change.

✖️ Division and Multiplication Properties

✖️ The critical reversal rule

Division and Multiplication Properties of Inequality:

• If a < b and c > 0 (positive), then a/c < b/c and ac < bc (inequality stays the same).
• If a < b and c < 0 (negative), then a/c > b/c and ac > bc (inequality reverses).

• When you multiply or divide by a positive number, the inequality stays the same.
• When you multiply or divide by a negative number, the inequality reverses direction.

⚠️ Why the sign reverses

The excerpt demonstrates this with numerical examples:

• Start with 10 > 5 (true).
• Divide both sides by −5: 10/(−5) = −2 and 5/(−5) = −1.
• Now −2 < −1 (the inequality reversed).
• This happens because negative numbers flip the order on the number line.

🔧 Solving with multiplication/division

Example: Solve −10a ≥ 50

• Divide both sides by −10.
• Since −10 < 0 (negative), reverse the inequality: a ≤ −5.
• Don't confuse: if you had divided by positive 10, the sign would stay the same.

Example: Solve 7y < 42

• Divide both sides by 7.
• Since 7 > 0 (positive), the inequality stays the same: y < 6.

🧮 Multi-step inequalities and special cases

🧮 Solving inequalities that require simplification

Follow the same steps as solving equations:

1. Simplify each side (distribute, combine like terms).
2. Collect variable terms on one side using addition/subtraction.
3. Collect constants on the other side.
4. Multiply or divide to isolate the variable (watch the sign!).

Example: Solve 4m ≤ 9m + 17

• Subtract 9m from both sides: −5m ≤ 17.
• Divide by −5 (negative), so reverse: m ≥ −17/5.

🔄 Rewriting when the variable ends up on the right

• If you get −20 < u, rewrite it as u > −20 (same meaning).
• Think: "If Xavier is taller than Alex, then Alex is shorter than Xavier."

✅ Identity inequalities

When simplification leaves a true statement with no variables (like −10 < 36):

• The inequality is an identity.
• Solution: all real numbers, written as (−∞, ∞).

❌ Contradiction inequalities

When simplification leaves a false statement with no variables (like 0 > 18):

• The inequality is a contradiction.
• There is no solution.

🗣️ Translating word problems

🗣️ Common inequality phrases

The excerpt provides key translation phrases:

Phrase	Symbol	Example meaning
is greater than, is more than, exceeds	>	x is more than 5 → x > 5
is at least, is no less than, is the minimum	≥	x is at least 21 → x ≥ 21
is less than, is smaller than, has fewer than	<	x is less than 10 → x < 10
is at most, is no more than, is the maximum	≤	x is at most 8 → x ≤ 8

🔧 Translation examples

Example: "Twelve times c is no more than 96"

• Translate: 12c ≤ 96.
• Solve: c ≤ 8.

Example: "Thirty less than x is at least 45"

• Translate: x − 30 ≥ 45.
• Solve: x ≥ 75.
• Don't confuse "less than" (subtraction) with "is less than" (inequality symbol).

🧭 Overview

🧠 One-sentence thesis

The Division and Multiplication Properties of Equality allow you to isolate a variable by dividing or multiplying both sides of an equation by the same non-zero number, preserving the equality.

📌 Key points (3–5)

• Division Property: dividing both sides of an equation by the same non-zero number keeps the equation balanced.
• Multiplication Property: multiplying both sides of an equation by the same number keeps the equation balanced.
• Why they work: these operations maintain equality because you perform the same operation on both sides.
• Common confusion: remember you cannot divide by zero; the Division Property requires c ≠ 0.
• When to use: use division when the variable has a coefficient you want to remove; use multiplication when the variable is in a fraction.

⚖️ The two core properties

⚖️ Division Property of Equality

Division Property of Equality: For any numbers a, b, and c, and c ≠ 0, if a = b, then a/c = b/c.

• What it means in plain language: if two quantities are equal, dividing both by the same non-zero number produces two new equal quantities.
• Why the restriction: you cannot divide by zero because division by zero is undefined.
• When to apply: use this when the variable has a coefficient (a number multiplied by it) and you want to isolate the variable.
• Example: if 8x = 72, divide both sides by 8 to get x = 9.

⚖️ Multiplication Property of Equality

Multiplication Property of Equality: For any numbers a, b, and c, if a = b, then ac = bc.

• What it means in plain language: if two quantities are equal, multiplying both by the same number produces two new equal quantities.
• No restriction: unlike division, you can multiply by any number, including zero (though that's rarely useful).
• When to apply: use this when the variable is divided by a number (in a fraction) and you want to clear the denominator.
• Example: if n/6 = 18, multiply both sides by 6 to get n = 108.

🔧 How to solve equations with these properties

🔧 Using division to isolate the variable

When the variable has a coefficient:

• Identify the coefficient (the number multiplied by the variable).
• Divide both sides of the equation by that coefficient.
• Simplify to find the variable's value.

Example: solve 13a = -65.

• The coefficient is 13.
• Divide both sides by 13: a = -65/13.
• Simplify: a = -5.

Don't confuse: dividing both sides is not the same as "moving" a number—you must divide the entire side, not just one term.

🔧 Using multiplication to clear fractions

When the variable is in a fraction:

• Identify the denominator (the number the variable is divided by).
• Multiply both sides by that denominator.
• Simplify to find the variable's value.

Example: solve y/(-10) = 30.

• The denominator is -10.
• Multiply both sides by -10: y = 30 × (-10).
• Simplify: y = -300.

Don't confuse: when the variable is divided by a negative number, multiplying by that negative will give a negative result if the other side is positive.

🔧 Handling special cases

Negative coefficients:

• If the coefficient is negative (e.g., -18m = -72), divide by the negative number.
• Negative divided by negative gives positive: m = -72/(-18) = 4.

Fractional coefficients:

• If the coefficient is a fraction (e.g., (3/4)x = 36), multiply both sides by the reciprocal of the fraction.
• The reciprocal of 3/4 is 4/3, so multiply both sides by 4/3.

Decimal coefficients:

• Treat decimals like whole numbers: if 0.25p = 5.25, divide both sides by 0.25.
• Result: p = 5.25/0.25 = 21.

✅ Checking your solution

✅ Why checking matters

• Arithmetic errors can happen during division or multiplication.
• Checking confirms that your solution makes the original equation true.

✅ How to check

1. Substitute your solution back into the original equation (not the simplified version).
2. Simplify both sides separately.
3. Verify that both sides are equal.

Example: if you solved 8x = 72 and got x = 9, check by substituting:

• Left side: 8(9) = 72.
• Right side: 72.
• Both sides equal 72, so the solution is correct.

🔄 Equations requiring simplification first

🔄 Combining like terms before applying properties

Some equations have multiple terms with the variable that must be combined first.

Example: solve 5r - 3r + 9r = 35 - 2.

• Combine like terms on the left: (5 - 3 + 9)r = 11r.
• Combine constants on the right: 35 - 2 = 33.
• Now the equation is 11r = 33.
• Divide both sides by 11: r = 3.

Don't confuse: you must simplify each side fully before applying the Division or Multiplication Property.

🔄 Distributing before solving

If parentheses are present, use the distributive property first.

Example: solve -9(d - 2) - 15 = -24.

• Distribute: -9d + 18 - 15 = -24.
• Combine constants: -9d + 3 = -24.
• Subtract 3 from both sides: -9d = -27.
• Divide by -9: d = 3.

🧭 Overview

🧠 One-sentence thesis

Word problems involving numbers can be systematically solved by translating English sentences into algebraic equations with one variable, then solving step-by-step to find unknown values.

📌 Key points (3–5)

• Seven-step problem-solving strategy: Read, identify what you're looking for, name variables, translate to an equation, solve, check, and answer the question.
• Defining multiple unknowns with one variable: When a problem involves two or more related numbers, express all of them in terms of a single variable (e.g., if one number is "five more than another," use n and n + 5).
• Consecutive integer patterns: Consecutive integers differ by 1 (n, n+1, n+2); consecutive even or odd integers differ by 2 (n, n+2, n+4).
• Common confusion: Whether dealing with even or odd consecutive integers, always add 2 to get the next one—adding 2 to an odd number gives the next odd number.
• Real-world applications: The same translation method works for earnings, costs, and other practical scenarios where relationships between quantities are described in words.

🔢 The seven-step problem-solving process

📖 Step 1–3: Read, identify, and name

• Read the problem carefully to understand what is being asked.
• Identify exactly what you are looking for (e.g., "the number," "two numbers," "three consecutive integers").
• Name your variable(s): choose a letter (commonly n or x) to represent the unknown.
• If multiple unknowns are related, express them all in terms of one variable using the relationships given in the problem.

🔄 Step 4–5: Translate and solve

• Translate: Restate the problem as one clear sentence containing all key information, then convert that sentence into an algebraic equation.
• Solve: Use algebraic techniques (combine like terms, add/subtract from both sides, divide/multiply) to isolate the variable.
• Example: "The sum of twice a number and seven is 15" becomes 2n + 7 = 15; subtract 7 from both sides to get 2n = 8, then divide by 2 to find n = 4.

✅ Step 6–7: Check and answer

• Check: Substitute your solution back into the original problem statement to verify it satisfies all conditions.
• Answer: Write a complete sentence answering the question asked in the problem.
• Don't confuse: checking means using the original word problem, not just the equation you wrote.

🔗 Defining related numbers with one variable

🔗 Expressing the second number in terms of the first

When a problem describes two numbers with a relationship, define the first as n and express the second using that same variable.

Relationship description	First number	Second number
"One number is five more than another"	n	n + 5
"One number is four less than the other"	n	n − 4
"One number is ten more than twice another"	x	2x + 10

• This avoids using multiple variables, keeping the equation simpler.
• Example: If one number is five more than another and their sum is 21, write n + (n + 5) = 21, which simplifies to 2n + 5 = 21, giving n = 8 and the second number as 13.

💰 Real-world relationship problems

The same technique applies to earnings, prices, and other quantities.

• Example: A couple earns $110,000 together; the wife earns $16,000 less than twice what the husband earns.
• Let h = husband's earnings.
• Then 2h − 16,000 = wife's earnings.
• Equation: h + (2h − 16,000) = 110,000.
• Solve: 3h − 16,000 = 110,000 → 3h = 126,000 → h = 42,000.
• The husband earns $42,000; the wife earns $68,000.

🔢 Consecutive integer problems

🔢 Consecutive integers (differ by 1)

Consecutive integers: integers that immediately follow each other.

• Examples: 1, 2, 3, 4 or −10, −9, −8, −7.
• Pattern: if the first integer is n, the next is n + 1, then n + 2, and so on.
• Example: The sum of two consecutive integers is 47. Let n = first integer, n + 1 = second. Equation: n + (n + 1) = 47 → 2n + 1 = 47 → 2n = 46 → n = 23. The integers are 23 and 24.

➕ Consecutive even integers (differ by 2)

Consecutive even integers: even integers that immediately follow one another.

• Examples: 18, 20, 22 or −12, −10, −8.
• Pattern: n, n + 2, n + 4 (where n is even).
• Example: Find three consecutive even integers whose sum is 84. Let n = first, n + 2 = second, n + 4 = third. Equation: n + (n + 2) + (n + 4) = 84 → 3n + 6 = 84 → 3n = 78 → n = 26. The integers are 26, 28, and 30.

➕ Consecutive odd integers (also differ by 2)

Consecutive odd integers: odd integers that immediately follow one another.

• Examples: 77, 79, 81.
• Pattern: n, n + 2, n + 4 (where n is odd).
• Common confusion: It may seem strange to add 2 (an even number) to get from one odd integer to the next, but adding 2 to any odd number gives the next odd number (e.g., 3 + 2 = 5, 11 + 2 = 13).
• The algebraic pattern is identical for both consecutive even and consecutive odd integers—always add 2.

🧮 Worked example patterns

🧮 Single unknown number

• "The difference of a number and eight is 17" → n − 8 = 17 → n = 25.
• "The sum of four times a number and two is 14" → 4n + 2 = 14 → 4n = 12 → n = 3.

🧮 Two related numbers

• "One number is six more than another; their sum is twenty-four" → n + (n + 6) = 24 → 2n + 6 = 24 → 2n = 18 → n = 9; the numbers are 9 and 15.
• "The sum of two numbers is negative fourteen; one number is four less than the other" → n + (n − 4) = −14 → 2n − 4 = −14 → 2n = −10 → n = −5; the numbers are −5 and −9.

🧮 Three consecutive integers

• "Find three consecutive integers whose sum is −42" → n + (n + 1) + (n + 2) = −42 → 3n + 3 = −42 → 3n = −45 → n = −15; the integers are −15, −14, and −13.

🧭 Overview

🧠 One-sentence thesis

Percent problems in everyday life—tips, discounts, interest, and mark-ups—can be solved by translating English sentences into algebraic equations using the basic percent relationship "part = percent × whole."

📌 Key points (3–5)

• Core translation pattern: "is" means equals, "of" means multiply, and "what" represents the unknown variable.
• Three basic question types: finding the part (What number is 35% of 90?), finding the whole (6.5% of what number is $1.17?), or finding the percent (144 is what percent of 96?).
• Real-world applications: tips, sales tax, interest, discounts, mark-ups, and percent increase/decrease all use the same translation method.
• Common confusion: Percent increase/decrease requires first finding the amount of change (difference between original and new), then calculating what percent that change is of the original amount.
• Simple interest formula: I = P × r × t, where interest equals principal times rate times time.

🔤 Translating percent sentences into equations

🔤 The basic translation rules

• "is" translates to the equals sign (=)
• "of" translates to multiplication (×)
• "what" or the unknown quantity becomes a variable

📝 Three standard question formats

Question type	Example	What you're finding
Find the part	What number is 35% of 90?	The result of the percent calculation
Find the whole	6.5% of what number is $1.17?	The original total amount
Find the percent	144 is what percent of 96?	The percent itself (must convert final answer to percent form)

⚠️ Key reminder

Always convert the percent to a decimal before multiplying (e.g., 35% becomes 0.35).

💰 Real-world percent applications

💰 Tips and restaurant bills

• Pattern: Tip amount = tip rate × total bill
• Example: For an $68.50 bill with 18% tip, translate "tip is 18% of $68.50" → t = 0.18 × 68.50 = $12.33
• The tip is always a portion of the bill, so the answer should be less than the total.

🥣 Nutrition labels and daily values

• Pattern: Given amount = percent × total recommended amount
• Example: 85 mg of potassium is 2% of the daily amount → 85 = 0.02 × a → a = 4,250 mg
• When finding the total recommended amount, you're finding the whole, so divide the given amount by the decimal percent.

🍫 Finding what percent one quantity is of another

• Pattern: Part is what percent of whole?
• Example: 240 calories of fat out of 480 total calories → 240 = p × 480 → p = 0.5 = 50%
• Don't forget to convert the final decimal answer to percent form.

📈 Percent increase and decrease

📈 Percent increase

Percent increase: measures how much a quantity has grown as a percent of the original amount.

Two-step process:

1. Find the amount of increase: new amount − original amount = increase
2. Find what percent the increase is of the original: increase = percent × original amount

Example: Fees rise from $26 to $36

• Increase = 36 − 26 = $10
• 10 = p × 26 → p = 0.384 = 38.4%

📉 Percent decrease

Percent decrease: measures how much a quantity has fallen as a percent of the original amount.

Two-step process:

1. Find the amount of decrease: original amount − new amount = decrease
2. Find what percent the decrease is of the original: decrease = percent × original amount

Example: Gas price drops from $3.71 to $3.64

• Decrease = 3.71 − 3.64 = $0.07
• 0.07 = p × 3.71 → p = 0.019 = 1.9%

⚠️ Critical distinction

Always calculate the percent change based on the original amount, not the new amount—this is a common error.

💵 Simple interest applications

💵 The simple interest formula

Simple interest formula: I = P × r × t

• I = interest earned
• P = principal (amount invested or borrowed)
• r = annual interest rate (as a decimal)
• t = time in years

🏦 Finding interest earned

When you know the principal, rate, and time, multiply them together.

Example: $12,500 at 4% for 5 years

• I = 12,500 × 0.04 × 5 = $2,500

📊 Finding the rate

When you know the interest, principal, and time, substitute and solve for r.

Example: $3,000 loan paid back with $660 interest after 4 years

• 660 = 3,000 × r × 4
• 660 = 12,000r
• r = 0.055 = 5.5%

💰 Finding the principal

When you know the interest, rate, and time, substitute and solve for P.

Example: $6,596.25 interest at 7.5% over 5 years

• 6,596.25 = P × 0.075 × 5
• 6,596.25 = 0.375P
• P = $17,590

🏷️ Discount and mark-up applications

🏷️ Discount problems

Discount: the amount subtracted from the original price.

• Amount of discount = discount rate × original price
• Sale price = original price − amount of discount

Two common scenarios:

1. Given original price and discount rate, find sale price: Calculate the discount amount, then subtract from original.
2. Given original and sale price, find discount rate: First find discount amount (original − sale), then calculate what percent that is of the original.

Example: $140 dress discounted 35%

• Discount = 0.35 × 140 = $49
• Sale price = 140 − 49 = $91

🏪 Mark-up problems

Mark-up: the amount added to the original cost to get the selling price.

• Amount of mark-up = mark-up rate × original cost
• List price = original cost + amount of mark-up

Example: $250 photograph marked up 40%

• Mark-up = 0.40 × 250 = $100
• List price = 250 + 100 = $350

⚠️ Don't confuse

• Discount makes the price go down (sale price < original price)
• Mark-up makes the price go up (list price > original cost)

🧭 Overview

🧠 One-sentence thesis

Mixture problems combine two or more items with different values—such as coins, tickets, or stamps—by organizing information in a table and translating the total value relationship into an equation.

📌 Key points (3–5)

• Core model: For any type of item, number × value = total value; add up all total values to get the overall total.
• Coin problems setup: Organize coins by type in a table with columns for type, number, value, and total value.
• Variable strategy: When one quantity is defined in terms of another (e.g., "nine more nickels than dimes"), let the variable represent the base quantity and write an expression for the other.
• Common confusion: Don't confuse the number of coins with their value—17 dimes is 17 coins, but their total value is 17 × $0.10 = $1.70.
• Why it matters: This model applies to grocers, bartenders, chemists, investment bankers, landscapers, and anyone combining items with different unit values.

🪙 The fundamental mixture model

🪙 Number × value = total value

Total Value of Coins: For the same type of coin, the total value of a number of coins is found by using the model number · value = total value, where number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.

• This is the backbone of all mixture problems.
• Example: 17 dimes → 17 (number) × $0.10 (value per dime) = $1.70 (total value).
• The same logic applies to any item with a unit value: tickets, stamps, ingredients, etc.

📊 Organizing with a table

• Create a table with four columns: Type | Number | Value | Total Value.
• Fill in what you know: the types of items, the value per item, and the overall total.
• Use variables and expressions for unknown quantities.
• Multiply across each row: number × value = total value.
• The sum of all "total value" entries equals the grand total given in the problem.

🔢 Solving coin word problems step-by-step

🔢 Seven-step method

1. Read the problem and identify the types of coins; create the table.
2. Identify what you are looking for (e.g., number of dimes and nickels).
3. Name the unknowns with variables; use expressions when one quantity is defined in terms of another.
4. Translate into an equation by adding the total values.
5. Solve the equation using algebra.
6. Check that the answer makes sense and satisfies the problem.
7. Answer with a complete sentence.

🧩 Choosing variables wisely

• When the problem says "nine more nickels than dimes," let d = number of dimes, then d + 9 = number of nickels.
• When it says "twice as many pennies as quarters," let q = number of quarters, then 2q = number of pennies.
• Always define the variable for the base quantity, then write an expression for the quantity defined in terms of it.
• Don't confuse: if you let d = dimes, then d + 9 is the number of nickels, not their value.

🧮 Worked example patterns

🧮 Example: "Nine more nickels than dimes"

• Problem: Adalberto has $2.25 in dimes and nickels; he has nine more nickels than dimes.
• Let d = number of dimes, so d + 9 = number of nickels.
• Table row for dimes: d (number) × $0.10 (value) = 0.10d (total value).
• Table row for nickels: (d + 9) × $0.05 = 0.05(d + 9).
• Equation: 0.10d + 0.05(d + 9) = 2.25.
• Solve: 0.10d + 0.05d + 0.45 = 2.25 → 0.15d = 1.80 → d = 12 dimes, so 21 nickels.
• Check: 12(0.10) + 21(0.05) = 1.20 + 1.05 = 2.25 ✓

🧮 Example: "Twice as many pennies as quarters"

• Problem: Maria has $2.43 in quarters and pennies; she has twice as many pennies as quarters.
• Let q = number of quarters, so 2q = number of pennies.
• Equation: 0.25q + 0.01(2q) = 2.43.
• Solve: 0.25q + 0.02q = 2.43 → 0.27q = 2.43 → q = 9 quarters, so 18 pennies.
• Check: 9(0.25) + 18(0.01) = 2.25 + 0.18 = 2.43 ✓

🧮 Example: "Two more than ten times"

• Problem: Danny has $2.14 in pennies and nickels; the number of nickels is two more than ten times the number of pennies.
• Let p = number of pennies, so 10p + 2 = number of nickels.
• Equation: 0.01p + 0.05(10p + 2) = 2.14.
• Solve: 0.01p + 0.50p + 0.10 = 2.14 → 0.51p = 2.04 → p = 4 pennies, so 42 nickels.
• Check: 4(0.01) + 42(0.05) = 0.04 + 2.10 = 2.14 ✓

🎟️ Extension to tickets and stamps

🎟️ Same model, different items

• The excerpt notes: "Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a [value]."
• Use the identical table structure: Type | Number | Value | Total Value.
• Example: adult tickets at $12 each, child tickets at $8 each, total revenue $200 → same setup, same equation-building process.
• The mixture model applies to any scenario where you combine items with different unit values.

🧭 Overview

🧠 One-sentence thesis

The excerpt demonstrates systematic problem-solving techniques for coin, ticket, stamp, and mixture word problems by organizing information in tables, defining variables based on relationships between quantities, and writing equations from total values.

📌 Key points (3–5)

• Core method: Use a table to organize types, quantities, values, and total values; define one unknown variable and express the other in terms of it; write an equation by adding total values.
• Two variable-definition strategies: (1) when one quantity is described in terms of another (e.g., "two more than ten times"), define that relationship directly; (2) when the total is known, define the second quantity as "total minus first quantity."
• Common confusion: Don't mix up "number of items" with "total value"—always multiply number times unit value to get total value for each type.
• Application domains: The same table-and-equation structure works for coins, tickets, stamps, trail mix ingredients, and investment accounts with different interest rates.
• Why it matters: This structured approach prevents errors, ensures all information is accounted for, and makes checking answers straightforward.

🪙 Coin problems: organizing and defining variables

🪙 The table structure

Every coin problem uses a table with columns:

• Type (pennies, nickels, dimes, quarters, etc.)
• Number (the variable or expression)
• Value (unit value: $0.01, $0.05, $0.10, $0.25, etc.)
• Total Value (number × value)

The equation comes from adding the total values in the last column.

🔢 Strategy 1: One quantity defined in terms of another

When the problem states a relationship like "the number of nickels is two more than ten times the number of pennies":

• Let p = number of pennies (the base quantity).
• Then 10p + 2 = number of nickels (the derived quantity).
• Multiply each by its unit value to get total values.
• Write the equation: 0.01p + 0.05(10p + 2) = total dollar amount.

Example: Danny has $2.14 in pennies and nickels; nickels are two more than ten times pennies. Let p = pennies, so 10p + 2 = nickels. Equation: 0.01p + 0.05(10p + 2) = 2.14. Solving gives p = 4 pennies and 42 nickels.

🔢 Strategy 2: Total number known, subtract to find the other

When the problem gives the total number of coins (or tickets, stamps, etc.):

• Let c = number of one type (e.g., children's tickets).
• Then (total − c) = number of the other type (e.g., adult tickets).
• This ensures the two quantities add up to the known total.

The excerpt explains: "If Bianca sold a total of 100 tickets and sold x child tickets, then she sold 100 − x adult tickets."

Example: Galen sold 810 tickets (children at $3, adults at $5) for $2,820 total. Let c = children's tickets, so 810 − c = adult tickets. Equation: 3c + 5(810 − c) = 2,820. Solving gives c = 615 children's tickets and 195 adult tickets.

⚠️ Don't confuse number and value

• The number is how many coins (or tickets, stamps) you have.
• The value is the worth of one coin (e.g., $0.05 for a nickel).
• The total value is number × value.
• The equation always sums total values, not numbers.

🎟️ Ticket and stamp problems: same structure, different items

🎟️ Tickets with one quantity defined by the other

Ticket problems follow the coin-problem table exactly, but with ticket types and prices.

Example: A concert sold $1,506 in tickets; student tickets $6 each, adult tickets $9 each; adults were five less than three times students. Let s = student tickets, so 3s − 5 = adult tickets. Equation: 6s + 9(3s − 5) = 1,506. Solving gives s = 47 students and 136 adults.

🎟️ Tickets with total number known

When the total number of tickets is given, use the subtraction strategy.

Example: 115 museum tickets sold for $1,163; adults $12, students $5. Let a = adult tickets, so 115 − a = student tickets. Equation: 12a + 5(115 − a) = 1,163.

📮 Stamp problems

Stamp problems are identical in structure; the "value" is the stamp denomination (e.g., $0.41, $0.02, $0.20).

Example: Monica paid $8.36 for stamps; 41-cent stamps were four more than twice the number of 2-cent stamps. Let x = number of 2-cent stamps, so 2x + 4 = number of 41-cent stamps. Equation: 0.02x + 0.41(2x + 4) = 8.36. Solving gives x = 8 two-cent stamps and 20 41-cent stamps.

🥜 Mixture problems: ingredients and investments

🥜 Trail mix and ingredient mixtures

Mixture problems use the same table structure:

• Type (raisins, nuts, juice, soda, etc.)
• Number (pounds, gallons, liters)
• Value (price per unit)
• Total Value (number × value)

The key difference: the total row represents the final mixture with its own per-unit cost.

Example: Henning makes 10 pounds of trail mix; raisins $2/lb, nuts $6/lb; he wants the mix to cost $5.20/lb. Let x = pounds of raisins, so 10 − x = pounds of nuts. Equation: 2x + 6(10 − x) = 5.20(10). Solving gives x = 2 pounds raisins and 8 pounds nuts.

💰 Investment problems with simple interest

For investments earning interest for one year, use the simple interest formula I = Pr (where t = 1 year, so I = principal × rate).

The table structure:

• Type (account at 3%, account at 5%, etc.)
• Principal (amount invested)
• Rate (interest rate as a decimal)
• Interest (principal × rate)

The equation sums the interest from each account to equal the desired total interest.

Example: Stacey invests $20,000 in two accounts (3% and 5%) and wants 4.5% overall interest. Let x = amount at 3%, so 20,000 − x = amount at 5%. Equation: 0.03x + 0.05(20,000 − x) = 0.045(20,000). Solving gives x = $5,000 at 3% and $15,000 at 5%.

🔍 Why the mixture model works

The excerpt states: "Grocers and bartenders use the mixture model to set a fair price for a product made from mixing two or more ingredients. Financial planners use the mixture model when they invest money in a variety of accounts and want to find the overall interest rate."

The common thread: you have two (or more) components with different unit values, and you want the combined result to have a specific average or total value.

✅ The seven-step problem-solving process

✅ Steps 1–3: Read, Identify, Name

1. Read the problem and determine the types involved (coins, tickets, ingredients, accounts).
2. Identify what you are looking for (number of each type).
3. Name the unknowns using variables; choose the base variable for the quantity that others are defined in terms of, or use subtraction if the total is known.

✅ Steps 4–5: Translate, Solve

4. Translate by writing the equation from the total-value row of the table.
5. Solve the equation using algebra (distribute, combine like terms, isolate the variable).

✅ Steps 6–7: Check, Answer

6. Check by substituting your solution back into the original problem statement and verifying the total value or total number.
7. Answer the question in a complete sentence with units.

Example check (from Danny's coin problem): "Danny has four pennies and 42 nickels. Is the total value $2.14? 4(0.01) + 42(0.05) = 0.04 + 2.10 = 2.14 ✓"

⚠️ Don't skip the check step

The excerpt consistently shows a check calculation for every example. This catches arithmetic errors and confirms that the solution satisfies both the relationship between quantities and the total value.

📊 Summary table: problem types and strategies

Problem type	What varies	What's constant	Variable strategy	Equation source
Coin (relationship given)	Number of each coin type	Unit values ($0.01, $0.05, etc.)	Define base variable; express other as formula (e.g., 10p + 2)	Sum of (number × value) = total dollars
Coin/Ticket (total number given)	Number of each type	Total number, unit values	Let x = one type; other = total − x	Sum of (number × value) = total dollars
Stamp	Number of each stamp type	Stamp denominations	Same as coin problems	Sum of (number × value) = total cost
Trail mix / ingredient mixture	Pounds/gallons of each ingredient	Price per unit, total amount, desired per-unit cost	Let x = amount of one; total − x = amount of other	Sum of (amount × price) = total cost of mixture
Investment (simple interest)	Amount in each account	Interest rates, total principal, desired overall rate	Let x = amount in one account; total − x = amount in other	Sum of (principal × rate) = total interest

🧭 Overview

🧠 One-sentence thesis

Geometry applications use formulas for triangles, rectangles, and the Pythagorean Theorem to solve real-world problems by translating word problems into equations, solving them algebraically, and checking that solutions make sense in context.

📌 Key points (3–5)

• Problem-solving strategy: Read, draw and label a figure, identify what you're looking for, choose a variable, translate to an equation using the appropriate formula, solve, check, and answer with a complete sentence.
• Triangle properties: The sum of angles equals 180°; perimeter is the sum of all sides; area is one-half base times height.
• Pythagorean Theorem: In right triangles, the square of the hypotenuse equals the sum of the squares of the two legs (a² + b² = c²).
• Rectangle properties: Opposite sides are equal; perimeter is 2L + 2W; area is L × W.
• Common confusion: Don't confuse the "number" of something with its "value"—when one dimension is defined in terms of another (e.g., "width is two feet less than length"), wait to draw the figure until you write expressions for all unknowns.

🔺 Triangle applications

🔺 Angle relationships

Triangle angle sum: The sum of the measures of the angles of a triangle is 180°.

• For any triangle ABC: m∠A + m∠B + m∠C = 180°
• This property applies to all triangles, not just right triangles
• Example: If two angles measure 55° and 82°, the third angle is 180 - 55 - 82 = 43°

📐 Right triangles

Right triangle: A triangle with one 90° angle, often marked with a small square at the vertex.

• One angle is always 90°, so the other two angles must sum to 90°
• Example: If one angle of a right triangle is 28°, the third angle is 180 - 90 - 28 = 62°
• When one angle is defined in terms of another (e.g., "20 degrees more than the smallest"), set up an equation using the angle sum property

📏 Perimeter and area

Triangle perimeter: P = a + b + c (sum of all three sides) Triangle area: A = ½bh, where b is the base and h is the height

• The height must make a 90° angle with the base
• Example: If a triangular window has area 90 square meters and base 15 meters, solve 90 = ½(15)h to get h = 12 meters
• Don't confuse: Perimeter adds lengths; area multiplies base times height (then divides by 2)

📐 Pythagorean Theorem applications

📐 The theorem

Pythagorean Theorem: In any right triangle, a² + b² = c², where a and b are the lengths of the legs and c is the length of the hypotenuse.

• The hypotenuse is the side opposite the 90° angle
• The two legs are the sides that form the right angle
• This only works for right triangles

🔍 Finding the hypotenuse

• When you know both legs, substitute into a² + b² = c²
• Example: If legs are 3 and 4, then 3² + 4² = c², so 9 + 16 = c², so 25 = c², so c = 5
• Take the square root of both sides to find c

🔍 Finding a leg

• When you know the hypotenuse and one leg, substitute and solve for the unknown leg
• Example: If one leg is 5 and hypotenuse is 13, then 5² + b² = 13², so 25 + b² = 169, so b² = 144, so b = 12
• Isolate the variable term before taking the square root

🧮 Square roots

Square root definition: If m = n², then √m = n, for n ≥ 0

• Example: √25 = 5 because 25 = 5²
• When the result is not a perfect square, approximate to the nearest tenth
• Example: If 2x² = 100, then x² = 50, so x = √50 ≈ 7.1

🏗️ Real-world applications

• Ladders leaning against walls form right triangles
• Diagonal braces in construction create right triangles
• Example: A 10-inch diagonal brace with equal legs means x² + x² = 10², so 2x² = 100, so x² = 50, so x ≈ 7.1 inches from the corner

🟦 Rectangle applications

🟦 Rectangle properties

Rectangle: A four-sided figure with four right (90°) angles and opposite sides of equal length. Perimeter: P = 2L + 2W (twice the length plus twice the width) Area: A = L × W (length times width)

• L represents length, W represents width
• Opposite sides are equal, so you only need two measurements
• Example: A rectangle 32 meters by 20 meters has perimeter 2(32) + 2(20) = 104 meters

🔢 Direct applications

• When given three of the four values (P, L, W, A), substitute into the appropriate formula
• Example: If area is 168 square feet and length is 14 feet, solve 168 = 14W to get W = 12 feet
• Example: If perimeter is 50 inches and width is 10 inches, solve 50 = 2L + 2(10) to get L = 15 inches

🔤 Variable expressions

• When one dimension is defined in terms of the other, write an expression before drawing the figure
• Example: "Width is two feet less than length" means if L = length, then W = L - 2
• Example: "Length is four more than twice the width" means if W = width, then L = 2W + 4

🧩 Multi-step problems

Given information	Strategy
Width defined in terms of length	Let L = length, express width as an expression in L
Length defined in terms of width	Let W = width, express length as an expression in W
Perimeter and relationship	Substitute both expressions into P = 2L + 2W and solve

• Example: "Width is two feet less than length, perimeter is 52 feet"
  • Let L = length, then W = L - 2
  • Substitute: 52 = 2L + 2(L - 2)
  • Solve: 52 = 2L + 2L - 4, so 56 = 4L, so L = 14 feet
  • Then W = 14 - 2 = 12 feet

🛠️ Problem-solving strategy

🛠️ Seven-step process

1. Read the problem and make sure all words and ideas are understood; draw the figure and label it with given information
2. Identify what you are looking for
3. Label what you are looking for by choosing a variable to represent it
4. Translate into an equation by writing the appropriate formula and substituting given information
5. Solve the equation using algebra
6. Check the answer by substituting back and making sure it makes sense
7. Answer the question with a complete sentence

✏️ Drawing and labeling

• Always draw a diagram first—it helps you visualize the problem
• Label known values directly on the figure
• Use variables for unknown values
• When one quantity is defined in terms of another, write the expression before completing the diagram

✅ Checking your work

• Substitute your answer back into the original equation
• Verify that the answer makes sense in the real-world context
• Example: If you find a width of 12 feet and length of 14 feet with perimeter 52, check that 12 + 14 + 12 + 14 = 52 ✓

🧭 Overview

🧠 One-sentence thesis

A systematic seven-step problem-solving strategy helps translate word problems into solvable equations by organizing information, choosing variables, translating to algebra, solving, checking, and answering in complete sentences.

📌 Key points (3–5)

• The seven-step strategy: Read, Identify, Name, Translate, Solve, Check, Answer—a structured approach to any word problem.
• Consecutive integer patterns: consecutive integers differ by 1 (n, n+1, n+2); consecutive even or odd integers differ by 2 (n, n+2, n+4).
• Variable naming matters: choose a variable to represent the unknown quantity and use expressions for related quantities.
• Common confusion: "consecutive even" vs "consecutive odd" use the same algebraic pattern (n, n+2, n+4) because both skip one integer.
• Why it matters: the strategy turns confusing word problems into manageable algebra by breaking the process into clear steps.

🛠️ The Seven-Step Strategy

📖 Step 1: Read the problem

• Make sure all words and ideas are understood before proceeding.
• This is not just skimming—ensure you grasp what the problem is asking.

🔍 Step 2: Identify what we are looking for

• Pinpoint the unknown quantity the problem asks you to find.
• Example: "How many hours must she tutor?" → looking for number of hours.

🏷️ Step 3: Name what we are looking for

Choose a variable to represent the unknown quantity.

• Pick a letter (often h, n, x) and state what it represents.
• Example: Let h = number of hours tutoring.

🔄 Step 4: Translate into an equation

• Restate the problem in one sentence with all important information.
• Then convert that sentence into an algebraic equation.
• This is the bridge from English to algebra.

⚙️ Step 5: Solve the equation

• Use algebra techniques (isolate the variable, combine like terms, etc.).
• Apply the methods learned in earlier chapters.

✅ Step 6: Check the answer

• Substitute your solution back into the original problem context.
• Does it make sense? Does it satisfy all conditions?
• Don't confuse: checking in the equation vs checking in the problem—both are important.

💬 Step 7: Answer the question with a complete sentence

• State the result in words, not just a number.
• Example: "She must tutor 12 hours" rather than just "12."

🔢 Consecutive Integer Patterns

🔢 Consecutive integers

Consecutive integers are integers that immediately follow each other.

• Pattern: n, n+1, n+2, n+3, ...
• Each integer is 1 more than the previous.
• Example: if the first integer is n, the next is n+1, then n+2.

➕ Consecutive even integers

Consecutive even integers are even integers that immediately follow one another.

• Pattern: n, n+2, n+4, n+6, ...
• Each even integer is 2 more than the previous even integer.
• Example: 8, 10, 12 can be represented as n, n+2, n+4.

➕ Consecutive odd integers

Consecutive odd integers are odd integers that immediately follow one another.

• Pattern: n, n+2, n+4, n+6, ...
• Each odd integer is 2 more than the previous odd integer.
• Example: 7, 9, 11 can be represented as n, n+2, n+4.
• Don't confuse: consecutive odd uses the same +2 pattern as consecutive even because you skip the even (or odd) numbers in between.

📝 Applying the Strategy

📝 Setting up the equation

• Use the variable expressions to represent all quantities in the problem.
• Write relationships as equations using the information given.
• Example: "The sum of a number and three is forty-one" → n + 3 = 41.

📝 Organizing information

• Sometimes a table or diagram helps organize what you know.
• List knowns and unknowns clearly before translating.

📝 Common problem types

The excerpt mentions several contexts where this strategy applies:

Problem type	What to look for
Number problems	Relationships between unknown numbers
Consecutive integers	Patterns like n, n+1, n+2 or n, n+2, n+4
Money/savings	Total amounts, costs, and budgets

📝 Why complete sentences matter

• The final answer step forces you to return to the original question.
• It ensures you answered what was asked, not just solved an equation.
• Example: solving for n doesn't help if the question asked for n+1.

🧭 Overview

🧠 One-sentence thesis

Linear equations in two variables have infinitely many solutions represented as ordered pairs on the coordinate plane, where each solution makes the equation a true statement when both coordinates are substituted.

📌 Key points (3–5)

• Plotting and reading points: ordered pairs (x, y) are plotted on a rectangular coordinate system using x-axis and y-axis values; reading coordinates requires identifying the correct axis position.
• Linear equations in two variables: equations of the form Ax + By = C (where A and B are not both zero) have infinitely many solutions, unlike one-variable equations that typically have one solution.
• Standard form: a linear equation is in standard form when written as Ax + By = C, with A, B, and C preferably integers and A ≥ 0.
• Common confusion: an ordered pair is a solution only when substituting both x and y values into the equation produces a true statement; not all ordered pairs satisfy a given equation.
• Special axis points: when x = 0, the point lies on the y-axis; when y = 0, the point lies on the x-axis; when both equal 0, the point is the origin.

📍 Plotting and identifying points

📍 How to plot ordered pairs

• An ordered pair (x, y) represents a point on the rectangular coordinate system.
• The first number (x-coordinate) tells you the horizontal position; the second number (y-coordinate) tells you the vertical position.
• Always write coordinates in the correct order: (x, y).

🎯 Special positions on axes

Condition	Location	Example
y = 0	Point is on the x-axis	(4, 0), (−3, 0)
x = 0	Point is on the y-axis	(0, −1), (0, 2)
x = 0 and y = 0	Point is the origin	(0, 0)

🔍 How to read coordinates from a graph

• For the x-coordinate: read the number on the x-axis directly above or below the point.
• For the y-coordinate: read the number on the y-axis directly to the left or right of the point.
• Example: Point A is above −3 on the x-axis and to the left of 3 on the y-axis → coordinates are (−3, 3).

Don't confuse: "above/below" refers to the x-axis position (giving the x-coordinate), while "left/right" refers to the y-axis position (giving the y-coordinate).

🧮 Linear equations in two variables

🧮 What makes an equation linear

Linear equation in two variables: an equation of the form Ax + By = C, where A and B are not both zero.

• The word "linear" relates to "line"—these equations graph as straight lines.
• Unlike one-variable equations (which typically have one solution, like x = 4), linear equations in two variables have infinitely many solutions.
• Example: The equation 3x + 5 = 17 has one solution (x = 4), but x + 4y = 8 has infinitely many solutions.

📐 Standard form

Standard form of linear equation: Ax + By = C

• Most people prefer A, B, and C to be integers and A ≥ 0, although this is not strictly necessary.
• Some equations need rewriting to appear in standard form.
• Example: y = −3x + 5 can be rewritten as 3x + y = 5 by adding 3x to both sides and using the Commutative Property.

🔄 Converting to standard form

The excerpt shows how to rewrite y = −3x + 5:

1. Add 3x to both sides: y + 3x = −3x + 5 + 3x
2. Simplify: y + 3x = 5
3. Use Commutative Property: 3x + y = 5

Now it is clearly in Ax + By = C form.

✅ Solutions to linear equations

✅ What is a solution

Solution of a linear equation in two variables: an ordered pair (x, y) is a solution if the equation is a true statement when the x- and y-values are substituted into the equation.

• For every number substituted for x, there is a corresponding y value.
• This pair of values is represented by the ordered pair (x, y).
• When you substitute these values into the equation, the result must be a true statement (left side equals right side).

🧪 How to verify a solution

The excerpt demonstrates checking if (0, 2) is a solution to x + 4y = 8:

1. Substitute x = 0 and y = 2 into the equation.
2. Check if the resulting statement is true.
3. If both sides are equal, the ordered pair is a solution; if not, it is not a solution.

Don't confuse: just because an ordered pair looks reasonable doesn't mean it's a solution—you must substitute and verify that the equation becomes a true statement.

🔢 Multiple solutions

• Linear equations in two variables have infinitely many solutions, not just one.
• Each solution corresponds to a point on the graph of the equation.
• Example: For x + 4y = 8, you can test multiple ordered pairs like (0, 2), (2, −4), and (−4, 3) to see which ones satisfy the equation.

🧭 Overview

🧠 One-sentence thesis

Mixture applications use algebraic equations to solve real-world problems involving combinations of items with different values, such as coins, tickets, stamps, and blended products.

📌 Key points (3–5)

• Core model for coins: number · value = total value, where you multiply the count of each coin type by its denomination to find its contribution to the total.
• Problem structure: mixture problems typically involve two or more items with different unit values that combine to reach a specific total value or quantity.
• Variable strategy: choose one unknown quantity as your variable, then express other quantities in terms of that variable using relationships given in the problem.
• Common confusion: don't mix up the number of items with their value—always distinguish between "how many" (count) and "how much each is worth" (unit value).
• Applications span multiple contexts: the same algebraic approach works for coin problems, ticket sales, stamp purchases, and ingredient mixtures.

💰 Coin word problems

💰 The fundamental coin equation

For the same type of coin, the total value of a number of coins is found by using the model: number · value = total value

• number = how many coins of that type
• value = the denomination of each coin (e.g., $0.25 for quarters)
• total value = the combined worth of all coins of that type

Example: If you have 8 quarters, the total value is 8 · $0.25 = $2.00.

📋 Organizing coin problems with tables

The excerpt recommends creating a table with these columns:

• Type: the kind of coin (pennies, nickels, dimes, quarters, etc.)
• Number: how many of each coin
• Value: the worth of one coin
• Total value: number × value for that coin type

This visual organization helps you see all the information and identify what's missing.

🔢 Setting up the equation

• Write expressions for the number of each coin type using your variable
• Multiply each count by its value to get each coin type's total value
• Add up all the total values to equal the overall total given in the problem

Don't confuse: The "number" column contains counts (whole numbers of coins), while the "value" and "total value" columns contain dollar amounts.

🎟️ Ticket and stamp problems

🎟️ Same structure, different context

Ticket and stamp problems follow the identical model as coin problems:

• number of tickets/stamps · price per ticket/stamp = total revenue/cost

Example: If 125 tickets were sold, some at $6 (student) and some at $10 (adult), you'd set up two categories just like two coin types.

🎫 Typical ticket problem setup

• Two or more ticket types at different prices
• Given: total number sold and total revenue
• Find: how many of each type were sold

The relationship between ticket types often involves phrases like "12 more than twice the number of adults" which you translate into algebraic expressions.

📮 Stamp problem variations

Stamp problems work identically but use stamp denominations (e.g., $0.41 stamps, $0.56 stamps).

The excerpt shows examples where the number of one stamp type is expressed in terms of another (e.g., "five more than the number of $0.26 stamps").

🧪 Mixture word problems (ingredients and investments)

🧪 Blending items with different unit values

Mixture problems involve combining two components where:

• Each component has a different price per unit
• You want a specific total quantity at a target average price

Example: Making 10 pounds of trail mix from raisins ($3.45/lb) and nuts ($7.95/lb) to achieve $6.96/lb overall.

💵 Investment mixture problems

The same model applies to splitting money between accounts with different interest rates:

• Total amount invested = sum of amounts in each account
• Weighted average interest rate = desired overall return

Example: Investing $15,000 split between 4.5% and 1.8% accounts to achieve 4.05% overall interest.

🎯 The mixture equation pattern

For two components:

• (amount₁ · rate₁) + (amount₂ · rate₂) = (total amount · target rate)

Don't confuse: The "rate" here means price per unit or interest rate, not speed—context determines the meaning.

🔧 Problem-solving strategy for all mixture applications

📝 Step-by-step approach

1. Read carefully and identify what types of items are being combined
2. Identify what you're looking for (usually quantities of each type)
3. Name your variable (choose one unknown quantity)
4. Create a table to organize type, number/amount, value/rate, and total value
5. Translate relationships into expressions using your variable
6. Write the equation by adding total values across all types
7. Solve using algebra techniques
8. Check that your answer makes sense in the original problem context
9. Answer with a complete sentence stating what you found

⚠️ Common pitfalls to avoid

• Mixing up the number of items with their individual values
• Forgetting to multiply number × value before adding
• Using the wrong units (keep dollars as dollars, pounds as pounds, etc.)
• Not expressing all quantities in terms of your chosen variable

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Note: The excerpt provided contains primarily exercise problems (numbered 319-338) rather than instructional content. The review notes above are based on the "3.3 Solve Mixture Applications" key concepts section found in the Chapter 3 Review portion of the document, which summarizes the methodology for coin word problems and references mixture problems generally.

🧭 Overview

🧠 One-sentence thesis

The rectangular coordinate system uses ordered pairs to locate points on a grid formed by perpendicular number lines, enabling visual representation of relationships between two variables.

📌 Key points (3–5)

• What the system is: a grid formed by a horizontal x-axis and vertical y-axis that divides the plane into four quadrants.
• How points are located: every point is represented by an ordered pair (x, y), where the first number is the x-coordinate and the second is the y-coordinate.
• Quadrant sign patterns: the signs of coordinates determine which quadrant a point occupies (I: +,+; II: −,+; III: −,−; IV: +,−).
• Common confusion: order matters in ordered pairs—(1, 3) and (3, 1) are different points; don't mix up which coordinate comes first.
• Special cases: points on the axes have one coordinate equal to zero; the origin (0, 0) is where both axes intersect.

📐 The coordinate plane structure

📐 What makes up the system

Rectangular coordinate system: a grid system used in algebra to show relationships between two variables, also called the xy-plane or coordinate plane.

• The x-axis is the horizontal number line.
• The y-axis is the vertical number line.
• These two axes intersect at right angles and divide the plane into four regions.

🔢 The four quadrants

• The axes divide the plane into four quadrants, numbered with Roman numerals I, II, III, IV.
• Numbering begins in the upper right and proceeds counterclockwise.
• Each quadrant has a distinct sign pattern for its coordinates.

⭐ The origin

Origin: the point (0, 0) where the x-axis and y-axis intersect.

• This is the only point where both coordinates are zero.
• It serves as the reference point for the entire coordinate system.

🎯 Ordered pairs and coordinates

🎯 What an ordered pair represents

Ordered pair: a pair (x, y) that gives the coordinates of a point in a rectangular coordinate system.

• The first number is the x-coordinate (horizontal position).
• The second number is the y-coordinate (vertical position).
• The phrase "ordered pair" emphasizes that order is important.

📍 How to plot a point

To plot a point like (1, 3):

1. Locate the x-coordinate (1) on the x-axis and sketch a vertical line through x = 1.
2. Locate the y-coordinate (3) on the y-axis and sketch a horizontal line through y = 3.
3. The point is where these two lines meet.
4. The sketch lines are helpers only—they are not part of the graph.

🧭 Sign patterns and location

🧭 Quadrant sign patterns

Quadrant	Sign pattern	Description
I	(+, +)	Right of y-axis, above x-axis
II	(−, +)	Left of y-axis, above x-axis
III	(−, −)	Left of y-axis, below x-axis
IV	(+, −)	Right of y-axis, below x-axis

🔍 Reading signs to find location

• Negative x-coordinate: point is to the left of the y-axis.
• Positive x-coordinate: point is to the right of the y-axis.
• Negative y-coordinate: point is below the x-axis.
• Positive y-coordinate: point is above the x-axis.

Example: The point (−5, 4) has negative x and positive y, so it is in Quadrant II.

⚠️ Don't confuse

• (−2, 5) is in Quadrant II, but (2, −5) is in Quadrant IV—switching signs changes the quadrant.
• The order matters: (2, 3) is not the same as (3, 2).

🎲 Special cases: points on the axes

🎲 Points with zero coordinates

Points on the axes: Points with y-coordinate equal to 0 are on the x-axis and have coordinates (a, 0). Points with x-coordinate equal to 0 are on the y-axis and have coordinates (0, b).

• On the x-axis: the y-coordinate is zero, e.g., (4, 0) or (−2, 0).
• On the y-axis: the x-coordinate is zero, e.g., (0, 5) or (0, −1).
• Points on the axes are not in any quadrant—they lie on the boundary between quadrants.

🔢 Fractional or decimal coordinates

• Coordinates can be fractions or decimals, e.g., (3, 5/2).
• It may help to convert fractions to mixed numbers or decimals for easier plotting.
• Example: 5/2 = 2.5, so (3, 5/2) is at x = 3 and y = 2.5, placing it in Quadrant I.

🧭 Overview

🧠 One-sentence thesis

Points in a rectangular coordinate system are located by ordered pairs (x, y), and linear equations in two variables describe relationships where both coordinates can vary.

📌 Key points (3–5)

• Quadrant identification: The signs of x and y coordinates determine which quadrant a point occupies (positive/negative patterns differ by quadrant).
• Special axis points: Points with one coordinate equal to zero lie on an axis rather than in a quadrant.
• Reading coordinates: To identify a point's location, read the x-value from the x-axis and the y-value from the y-axis, always writing them as (x, y).
• Common confusion: Don't mix up the order—ordered pairs must be written as (x, y), not (y, x).
• Two-variable equations: Linear equations can have two variables (not just one), taking the form Ax + By = C.

📍 Understanding quadrants and sign patterns

📍 How signs determine location

The excerpt explains that coordinate signs follow predictable patterns:

Quadrant	Sign pattern (x, y)	Example
I	(+, +)	Both positive
II	(−, +)	x negative, y positive
III	(−, −)	Both negative
IV	(+, −)	x positive, y negative

• You can determine the quadrant just by looking at the signs before plotting.
• Example: The point (−2, 5) has negative x and positive y, so it must be in Quadrant II.
• Example: The point (2, −5) has positive x and negative y, so it must be in Quadrant IV.

🎯 Points on the axes

Points on the Axes: Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates (a, 0). Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates (0, b).

• When x equals 0, the point sits on the y-axis.
• When y equals 0, the point sits on the x-axis.
• When both equal 0, the point is at the origin (0, 0).
• Don't confuse: Axis points are not "in" any quadrant—they lie on the boundary between quadrants.

🔍 Reading and writing coordinates

🔍 How to identify coordinates from a graph

The excerpt provides a clear method:

• For the x-coordinate: Read the number on the x-axis directly above or below the point.
• For the y-coordinate: Read the number on the y-axis directly to the left or right of the point.
• Always write the result as an ordered pair in the correct order: (x, y).

✍️ Order matters

• The excerpt emphasizes: "Remember, when you write the ordered pair use the correct order, (x, y)."
• The first number is always the x-coordinate (horizontal position).
• The second number is always the y-coordinate (vertical position).
• Example: Point A above −3 on the x-axis and left of 3 on the y-axis has coordinates (−3, 3), not (3, −3).

🧮 Linear equations in two variables

🧮 What makes an equation linear

Linear Equation: An equation of the form Ax + By = C, where A and B are not both zero, is called a linear equation in two variables.

• Unlike one-variable equations (which have one solution like x = 4), two-variable equations involve both x and y.
• The word "line" in "linear" hints at how these equations behave graphically.
• Example: The equation 3x + 5 = 17 has one variable and one solution (x = 4).

🔄 Rewriting equations into standard form

The excerpt shows that equations can be rearranged into Ax + By = C form:

• Starting equation: y = −3x + 5
• Add 3x to both sides: y + 3x = −3x + 5 + 3x
• Simplify: y + 3x = 5
• Use the Commutative Property to arrange as: 3x + y = 5
• This demonstrates that different-looking equations may be equivalent linear equations in two variables.

🧭 Overview

🧠 One-sentence thesis

Linear equations in two variables have infinitely many solutions represented as ordered pairs, and any solution can be verified by substituting its coordinates into the equation to check if the result is a true statement.

📌 Key points (3–5)

• What a linear equation in two variables is: an equation of the form Ax + By = C, where A and B are not both zero.
• What a solution means: an ordered pair (x, y) is a solution when substituting those values into the equation produces a true statement.
• How to verify solutions: substitute the x- and y-values from the ordered pair into the equation and check if both sides are equal.
• Common confusion: linear equations in two variables have infinitely many solutions (not just one like single-variable equations); for every x-value there is a corresponding y-value.
• How to find solutions: pick any value for x and solve for y, or pick any value for y and solve for x.

📐 Understanding linear equations in two variables

📐 Definition and form

A linear equation in two variables: an equation of the form Ax + By = C, where A and B are not both zero.

• The word "linear" contains "line"—this hints at the graphical nature of these equations.
• Example: x + 4y = 8 is a linear equation in two variables.
• Some equations need to be rewritten to see they are linear. For instance, y = -3x + 5 can be rewritten as 3x + y = 5.

📋 Standard form

Standard form of a linear equation: Ax + By = C

• To rewrite an equation in standard form, use properties of equality to move terms.
• Example: Starting with y = -3x + 5, add 3x to both sides to get y + 3x = 5, then use the Commutative Property to write 3x + y = 5.
• Most people prefer A, B, and C to be integers and A ≥ 0, though this is not strictly required.

🔄 Contrast with one-variable equations

Feature	One-variable equations	Two-variable equations
Typical form	3x + 5 = 17	Ax + By = C
Number of solutions	Usually exactly one (e.g., x = 4)	Infinitely many
Solution representation	Single number	Ordered pair (x, y)

✅ Verifying solutions

✅ What makes an ordered pair a solution

An ordered pair (x, y) is a solution of the linear equation Ax + By = C if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.

• A solution is not just any ordered pair—it must make the equation true.
• When you substitute the values, the left side must equal the right side.

🔍 Step-by-step verification process

1. Take the ordered pair (x, y) you want to test.
2. Substitute the x-value for x in the equation.
3. Substitute the y-value for y in the equation.
4. Simplify both sides.
5. Check if the result is a true statement (left side = right side).

Example: To check if (0, 2) is a solution to x + 4y = 8:

• Substitute: 0 + 4(2) = 8
• Simplify: 0 + 8 = 8
• Result: 8 = 8 ✓ (true statement, so it is a solution)

Example: To check if (2, -4) is a solution to x + 4y = 8:

• Substitute: 2 + 4(-4) = 8
• Simplify: 2 + (-16) = 8
• Result: -14 = 8 ✗ (false statement, so it is not a solution)

🔢 Finding and organizing solutions

🔢 How to find solutions

• You can pick any value for x and solve the equation for y.
• Or you can pick any value for y and solve the equation for x.
• Each choice gives you a valid ordered pair solution.

Example: For y = 5x - 1, if you let x = 2:

• Substitute: y = 5(2) - 1
• Simplify: y = 10 - 1
• Result: y = 9
• Solution: (2, 9)

📊 Tables of solutions

• Solutions can be organized in a table with columns for x, y, and (x, y).
• A table helps visualize multiple solutions at once.
• Since there are infinitely many solutions, a table shows only a sample of them.

Example table for y = 5x - 1:

x	y	(x, y)
0	-1	(0, -1)
1	4	(1, 4)
2	9	(2, 9)

♾️ Infinite solutions

• Don't confuse: Unlike one-variable equations that typically have one solution, linear equations in two variables have infinitely many solutions.
• For every number substituted for x, there is a corresponding y-value.
• You can keep finding new solutions by choosing different x- or y-values.

🧭 Overview

🧠 One-sentence thesis

Linear equations in two variables have infinitely many solutions represented as ordered pairs that can be verified by substitution and organized systematically in tables.

📌 Key points (3–5)

• Linear equations in two variables: equations of the form Ax + By = C (where A and B are not both zero) have infinitely many solutions, each represented as an ordered pair (x, y).
• Verifying solutions: substitute the x- and y-values from an ordered pair into the equation; if the result is a true statement, the pair is a solution.
• Finding solutions systematically: pick any value for x (or y), substitute it into the equation, solve for the other variable, and record the resulting ordered pair.
• Common confusion: unlike one-variable equations that typically have one solution, two-variable linear equations have infinitely many solutions—every valid (x, y) pair is a solution.
• Standard form: the form Ax + By = C is called standard form, and any linear equation can be rewritten this way.

📐 Understanding linear equations in two variables

📐 What makes an equation linear in two variables

A linear equation in two variables: an equation of the form Ax + By = C, where A and B are not both zero.

• The word "linear" contains "line"—this hints at the geometric meaning.
• Unlike one-variable equations (like 3x + 5 = 17, which has one solution x = 4), these equations involve two variables.
• Example: x + 4y = 8 is a linear equation in two variables.

🔄 Standard form

Standard form of a linear equation: Ax + By = C.

• Most people prefer A, B, and C to be integers and A to be greater than or equal to zero.
• Not all linear equations look like standard form at first.
• Example: y = -3x + 5 can be rewritten as 3x + y = 5 by adding 3x to both sides and using the Commutative Property.

♾️ Infinitely many solutions

• For every number substituted for x, there is a corresponding y-value.
• Each valid pair of values is a solution represented by the ordered pair (x, y).
• Don't confuse: one-variable equations usually have one solution; two-variable linear equations have infinitely many solutions.

✅ Verifying solutions

✅ What makes an ordered pair a solution

Solution of a linear equation in two variables: an ordered pair (x, y) is a solution if the equation is a true statement when the x- and y-values are substituted into the equation.

• The key test: does substituting the values make both sides of the equation equal?
• If yes → the ordered pair is a solution.
• If no → the ordered pair is not a solution.

🧪 Testing ordered pairs

The excerpt shows how to verify whether given ordered pairs are solutions:

• For x + 4y = 8: test pairs like (0, 2), (2, -4), (-4, 3) by substituting.
• For y = 5x - 1: test pairs like (0, -1), (1, 4), (-2, -7) by substituting.

Example process:

• To test if (0, 2) is a solution to x + 4y = 8:
  • Substitute: 0 + 4(2) = 8
  • Simplify: 0 + 8 = 8
  • Result: 8 = 8 is true → (0, 2) is a solution

🔍 Finding solutions systematically

🔍 The pick-and-solve method

• You can pick any value for x (or y) and solve for the other variable.
• This generates a valid ordered pair solution.
• Since you can pick infinitely many values, there are infinitely many solutions.

📋 Completing solution tables

The excerpt demonstrates organizing solutions in tables:

Given	Process	Result
Pick x = 0	Substitute into equation, solve for y	Get ordered pair (0, y)
Pick x = -1	Substitute into equation, solve for y	Get ordered pair (-1, y)
Pick x = 2	Substitute into equation, solve for y	Get ordered pair (2, y)

🧮 Working with standard form equations

For equations like 5x - 4y = 20:

• If you're given x = 0, substitute and solve: 5(0) - 4y = 20 → -4y = 20 → y = -5.
• If you're given y = 0, substitute and solve: 5x - 4(0) = 20 → 5x = 20 → x = 4.
• If you're given x = 8, substitute and solve: 5(8) - 4y = 20 → 40 - 4y = 20 → -4y = -20 → y = 5.

💡 Choosing smart values

The excerpt notes that while you can pick any number to substitute, it's a good idea to choose a number that makes solving easier (though the excerpt cuts off before completing this thought).

🧭 Overview

🧠 One-sentence thesis

Linear equations in two variables produce straight-line graphs where every point on the line is a solution to the equation, and every solution to the equation is a point on that line.

📌 Key points (3–5)

• What a linear equation in two variables is: an equation of the form Ax + By = C (where A and B are not both zero), also called standard form.
• The graph-solution relationship: every point on the line represents a solution (ordered pair) to the equation, and every solution appears as a point on the line.
• How to graph by plotting points: find three solutions by choosing values for x or y, plot them, check they line up, then draw the line through them.
• Common confusion: equations like y = 4x versus y = 4—the first has both variables and produces a slanted line; the second has only y and produces a horizontal line.
• Special cases: vertical lines (x = a) and horizontal lines (y = b) are graphs of equations with only one variable.

📐 Linear equations in two variables

📐 What makes an equation linear

Linear equation in two variables: an equation of the form Ax + By = C, where A and B are not both zero.

• The word "linear" contains "line"—these equations graph as straight lines.
• Example: y = -3x + 5 can be rewritten as 3x + y = 5 by adding 3x to both sides and using the Commutative Property.

📋 Standard form

Standard form of a linear equation: Ax + By = C

• Most people prefer A, B, and C to be integers with A ≥ 0, though this is not strictly required.
• Standard form makes it easy to identify linear equations.

✅ What counts as a solution

An ordered pair (x, y) is a solution of the linear equation Ax + By = C if the equation is a true statement when the x- and y-values are substituted into the equation.

• Linear equations have infinitely many solutions.
• For every x-value substituted, there is a corresponding y-value.
• Example: To check if (2, -4) is a solution to x + 4y = 8, substitute: 2 + 4(-4) = 2 - 16 = -14 ≠ 8, so it is not a solution.

🔗 The relationship between solutions and graphs

🔗 Every point on the line is a solution

• When you plot solutions to a linear equation, the points line up perfectly in a straight line.
• Example: For 3x + 2y = 6, the solutions (0, 3), (2, 0), and (1, 3/2) all lie on the same line.
• Arrows on the ends of the line indicate it continues infinitely in both directions.

🔗 Every solution is a point on the line

Graph of a linear equation: The graph of a linear equation Ax + By = C is a line where every point on the line is a solution of the equation, and every solution of this equation is a point on this line.

• Points not on the line are not solutions.
• Example: If point (-2, 6) is on the line for 3x + 2y = 6, then substituting gives 3(-2) + 2(6) = -6 + 12 = 6 ✓
• If point (4, 1) is not on the line, then 3(4) + 2(1) = 12 + 2 = 14 ≠ 6

📸 "A picture is worth a thousand words"

• The line shows you all the solutions at once.
• This visual representation makes it easier to understand the infinite set of solutions.

📊 How to graph by plotting points

📊 The point-plotting method steps

Step 1: Find three points whose coordinates are solutions to the equation (organize them in a table).

Step 2: Plot the points in a rectangular coordinate system.

Step 3: Check that the points line up—if they don't, carefully check your work.

Step 4: Draw the line through the three points, extend it to fill the grid, and put arrows on both ends.

🎯 Why use three points instead of two

• Technically, two points determine a line.
• But if one of two points is incorrect, you'll still draw a line—just the wrong one.
• With three points, if one is wrong, they won't line up, alerting you to check your work.

🧮 Choosing smart values

• When the equation is in y-form (y isolated), it's easier to choose x-values and solve for y.
• When coefficients are fractions (like y = (1/2)x + 3), choose x-values that are multiples of the denominator to avoid fraction answers.
• Example: For y = (1/2)x + 3, use x = 0, 2, 4 rather than x = 1, 3, 5 to avoid fractions.

🔄 Working with standard form

• For equations like 2x + 3y = 6, you can either solve for y first or work directly with the equation.
• A good strategy: let x = 0 to find one point, let y = 0 to find another, then choose a third value.
• Example: For 2x - 3y = 6, when x = 0, y = -2; when y = 0, x = 3; when x = 6, y = 2.

📏 Vertical and horizontal lines

📏 Vertical lines (x = a)

Vertical line: the graph of an equation of the form x = a. The line passes through the x-axis at (a, 0).

• Contains only the variable x.
• The x-value is constant; y can be any value.
• Example: For x = -3, all points have x-coordinate -3: (-3, 1), (-3, 2), (-3, 3), etc.
• The graph is a vertical line passing through -3 on the x-axis.

📏 Horizontal lines (y = b)

Horizontal line: the graph of an equation of the form y = b. The line passes through the y-axis at (0, b).

• Contains only the variable y.
• The y-value is constant; x can be any value.
• Example: For y = 4, all points have y-coordinate 4: (0, 4), (2, 4), (4, 4), etc.
• The graph is a horizontal line passing through 4 on the y-axis.

⚠️ Don't confuse: y = 4x versus y = 4

Equation	Variables	What it means	Graph
y = 4x	Both x and y	y depends on x; y changes as x changes	Slanted line through origin
y = 4	Only y	y is constant; does not depend on x	Horizontal line at y = 4

• Example: For y = 4x, when x = 0, y = 0; when x = 1, y = 4; when x = 2, y = 8 (different y-values).
• For y = 4, when x = 0, y = 4; when x = 1, y = 4; when x = 2, y = 4 (same y-value).

🧭 Overview

🧠 One-sentence thesis

Graphing a line using intercepts provides a systematic method by finding where the line crosses the x-axis and y-axis, then plotting a third point to verify the line.

📌 Key points (3–5)

• What intercepts are: the x-intercept is where the line crosses the x-axis (when y = 0); the y-intercept is where the line crosses the y-axis (when x = 0).
• How to find intercepts from an equation: set y = 0 and solve for x to get the x-intercept; set x = 0 and solve for y to get the y-intercept.
• Why a third point matters: after finding both intercepts, plot a third solution point to check that all three points line up before drawing the line.
• Common confusion: don't mix up which variable to set to zero—x-intercept requires y = 0, y-intercept requires x = 0.
• When to use this method: the intercept method is most convenient when the equation is in the form ax + by = c.

📍 Understanding intercepts

📍 What the x-intercept is

The x-intercept: the point (a, 0) where the line crosses the x-axis; the x-intercept occurs when y is zero.

• This is a specific point on the graph, not just a number.
• The y-coordinate is always 0 at the x-intercept.
• Example: if a line crosses the x-axis at x = 3, the x-intercept is the point (3, 0).

📍 What the y-intercept is

The y-intercept: the point (0, b) where the line crosses the y-axis; the y-intercept occurs when x is zero.

• This is also a specific point, with x-coordinate always 0.
• The notation (0, b) shows that x must be zero.
• Example: if a line crosses the y-axis at y = -2, the y-intercept is the point (0, -2).

🔄 Don't confuse which variable to zero

• For x-intercept: set y = 0, then solve for x.
• For y-intercept: set x = 0, then solve for y.
• The variable you set to zero is the opposite of the intercept you're finding.

🛠️ How to find intercepts from an equation

🛠️ Finding the x-intercept

• Use the equation of the line.
• Let y = 0 in the equation.
• Solve for x.
• The result gives you the point (x, 0).

🛠️ Finding the y-intercept

• Use the equation of the line.
• Let x = 0 in the equation.
• Solve for y.
• The result gives you the point (0, y).

📐 Graphing a line using intercepts

📐 The complete procedure

The excerpt provides a step-by-step method:

1. Find the x- and y-intercepts of the line: let y = 0 and solve for x; let x = 0 and solve for y.
2. Find a third solution to the equation: choose any other value for one variable and solve for the other.
3. Plot the three points and check that they line up: if they don't line up, there is an error in your work.
4. Draw the line: extend the line to fill the grid and put arrows on both ends.

✅ Why the third point is necessary

• The two intercepts alone define the line mathematically.
• But plotting a third point serves as a check: if all three points line up, your intercepts are correct.
• If they do not line up, carefully check your work—there is a calculation error.

🎯 When to use the intercept method

🎯 Strategy for choosing graphing methods

The excerpt provides guidance on which method is most convenient:

Equation form	Best method	Why
x = a (one variable only)	Recognize as vertical line	Passes through x-axis at a
y = b (one variable only)	Recognize as horizontal line	Passes through y-axis at b
y is isolated (e.g., y = mx + b)	Plot points	Choose any three x-values and solve for y
ax + by = c	Use intercepts	Finding intercepts is straightforward

🎯 Why ax + by = c works well with intercepts

• When the equation is in the form ax + by = c, setting one variable to zero makes the equation simple to solve.
• Example: for 2x + 3y = 6, setting y = 0 gives 2x = 6, so x = 3; setting x = 0 gives 3y = 6, so y = 2.
• This form does not require rearranging the equation first.

🧭 Overview

🧠 One-sentence thesis

Slope quantifies how steeply a line rises or falls by comparing its vertical change to its horizontal change, and it can be calculated from two points or read directly from a graph.

📌 Key points (3–5)

• What slope measures: the steepness of a line by comparing vertical change (rise) to horizontal change (run).
• Two ways to find slope: from a graph using rise over run, or from two points using the formula (y₂ − y₁) / (x₂ − x₁).
• Slope-intercept form: the equation y = mx + b encodes both the slope (m) and the y-intercept (b).
• Common confusion: rise is vertical change, run is horizontal change—don't mix them up in the fraction.
• Related concepts: slope connects to intercepts, vertical/horizontal lines, and the coordinate plane structure.

📐 What slope is

📏 Rise and run

The rise of a line is its vertical change.

The run of a line is its horizontal change.

• Rise measures how much the line goes up or down (change in y-direction).
• Run measures how much the line goes left or right (change in x-direction).
• These are the building blocks of slope.

📊 Slope definition

The slope of a line is m = rise / run.

• Slope (denoted m) is the ratio of vertical change to horizontal change.
• It tells you how steep the line is.
• Example: if a line rises 3 units for every 2 units it runs to the right, the slope is 3/2.

🧮 How to calculate slope

🧮 From two points

The slope of the line between two points (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁) / (x₂ − x₁).

• Subtract the y-coordinates to get the rise: y₂ − y₁.
• Subtract the x-coordinates to get the run: x₂ − x₁.
• Divide rise by run to get the slope.
• Example: for points (1, 2) and (4, 8), rise = 8 − 2 = 6, run = 4 − 1 = 3, so m = 6/3 = 2.

📈 From a graph

• The excerpt states: "Find the Slope of a Line from its Graph using m = rise / run."
• Locate two points on the line.
• Count the vertical distance between them (rise).
• Count the horizontal distance between them (run).
• Divide rise by run.

📝 Slope-intercept form

📝 The equation

The slope–intercept form of an equation of a line with slope m and y-intercept (0, b) is y = mx + b.

• m is the slope of the line.
• b is the y-coordinate where the line crosses the y-axis.
• This form makes it easy to see both the steepness and the starting point on the y-axis.
• Example: in y = 3x + 5, the slope is 3 and the y-intercept is (0, 5).

🔗 Connection to intercepts

The y-intercept occurs when x is zero.

• The y-intercept is the point (0, b) where the line crosses the y-axis.
• In slope-intercept form, b is directly visible.
• Don't confuse: the x-intercept (where y = 0) is not directly shown in y = mx + b; you must solve for x when y = 0.

🗺️ Context: the coordinate plane

🗺️ The xy-plane structure

A grid system is used in algebra to show a relationship between two variables; also called the xy-plane or the 'coordinate plane'.

The x-axis and the y-axis divide a plane into four regions, called quadrants.

• The coordinate plane has four quadrants with different sign patterns:

Quadrant	Sign pattern (x, y)
I	(+, +)
II	(−, +)
III	(−, −)
IV	(+, −)

• Slope is calculated within this coordinate system.

🎯 Ordered pairs and solutions

An ordered pair (x, y) is a solution to a linear inequality [when] the inequality is true when we substitute the values of x and y.

• The first number in an ordered pair (x, y) is the x-coordinate.
• The second number in an ordered pair (x, y) is the y-coordinate.
• Points on a line satisfy the line's equation; substituting their coordinates makes the equation true.

↕️ Special lines

A vertical line is the graph of an equation of the form x = a. The line passes through the x-axis at (a, 0).

• Vertical lines have the form x = a (constant x-value).
• Horizontal lines have the form y = b (constant y-value).
• Don't confuse: vertical lines have undefined slope (run = 0), while horizontal lines have slope = 0 (rise = 0).

🧭 Overview

🧠 One-sentence thesis

The slope-intercept form y = mx + b provides the most direct way to graph a line by immediately revealing both the slope and the y-intercept, making it the preferred method when the equation is already solved for y.

📌 Key points (3–5)

• What the slope-intercept form is: an equation written as y = mx + b, where m is the slope and (0, b) is the y-intercept.
• How to graph using this form: plot the y-intercept first, then use the slope (rise over run) to find a second point and draw the line.
• When to use this method: choose slope-intercept form when y is already isolated on one side of the equation.
• Common confusion: distinguish between the three main graphing methods—slope-intercept (y = mx + b), intercepts (Ax + By = C), and special cases (vertical/horizontal lines).
• Why it matters: recognizing the form of an equation helps you choose the most efficient graphing strategy.

📐 The slope-intercept form

📝 Definition and components

The slope-intercept form of an equation of a line with slope m and y-intercept (0, b) is y = mx + b.

• m = the slope of the line (how steep it is; rise over run)
• b = the y-coordinate where the line crosses the y-axis
• The y-intercept point is always (0, b) because x = 0 on the y-axis
• This form immediately shows both key characteristics of the line

🎯 Why this form is useful

• When the equation is already in y = mx + b form, you can read the slope and y-intercept directly without calculation
• No need to solve for intercepts or pick arbitrary x-values
• Example: in y = 3x + 5, you instantly know the slope is 3 and the y-intercept is (0, 5)

🖊️ Graphing with slope and y-intercept

📍 Step-by-step process

The excerpt provides a clear procedure:

1. Find the slope-intercept form of the equation (if not already in that form)
2. Identify the slope and y-intercept from the equation
3. Plot the y-intercept point (0, b) on the graph
4. Use the slope formula m = rise/run to identify the rise and run
5. Starting at the y-intercept, count out the rise and run to mark the second point
6. Connect the points with a line

🔢 Using rise over run

• The slope m tells you how to move from one point to the next
• Rise = vertical change (up is positive, down is negative)
• Run = horizontal change (right is positive, left is negative)
• Example: if m = 3, you can write it as 3/1, meaning rise 3 units up and run 1 unit right from the y-intercept

🧭 Choosing the right graphing method

🗺️ Strategy overview

The excerpt emphasizes considering the form of the equation first:

Equation form	Method	Key steps
x = a (one variable)	Vertical line	Draw a vertical line through x = a on the x-axis
y = b (one variable)	Horizontal line	Draw a horizontal line through y = b on the y-axis
y = mx + b (y isolated)	Slope-intercept	Identify slope and y-intercept, then graph
Ax + By = C (standard form)	Intercepts	Find x- and y-intercepts plus a third point

⚠️ When to use slope-intercept

• Best choice: when y is already isolated on one side of the equation in the form y = mx + b
• You can immediately identify the slope and y-intercept without extra algebra
• Don't confuse: if the equation is in standard form (Ax + By = C), the intercepts method is usually more convenient than converting to slope-intercept form

🔄 Comparison with other methods

• Plotting points method: works for any equation where y is isolated, but requires choosing arbitrary x-values and calculating corresponding y-values
• Intercepts method: efficient for standard form equations; requires finding where the line crosses both axes
• Slope-intercept method: most direct when the equation is already in y = mx + b form; combines the starting point (y-intercept) with the direction (slope)

📏 Special cases

➖ Horizontal lines

• Form: y = b (only one variable)
• Slope of a horizontal line is 0
• The line passes through the y-axis at b and runs parallel to the x-axis

⬆️ Vertical lines

• Form: x = a (only one variable)
• Slope of a vertical line is undefined
• The line passes through the x-axis at a and runs parallel to the y-axis

🔀 Parallel lines

Parallel lines are lines in the same plane that do not intersect.

• The excerpt introduces this concept but does not elaborate further
• This suggests that parallel lines will be discussed in relation to slope-intercept form (lines with the same slope are parallel)

🧭 Overview

🧠 One-sentence thesis

The equation of a line can be found by choosing the appropriate form—slope-intercept, point-slope, or intercept-based—depending on what information is given (slope, points, or intercepts).

📌 Key points (3–5)

• Three main forms: slope-intercept form (y = mx + b), point-slope form (y − y₁ = m(x − x₁)), and intercept form (Ax + By = C).
• What determines which form to use: whether you are given slope and y-intercept, slope and a point, or two points.
• Parallel and perpendicular lines: parallel lines share the same slope; perpendicular lines have slopes whose product is −1.
• Common confusion: don't mix up when to use slope-intercept vs point-slope—use slope-intercept when you know the y-intercept directly; use point-slope when you know any point and the slope.
• Special cases: horizontal lines have slope 0; vertical lines have undefined slope and cannot be written in slope-intercept form.

📐 Finding equations with different starting information

📍 Given slope and a point

Point-slope form: y − y₁ = m(x − x₁), where m is the slope and (x₁, y₁) is a known point.

• Steps:
  1. Identify the slope.
  2. Identify the point.
  3. Substitute the values into y − y₁ = m(x − x₁).
  4. Write the equation in slope-intercept form (if needed).
• Example: if you know the slope is 2 and the line passes through (3, 5), substitute m = 2, x₁ = 3, y₁ = 5 into the point-slope form.

📍 Given two points

• Steps:
  1. Find the slope using the two given points (rise over run).
  2. Choose one of the two points.
  3. Substitute the slope and the chosen point into y − y₁ = m(x − x₁).
  4. Write the equation in slope-intercept form.
• Why this works: once you have the slope, you can treat it as "slope and a point" and use point-slope form.

📍 Given slope and y-intercept

Slope-intercept form: y = mx + b, where m is the slope and (0, b) is the y-intercept.

• When to use: if you already know the y-intercept (the point where the line crosses the y-axis), this is the most direct form.
• Simply plug in m and b.
• Don't confuse: if you are given a point that is not the y-intercept, use point-slope form instead.

🔀 Parallel and perpendicular lines

↔️ Parallel lines

Parallel lines: lines in the same plane that do not intersect; they have the same slope and different y-intercepts.

• Key relationship: if m₁ and m₂ are the slopes of two parallel lines, then m₁ = m₂.
• Steps to find an equation parallel to a given line:
  1. Find the slope of the given line.
  2. The parallel line has the same slope.
  3. Identify the point the new line passes through.
  4. Substitute into y − y₁ = m(x − x₁).
  5. Write in slope-intercept form.
• Special case: parallel vertical lines have different x-intercepts (both have undefined slope).

⊥ Perpendicular lines

Perpendicular lines: lines in the same plane that form a right angle; their slopes multiply to −1.

• Key relationship: if m₁ and m₂ are the slopes of two perpendicular lines, then m₁ · m₂ = −1, or equivalently m₁ = −1/m₂.
• Steps to find an equation perpendicular to a given line:
  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line (negative reciprocal).
  3. Identify the point the new line passes through.
  4. Substitute into y − y₁ = m(x − x₁).
  5. Write in slope-intercept form (if the excerpt continues to that step).
• Special case: vertical and horizontal lines are always perpendicular to each other (one has slope 0, the other has undefined slope).

📊 Choosing the right method

🧭 Strategy for writing an equation

What you are given	Which form to use	Why
Slope and y-intercept	Slope-intercept: y = mx + b	You already know b (the y-intercept)
Slope and any point	Point-slope: y − y₁ = m(x − x₁)	You know m and a point (x₁, y₁)
Two points	Point-slope (after finding slope)	First calculate slope from the two points, then use point-slope form

• The excerpt emphasizes: consider the form of the equation and what information is available.
• Don't confuse: if the equation is already in the form Ax + By = C, you may find intercepts to graph it, but to write a new equation, use the forms above based on what you know.

🧮 Graphing using slope and intercept

📈 Graph a line using slope and y-intercept

• Steps:
  1. Find the slope-intercept form of the equation.
  2. Identify the slope and y-intercept.
  3. Plot the y-intercept.
  4. Use the slope formula (m = rise/run) to identify the rise and run.
  5. Starting at the y-intercept, count out the rise and run to mark the second point.
  6. Connect the points with a line.
• This method is efficient when the equation is already in y = mx + b form.

📈 Graph a line given a point and the slope

• Steps:
  1. Plot the given point.
  2. Use m = rise/run to identify the rise and run.
  3. Starting at the given point, count out the rise and run to mark the second point.
  4. Connect the points with a line.
• Example: if the slope is 3/2, rise 3 units and run 2 units from the starting point.

🔍 Special slopes

• Horizontal line: y = b has slope 0 (no rise, only run).
• Vertical line: x = a has undefined slope (no run, only rise; cannot be written in slope-intercept form).
• Don't confuse: a horizontal line is not the same as "no slope"—it has slope 0; a vertical line has undefined slope.

🧭 Overview

🧠 One-sentence thesis

Graphing linear inequalities extends the concept of graphing linear equations by shading a region of the coordinate plane that represents all solutions to the inequality, with the boundary line style determined by whether the inequality includes equality.

📌 Key points (3–5)

• What a linear inequality graph shows: a shaded region representing all ordered pairs that satisfy the inequality, not just a single line.
• Boundary line style matters: solid lines for ≤ or ≥ (boundary included), dashed lines for < or > (boundary excluded).
• Test-point method: pick any point not on the boundary, test if it satisfies the inequality, then shade the correct side accordingly.
• Common confusion: the boundary line itself is only part of the solution when the inequality symbol includes "or equal to" (≤ or ≥).
• Verification step: you can check whether specific ordered pairs are solutions by substituting them into the inequality.

📐 Understanding the boundary line

📐 What the boundary line represents

The boundary line is the graph of the related linear equation (replacing the inequality symbol with an equals sign).

• The boundary divides the coordinate plane into two half-planes.
• One side contains all solutions to the inequality; the other side does not.
• Example: for the inequality y < x − 3, the boundary line is y = x − 3.

✏️ Solid vs dashed boundary lines

The inequality symbol determines whether the boundary line is part of the solution set:

Inequality symbol	Boundary line style	Meaning
≤ or ≥	Solid	Points on the line ARE solutions
< or >	Dashed	Points on the line are NOT solutions

• Don't confuse: a dashed line means "up to but not including"; a solid line means "up to and including."
• Example: y ≤ 2x uses a solid line because points where y equals 2x satisfy the inequality.

🧪 The test-point method

🧪 How to choose and use a test point

Step-by-step process from the excerpt:

1. Identify and graph the boundary line (solid or dashed based on the symbol).
2. Choose a test point that is not on the boundary line.
3. Substitute the test point into the inequality to see if it makes a true statement.
4. Shade the appropriate region based on the test result.

✅ Interpreting the test result

• If the test point IS a solution: shade the side of the boundary line that includes the test point.
• If the test point is NOT a solution: shade the opposite side (the side that does not include the test point).
• Tip: the point (0, 0) is often convenient to test, unless it lies on the boundary line.

Example: Testing whether (0, 1) is a solution to y < x − 3:

• Substitute: 1 < 0 − 3 → 1 < −3 (false).
• Since (0, 1) is not a solution, shade the side that does NOT include (0, 1).

✔️ Verifying solutions

✔️ Checking ordered pairs

Before graphing, you can verify whether specific ordered pairs satisfy the inequality by direct substitution.

• Substitute the x and y values into the inequality.
• If the resulting statement is true, the ordered pair is a solution.
• If false, it is not a solution.

Example from the excerpt: to determine if (5, 2) is a solution to y < x − 3:

• Substitute: 2 < 5 − 3 → 2 < 2 (false).
• Therefore (5, 2) is not a solution.

📝 Reading inequality graphs

The excerpt also covers the reverse process: given a shaded graph with a boundary line, write the inequality it represents.

• Identify the boundary line equation.
• Determine if the line is solid (≤ or ≥) or dashed (< or >).
• Check which side is shaded to determine the direction of the inequality.

🎨 Graphing procedure summary

🎨 Complete graphing steps

The excerpt provides a four-step procedure:

1. Identify and graph the boundary line: replace the inequality with equality and graph; use solid or dashed style.
2. Test a point not on the boundary: substitute into the original inequality.
3. Determine which side to shade: if the test point satisfies the inequality, shade its side; otherwise shade the opposite side.
4. (Implicit): label or indicate the inequality represented.

🔍 Special cases

• Horizontal inequalities: y < constant or y > constant produce horizontal boundary lines.
• Vertical inequalities: x < constant or x > constant produce vertical boundary lines.
• Example from the excerpt: y < 6 has a horizontal dashed line at y = 6, with shading below.

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🧭 Overview

🧠 One-sentence thesis

A system of linear equations can be solved by graphing both lines and finding their intersection point, and the number of solutions depends on whether the lines intersect, are parallel, or are identical.

📌 Key points (3–5)

• What the graphing method does: find where two lines intersect; that point is the solution to both equations.
• Three possible outcomes: one solution (lines intersect), no solution (lines are parallel), or infinitely many solutions (lines are the same).
• How to verify: check that the intersection point satisfies both original equations.
• Common confusion: parallel vs identical lines—parallel lines never meet (no solution), but identical lines overlap everywhere (infinite solutions).
• Shortcut without graphing: you can determine the number of solutions by examining slopes and intercepts before drawing anything.

📍 Finding solutions by graphing

📍 The intersection point is the solution

• Graph both equations on the same coordinate grid.
• Look for where the two lines cross.
• The coordinates of that intersection point satisfy both equations simultaneously.
• Example: if the lines meet at (2, 3), then x = 2 and y = 3 is the solution to the system.

✅ Always verify the solution

• Substitute the coordinates back into both original equations.
• If the point makes both equations true, it is confirmed as the solution.
• This check ensures no graphing errors were made.

🔢 How many solutions exist

🔢 Three scenarios

Situation	Graph appearance	Number of solutions	Classification
Lines intersect at one point	Two lines cross	Exactly one solution	Independent system
Lines are parallel	Lines never meet	No solution	Inconsistent system
Lines are the same	Lines overlap completely	Infinite solutions	Dependent system

🔍 Parallel lines mean no solution

• Parallel lines have the same slope but different intercepts.
• They never touch, so there is no point that satisfies both equations.
• Don't confuse: "parallel" means distinct lines that never meet, not lines that are far apart at one spot but might meet elsewhere.

♾️ Identical lines mean infinite solutions

• If both equations describe the exact same line, every point on that line is a solution.
• This happens when one equation is a multiple of the other.

🧮 Determining solutions without graphing

🧮 Use slopes and intercepts

• You can predict the number of solutions by comparing the equations algebraically.
• Look at the slopes and y-intercepts before drawing anything.
• This saves time and confirms what the graph will show.

🎯 The shortcut method

• Same slope, different intercepts → parallel lines → no solution.
• Same slope, same intercept → identical lines → infinite solutions.
• Different slopes → lines must intersect → exactly one solution.
• Example: if one line has slope 2 and the other has slope -1, they will cross at exactly one point.

🛠️ Problem-solving strategy

🛠️ Systematic approach

The excerpt outlines a general strategy for systems of equations:

1. Read the problem carefully and understand all terms.
2. Identify what you are looking for.
3. Name the unknowns with variables.
4. Translate the situation into a system of equations.
5. Solve using appropriate techniques (graphing, substitution, or elimination).
6. Check the answer in the original problem context.
7. Answer with a complete sentence.

📝 Verification is essential

• Always check that your solution makes sense in the problem context.
• Substitute back into both original equations to confirm accuracy.
• A solution that works mathematically but doesn't fit the real-world scenario is not valid.

🧭 Overview

🧠 One-sentence thesis

The substitution method solves systems of equations by isolating one variable in one equation and replacing it in the other equation, allowing you to find both variable values step by step.

📌 Key points (3–5)

• Core strategy: solve one equation for a single variable, then substitute that expression into the other equation.
• Sequential solving: first find one variable from the resulting single-variable equation, then use that value to find the other variable.
• Verification step: always check that the ordered pair satisfies both original equations.
• Common confusion: don't forget to substitute back into an original equation (not the manipulated one) to find the second variable.
• Output format: write the solution as an ordered pair and verify it works in both equations.

🔧 The substitution method step-by-step

🔧 Step 1: Isolate one variable

• What to do: Solve one of the equations for either variable.
• Choose whichever equation and variable makes the algebra simplest.
• The result is an expression like "x equals something in terms of y" or "y equals something in terms of x."
• Example: If you have x + 2y = 6, you might solve for x to get x = 6 - 2y.

🔄 Step 2: Substitute the expression

• What to do: Substitute the expression from Step 1 into the other equation.
• This replaces one variable completely, leaving you with an equation in only one variable.
• Example: If the second equation is y = -1/2 x - 4 and you found x = 6 - 2y, replace x with (6 - 2y) in the second equation.

🧮 Step 3: Solve for the first variable

• What to do: Solve the resulting equation.
• Now you have a single-variable equation (only x or only y), which you solve using standard algebra techniques.
• This gives you the numerical value of one variable.

🔍 Finding the complete solution

🔍 Step 4: Find the second variable

• What to do: Substitute the solution from Step 3 into one of the original equations to find the other variable.
• Important: use an original equation, not an intermediate manipulated form.
• This gives you the numerical value of the second variable.
• Don't confuse: substituting into the wrong equation can introduce errors; always go back to an original equation.

📝 Step 5: Write as an ordered pair

• What to do: Write the solution as an ordered pair (x, y).
• The ordered pair format is the standard way to express the solution to a system of two linear equations.

✅ Step 6: Verify the solution

• What to do: Check that the ordered pair is a solution to both original equations.
• Substitute both x and y values into each original equation and confirm both equations are true.
• This verification step catches algebraic mistakes and confirms the solution is correct.
• Example: If you found (x, y) = (2, 1), plug x = 2 and y = 1 into both original equations and verify both sides equal.

🎯 Why substitution works

🎯 The logic behind substitution

• By expressing one variable in terms of the other, you reduce a two-variable problem to a one-variable problem.
• The substitution preserves the equality because you are replacing a variable with an equivalent expression.
• Once you know one variable, the system's constraints force the other variable to a specific value.

🎯 When to use substitution

• The excerpt presents substitution as one method among others (graphing and elimination are also mentioned in the surrounding context).
• Substitution is particularly efficient when one equation is already solved for a variable or can be easily rearranged.

🧭 Overview

🧠 One-sentence thesis

The elimination method solves systems of linear equations by strategically adding equations together to cancel out one variable, making it possible to solve for the remaining variable.

📌 Key points (3–5)

• Core strategy: Make the coefficients of one variable opposites, then add the equations to eliminate that variable.
• Standard form requirement: Both equations must be in standard form (Ax + By = C) before starting.
• Coefficient manipulation: Multiply one or both equations by constants to create opposite coefficients.
• Common confusion: Don't confuse "opposites" with "equal"—you need coefficients like 3 and -3, not 3 and 3, so they cancel when added.
• Verification step: Always check the ordered pair solution in both original equations to confirm correctness.

🔧 The Elimination Process

🔧 Prepare equations in standard form

• Both equations must be written as Ax + By = C before you begin.
• If any coefficients are fractions, clear them first (multiply through to eliminate fractions).
• This standardization makes it easier to see and manipulate coefficients.

🎯 Choose which variable to eliminate

• Decide whether to eliminate x or y first.
• Look at the coefficients: choose the variable that will be easier to make into opposites.
• Example: If one equation has 2x and another has 3x, you might multiply to get 6x and -6x.

✖️ Create opposite coefficients

• Multiply one or both equations by constants so that one variable has opposite coefficients.
• "Opposites" means one positive and one negative with the same absolute value (e.g., 5 and -5).
• This is the key step that makes elimination work—when you add the equations, that variable will cancel out.

➕ Execute the elimination

➕ Add the equations

• Once you have opposite coefficients for one variable, add the two equations together.
• The variable with opposite coefficients will disappear (5x + (-5x) = 0).
• You're left with a single-variable equation that you can solve directly.

🔍 Solve for the remaining variable

• After elimination, you have one equation with one unknown.
• Solve this equation using standard algebra techniques.
• This gives you the value of one variable in the solution ordered pair.

🔄 Complete the solution

🔄 Substitute back to find the other variable

• Take the value you just found and substitute it into one of the original equations.
• Solve for the other variable.
• Now you have both values needed for the ordered pair solution.

✅ Write and verify the solution

• Write the solution as an ordered pair (x, y).
• Check this ordered pair in both original equations to ensure it satisfies both.
• Don't confuse: checking in only one equation is insufficient—a true solution must work in both equations of the system.

📝 Step-by-step summary

Step	Action	Purpose
1	Write both equations in standard form	Standardize format
2	Clear any fractions	Simplify coefficients
3	Make coefficients of one variable opposites	Prepare for cancellation
4	Add the equations	Eliminate one variable
5	Solve for remaining variable	Find first value
6	Substitute into original equation	Find second value
7	Check in both original equations	Verify solution

🧭 Overview

🧠 One-sentence thesis

Systems of equations can be translated from real-world word problems and solved to find unknown quantities in scenarios involving number relationships, geometry, and motion.

📌 Key points (3–5)

• Translation first: word problems must be converted into a system of equations before solving.
• Application categories: the excerpt covers direct translation (number/age problems), geometry (angles, perimeters), and uniform motion (catching up scenarios).
• Multiple solution methods available: substitution and elimination are both valid; choose based on convenience.
• Common confusion: don't confuse supplementary angles (sum to 180°) with complementary angles (sum to 90°).
• Motion problems require careful setup: distinguish between different rates, times, and the relationship between travelers.

🔢 Translating word problems into systems

🔢 What translation means

Translation: converting a word problem into a system of equations without solving it yet.

• The excerpt explicitly separates "translate to a system of equations" from "solve."
• You must identify the unknowns and express the relationships as equations.
• Example: "The sum of two numbers is −32. One number is two less than twice the other" becomes two equations with two variables.

📝 Direct translation applications

• These involve straightforward relationships between quantities.
• Common patterns:
  • Sum and difference of numbers
  • Age relationships (one person older/younger than another)
  • Investment or earnings split between two categories
• Example: "Pam is 3 years older than her sister, Jan. The sum of their ages is 99" translates to two equations relating their ages.

📐 Geometry applications

📐 Angle relationships

Two key angle types appear:

Angle type	Definition	Sum
Supplementary	Two angles that together form a straight line	180°
Complementary	Two angles that together form a right angle	90°

• Don't confuse: supplementary angles sum to 180°, complementary to 90°.
• Example from excerpt: "Two angles are complementary. The measure of the larger angle is five more than four times the measure of the smaller angle."

📏 Perimeter problems

• Rectangle perimeter problems give the total perimeter and a relationship between length and width.
• The perimeter formula (2 times length plus 2 times width equals perimeter) provides one equation.
• The relationship between dimensions provides the second equation.
• Example: "The perimeter of a city rectangular park is 1428 feet. The length is 78 feet more than twice the width."

🚗 Uniform motion applications

🚗 Catching-up scenarios

• One person/vehicle leaves first, another leaves later at a different speed.
• The question asks: when will the second catch up to the first?
• Key variables: distance, rate (speed), and time for each traveler.

⏱️ Setting up motion equations

• Both travelers cover the same distance when one catches up.
• Time relationships must account for the head start.
• Example: "Lenore left one hour after Sheila. Sheila drove at a rate of 45 mph, and Lenore drove at a rate of 60 mph. How long will it take for Lenore to catch up to Sheila?"
  • Sheila has a 1-hour head start
  • When Lenore catches up, both have traveled the same distance
  • Distance equals rate times time for each person

🛠️ Choosing a solution method

🛠️ Substitution vs elimination

The excerpt asks students to decide which method is more convenient before solving.

• Substitution is often easier when one equation is already solved for a variable (e.g., "y = 3x − 9").
• Elimination is often easier when coefficients can be easily matched or cancelled (e.g., "x + y = 12" and "x − y = −10").
• The choice depends on the form of the equations, not the application type.

🧭 Overview

🧠 One-sentence thesis

Mixture applications require setting up systems of equations to solve problems where two or more components are combined according to constraints on both total quantity and total value or concentration.

📌 Key points (3–5)

• What mixture problems involve: combining two or more items (tickets, coins, ingredients, solutions) where you know total quantity and total value/cost/concentration.
• Two key equations: one equation tracks the total amount of items; the other tracks the total value, cost, or concentration.
• Common types: ticket/coin problems (different prices/values), ingredient problems (different costs per unit), and solution problems (different concentrations).
• Common confusion: don't mix up the "number of items" equation with the "total value" equation—each tracks a different constraint.
• Related application: interest problems use the same two-equation structure (total principal and total interest earned).

🎟️ Ticket and coin mixture problems

🎟️ Structure of ticket problems

• You know the total number of tickets and the total cost paid.
• Each ticket type has a different price.
• Set up two equations:
  • One for the count: (number of type A) + (number of type B) = total tickets
  • One for the value: (price of A) × (number of A) + (price of B) × (number of B) = total cost

Example: Lynn paid $2,780 for 261 tickets. Student tickets cost $10, adult tickets cost $15. Find how many of each.

• Equation 1: student tickets + adult tickets = 261
• Equation 2: 10 × (student tickets) + 15 × (adult tickets) = 2,780

🪙 Structure of coin problems

• Similar to ticket problems but with coins of different denominations.
• One equation for the number of coins, one for the total value in dollars or cents.

Example: Priam has dimes and pennies totaling $4.21. The number of dimes is three less than four times the number of pennies.

• Equation 1: relates the number of dimes to the number of pennies (given relationship)
• Equation 2: 0.10 × (dimes) + 0.01 × (pennies) = 4.21

🍬 Ingredient and party mix problems

🍬 Mixing ingredients by cost

• You want a certain total quantity of a mixture at a target cost per unit.
• Each ingredient has a different cost per unit.
• Set up two equations:
  • One for total quantity: (amount of ingredient A) + (amount of ingredient B) = total quantity
  • One for total cost: (cost per unit of A) × (amount of A) + (cost per unit of B) × (amount of B) = (target cost per unit) × (total quantity)

Example: Yumi wants 12 cups of party mix costing $1.29 per cup. Candies cost $2.49 per cup, nuts cost $0.69 per cup.

• Equation 1: cups of candies + cups of nuts = 12
• Equation 2: 2.49 × (candies) + 0.69 × (nuts) = 1.29 × 12

🧪 Solution concentration problems

• A scientist or chemist needs a certain total volume at a target concentration.
• Available solutions have different concentrations.
• Set up two equations:
  • One for total volume: (liters of solution A) + (liters of solution B) = total liters needed
  • One for concentration: (concentration of A) × (liters of A) + (concentration of B) × (liters of B) = (target concentration) × (total liters)

Example: A scientist needs 70 liters of 40% alcohol solution. He has 30% and 60% solutions available.

• Equation 1: liters of 30% + liters of 60% = 70
• Equation 2: 0.30 × (liters of 30%) + 0.60 × (liters of 60%) = 0.40 × 70

Don't confuse: the concentration equation uses percentages as decimals (30% = 0.30) and multiplies by volume to get the amount of pure substance.

💰 Interest application problems

💰 Structure of interest problems

• Similar two-equation setup: total principal invested and total interest earned.
• You have a total amount to invest and want a target interest rate overall.
• Each account or investment has a different interest rate.

Example: Jack has $12,000 to invest and wants 7.5% interest per year. A savings account earns 4%, a CD earns 9%.

• Equation 1: (amount in savings) + (amount in CD) = 12,000
• Equation 2: 0.04 × (savings) + 0.09 × (CD) = 0.075 × 12,000

💳 Loan interest problems

• The same structure applies when someone owes money at different interest rates.

Example: Linda will owe $43,000 in student loans (excerpt cuts off, but the setup would be similar).

Problem type	First equation (quantity)	Second equation (value/concentration/interest)
Tickets/coins	Total number of items	Total cost or value
Ingredient mix	Total quantity (cups, liters)	Total cost = cost per unit × total quantity
Solution mix	Total volume	Total concentration = concentration × volume
Interest	Total principal	Total interest = rate × principal

Key insight: All mixture applications share the same logical structure—one constraint on "how much stuff" and one constraint on "what it's worth or what percentage it is."

🧭 Overview

🧠 One-sentence thesis

Graphing systems of linear inequalities extends equation-solving techniques to show all ordered pairs that satisfy multiple inequality constraints simultaneously.

📌 Key points (3–5)

• Testing solutions: An ordered pair must satisfy all inequalities in the system to be a solution.
• Graphing method: Solve systems by graphing each inequality and finding the overlapping region.
• Real-world applications: Systems of inequalities model constraints like budget limits and minimum requirements.
• Common confusion: Unlike equations (which have discrete solutions), inequality systems have solution regions containing infinitely many points.
• Verification: Check whether specific ordered pairs fall within the solution region to answer practical questions.

🔍 Determining Solutions

🔍 What makes an ordered pair a solution

A solution to a system of linear inequalities is an ordered pair that satisfies every inequality in the system.

• You must test the pair in each inequality separately.
• All inequalities must be true for that pair; if even one is false, the pair is not a solution.
• Example: For the system with inequalities "4x + y > 6" and "3x - y ≤ 12", test whether (2, -1) works by substituting x = 2 and y = -1 into both inequalities.

✅ Testing process

• Substitute the x and y values from the ordered pair into each inequality.
• Simplify and check whether the resulting statement is true.
• The excerpt shows exercises testing pairs like (2, -1) and (3, -2) against multiple inequalities.

📊 Graphing Systems

📊 How to graph a system

• Graph each linear inequality individually on the same coordinate plane.
• Each inequality creates a half-plane (region) of solutions.
• The solution to the system is where all individual regions overlap.

🎨 The solution region

• Unlike a system of equations (which typically has one point or no solution), a system of inequalities has a region of solutions.
• Every point in the overlapping shaded area is a valid solution.
• The excerpt includes systems like "y < 3x + 1 and y ≥ -x - 2" where students must graph both and identify the overlap.

🔄 Types of boundaries

The excerpt shows inequalities with different symbols:

• Strict inequalities (< or >) create dashed boundary lines (points on the line are not included).
• Non-strict inequalities (≤ or ≥) create solid boundary lines (points on the line are included).

🛠️ Real-World Applications

🛠️ Modeling constraints

Systems of inequalities represent multiple real-world limitations at once:

• Budget constraints (total cost must be below a limit)
• Minimum requirements (must sell at least a certain amount)
• Physical constraints (limited display space)

💍 Bracelet and necklace example

The excerpt presents: Roxana sells bracelets for $12 and necklaces for $18, with constraints:

• Room to display no more than 40 pieces total
• Needs to sell at least $500 worth for profit
• This translates to a system of inequalities modeling both the space limit and revenue requirement.
• Students must determine whether specific combinations (like 26 bracelets and 14 necklaces) satisfy both constraints.

📚 Book budget example

Annie has $600 to buy paperback ($4 each) and hardcover ($15 each) books with the requirement that hardcover count must be at least 5 more than three times the paperback count.

• Two inequalities: one for budget, one for the quantity relationship.
• Test whether specific purchases (like 8 paperback and 40 hardcover) meet both conditions.

🧮 Application workflow

1. Translate the word problem into a system of inequalities.
2. Graph the system to visualize the feasible region.
3. Check whether proposed solutions fall within that region.
4. The excerpt emphasizes this three-step process for practical decision-making.

🧭 Overview

🧠 One-sentence thesis

Polynomials are expressions built from monomials (terms with constants and variables raised to whole-number exponents), and adding or subtracting them requires combining like terms—those with identical variables and exponents.

📌 Key points (3–5)

• What polynomials are: monomials or combinations of monomials joined by addition/subtraction; special names exist for one term (monomial), two terms (binomial), and three terms (trinomial).
• Degree of a polynomial: determined by the highest sum of exponents in any single term; constants have degree 0.
• How to add/subtract: identify like terms (same variables, same exponents), rearrange them together, then combine by adding or subtracting coefficients.
• Common confusion: like terms must match both variable and exponent—x² and x are not like terms; u²v and uv² are not like terms.
• Evaluation: substitute the given value(s) for the variable(s) and simplify using order of operations.

🧩 Core vocabulary and structure

🧩 What a monomial is

A monomial is a term of the form a·x^m, where a is a constant and m is a positive whole number.

• Examples: 8, −2x², 4y³, 11z⁷.
• A monomial has exactly one term (no addition or subtraction separating parts).

🏗️ What a polynomial is

A polynomial is a monomial, or two or more monomials combined by addition or subtraction.

• Binomial: exactly two terms (e.g., a + 7, 4b − 5).
• Trinomial: exactly three terms (e.g., x² − 7x + 12).
• Polynomials with more than three terms have no special name; they are simply called polynomials.
• Every monomial, binomial, and trinomial is also a polynomial—they are special cases within the polynomial family.

📏 Degree of a polynomial

The degree of a term is the sum of the exponents of its variables. The degree of a constant is 0. The degree of a polynomial is the highest degree of all its terms.

• Example: 10y has degree 1 (y = y¹).
• Example: 4x³ − 7x + 5 has degree 3 (the term 4x³ has the highest degree).
• Example: 8xy² + 2y has degree 3 (xy² means x¹y², so 1 + 2 = 3).
• Example: −15 (a constant) has degree 0.
• Standard form: write terms in descending order of degree (highest degree first).

➕ Adding and subtracting monomials

➕ Combining like terms

• Like terms must have the same variables with the same exponents.
• Add or subtract the coefficients; the variable part stays the same.
• Example: 25y² + 15y² = 40y² (both terms are y²).
• Example: 16p − (−7p) = 16p + 7p = 23p.

🚫 When terms are not alike

• If variables or exponents differ, terms cannot be combined.
• Example: c² + 7d² − 6c² simplifies to −5c² + 7d² (c² and d² are not like terms).
• Example: u²v + 5u² − 3v² has no like terms, so it stays as is.
• Don't confuse: u²v and uv² are different (exponents are on different variables).

➕ Adding and subtracting polynomials

➕ Adding polynomials

• Identify like terms across both polynomials.
• Rearrange to group like terms together (using the Commutative Property).
• Combine coefficients of like terms.
• Example: (5y² − 3y + 15) + (3y² − 4y − 11)
  → Group: (5y² + 3y²) + (−3y − 4y) + (15 − 11)
  → Result: 8y² − 7y + 4.

➖ Subtracting polynomials

• Distribute the negative sign to every term in the second polynomial.
• Then follow the same steps as addition: identify, rearrange, combine.
• Example: (9w² − 7w + 5) − (2w² − 4)
  → Distribute: 9w² − 7w + 5 − 2w² + 4
  → Group: (9w² − 2w²) − 7w + (5 + 4)
  → Result: 7w² − 7w + 9.
• Watch the wording: "Subtract A from B" means B − A (the order matters).

🔀 Polynomials with multiple variables

• Like terms must match all variables and their exponents.
• Example: (u² − 6uv + 5v²) + (3u² + 2uv)
  → Group: (u² + 3u²) + (−6uv + 2uv) + 5v²
  → Result: 4u² − 4uv + 5v².
• Example: (p² + q²) − (p² + 10pq − 2q²)
  → Distribute: p² + q² − p² − 10pq + 2q²
  → Result: −10pq + 3q² (p² terms cancel).

🔢 Evaluating polynomials

🔢 Substitution and simplification

• Replace each variable with the given value.
• Simplify using order of operations: exponents first, then multiplication/division, then addition/subtraction.
• Example: Evaluate 5x² − 8x + 4 when x = 4
  → 5(4)² − 8(4) + 4
  → 5(16) − 32 + 4
  → 80 − 32 + 4 = 52.

🌍 Real-world applications

• Height of a falling object: The polynomial −16t² + 250 gives the height (in feet) of a ball t seconds after being dropped from 250 feet.
  Example: At t = 2, height = −16(2)² + 250 = −64 + 250 = 186 feet.
• Cost of production: The polynomial 6x² + 15xy gives the cost (in dollars) of producing a rectangular container with square top/bottom of side x feet and height y feet.
  Example: x = 4, y = 6 → 6(4)² + 15(4)(6) = 96 + 360 = 456 dollars.
• Don't confuse: the variable represents a specific quantity (time, length, etc.); substitute carefully and include units in your final answer.

🧭 Overview

🧠 One-sentence thesis

Exponent properties—Product, Power, and Product to a Power—allow us to simplify expressions and multiply monomials by adding or multiplying exponents when bases are the same.

📌 Key points (3–5)

• Product Property: When multiplying like bases, add the exponents: a^m · a^n = a^(m+n).
• Power Property: When raising a power to a power, multiply the exponents: (a^m)^n = a^(m·n).
• Product to a Power Property: When raising a product to a power, raise each factor to that power: (ab)^m = a^m · b^m.
• Common confusion: Distinguish (−5)^4 (parentheses mean raise −5 to the 4th power, result is positive) from −5^4 (raise 5 to the 4th power then take the opposite, result is negative).
• Why it matters: These properties are essential for simplifying algebraic expressions and multiplying monomials efficiently.

🔢 Exponential notation basics

🔢 What exponents mean

Exponential notation: In the expression a^m, the exponent m tells us how many times we use the base a as a factor.

• The expression a^m is read "a to the m-th power."
• The base is the repeated factor; the exponent counts how many times it appears.
• Example: 4^3 means 4 · 4 · 4 = 64.

⚠️ Parentheses and negative bases

• (−5)^4 means multiply four factors of −5: (−5)(−5)(−5)(−5) = 625 (positive result).
• −5^4 means the opposite of 5^4: −(5 · 5 · 5 · 5) = −625 (negative result).
• Don't confuse: Parentheses around a negative base change the outcome. Without parentheses, only the positive number is raised to the power, then the negative sign is applied.

🧮 Any base to the first power

• Any number raised to the exponent 1 equals itself: a^1 = a.
• Example: 7^1 = 7.

🔗 Product Property for Exponents

🔗 The rule

Product Property for Exponents: If a is a real number and m and n are counting numbers, then a^m · a^n = a^(m+n). To multiply with like bases, add the exponents.

• The base stays the same; only the exponents are added.
• Example: x^2 · x^3 = x^(2+3) = x^5.
• Verification with numbers: 2^2 · 2^3 = 4 · 8 = 32, and 2^(2+3) = 2^5 = 32 ✓.

🔗 Why it works

• x^2 · x^3 means (x · x) · (x · x · x) = x · x · x · x · x = x^5.
• Counting the total factors: 2 factors + 3 factors = 5 factors.

🔗 Extending to more than two factors

• The property applies to any number of factors with the same base.
• Example: d^4 · d^5 · d^2 = d^(4+5+2) = d^11.
• Add all the exponents when the bases match.

🔗 When the base is implicit

• If a variable appears without an exponent, it has an implicit exponent of 1.
• Example: a^7 · a = a^7 · a^1 = a^(7+1) = a^8.

🔋 Power Property for Exponents

🔋 The rule

Power Property for Exponents: If a is a real number and m and n are whole numbers, then (a^m)^n = a^(m·n). To raise a power to a power, multiply the exponents.

• The base stays the same; the exponents are multiplied.
• Example: (y^5)^9 = y^(5·9) = y^45.
• Verification with numbers: (3^2)^3 = 9^3 = 729, and 3^(2·3) = 3^6 = 729 ✓.

🔋 Why it works

• (x^2)^3 means x^2 · x^2 · x^2.
• Each x^2 contributes 2 factors of x, and there are 3 such groups: 2 · 3 = 6 factors total.
• So (x^2)^3 = x^6.

🔋 Applying to numbers

• Example: (4^4)^7 = 4^(4·7) = 4^28.
• The property holds for any real base.

📦 Product to a Power Property

📦 The rule

Product to a Power Property for Exponents: If a and b are real numbers and m is a whole number, then (ab)^m = a^m · b^m. To raise a product to a power, raise each factor to that power.

• The exponent applies to every factor inside the parentheses.
• Example: (2x)^3 = 2^3 · x^3 = 8x^3.
• Verification with numbers: (2 · 3)^2 = 6^2 = 36, and 2^2 · 3^2 = 4 · 9 = 36 ✓.

📦 Why it works

• (2x)^3 means 2x · 2x · 2x.
• Group like factors: (2 · 2 · 2) · (x · x · x) = 2^3 · x^3.

📦 Multiple variables

• The property extends to any number of factors.
• Example: (3mn)^3 = 3^3 · m^3 · n^3 = 27m^3n^3.

📦 Negative coefficients

• Example: (−9d)^2 = (−9)^2 · d^2 = 81d^2.
• The exponent applies to the negative sign when it is inside the parentheses.

🧩 Combining multiple properties

🧩 Summary of properties

Property	Rule	What to do
Product Property	a^m · a^n = a^(m+n)	Add exponents (same base)
Power Property	(a^m)^n = a^(m·n)	Multiply exponents
Product to a Power	(ab)^m = a^m · b^m	Raise each factor to the power

• All three properties can be used together in one problem.
• Apply properties step by step, following the order of operations.

🧩 Multi-step simplification

• Example: (y^3)^6 · (y^5)^4
  • First apply Power Property: y^18 · y^20.
  • Then apply Product Property: y^(18+20) = y^38.
• Example: (−6x^4y^5)^2
  • Apply Product to a Power: (−6)^2 · (x^4)^2 · (y^5)^2.
  • Apply Power Property: 36 · x^8 · y^10 = 36x^8y^10.

🧩 Order matters

• Simplify inside parentheses first (Power Property or Product to a Power).
• Then combine like bases using the Product Property.
• Example: (5m)^2 · (3m^3)
  • Raise 5m to the second power: 25m^2.
  • Multiply: 25m^2 · 3m^3 = 75m^5.

🧩 Complex expressions

• Example: (3x^2y)^4 · (2xy^2)^3
  • Apply Product to a Power: (81x^8y^4) · (8x^3y^6).
  • Use Commutative Property to group: 81 · 8 · x^8 · x^3 · y^4 · y^6.
  • Multiply constants and add exponents: 648x^11y^10.

🔢 Multiplying monomials

🔢 What is a monomial

• A monomial is an algebraic expression with one term (a number, a variable, or a product of numbers and variables with whole-number exponents).
• Examples: 3x^2, −4x^3, (5/6)x^3y.

🔢 Multiplication process

• Use the Commutative Property to rearrange terms: group coefficients together, then variables.
• Multiply the coefficients.
• Apply the Product Property to add exponents of like bases.
• Example: (3x^2)(−4x^3)
  • Rearrange: 3 · (−4) · x^2 · x^3.
  • Multiply: −12x^5.

🔢 Monomials with fractions

• Example: (5/6 · x^3y) · (12xy^2)
  • Rearrange: (5/6) · 12 · x^3 · x · y · y^2.
  • Multiply: 10x^4y^3.
• Simplify the fraction coefficients first, then apply exponent rules.

🔢 Multiple variables

• Example: (4a^3b)(9a^2b^6)
  • Rearrange: 4 · 9 · a^3 · a^2 · b · b^6.
  • Multiply: 36a^5b^7.
• Add exponents separately for each variable.

🧭 Overview

🧠 One-sentence thesis

Multiplying polynomials extends the Distributive Property to combine terms systematically, whether multiplying a monomial by a polynomial, two binomials using FOIL, or larger polynomials using vertical or distributive methods.

📌 Key points (3–5)

• Monomial × polynomial: Distribute the monomial to every term in the polynomial, then simplify.
• Binomial × binomial: Use the Distributive Property, the FOIL method (First, Outer, Inner, Last), or the Vertical Method to get four products, then combine like terms.
• FOIL only works for binomials: For trinomials or larger polynomials, use the Distributive Property or Vertical Method.
• Common confusion: FOIL is a shortcut for the Distributive Property applied twice—it is not a separate rule and does not apply to multiplying a trinomial by a binomial.
• All methods yield the same result: Choose the method that feels most natural; the Distributive Property is universal, FOIL is fast for binomials, and the Vertical Method mirrors arithmetic multiplication.

🔢 Multiplying a monomial by a polynomial

🔢 The core idea: distribute

Multiplying a polynomial by a monomial: multiply the monomial by each term in the polynomial using the Distributive Property.

• This is the same process as distributing a number over parentheses, e.g., 2(x + 3) = 2x + 6.
• Now the "number" can be a variable or a product of variables and coefficients.
• Example: 4(x + 3) = 4x + 12; y(y − 2) = y² − 2y.

🧮 Multiplying with exponents

• When the monomial contains a variable that also appears in the polynomial, use the Product Property for Exponents: multiply coefficients and add exponents of like bases.
• Example: 7x(2x + y) = 14x² + 7xy.
• Example: −2y(4y² + 3y − 5) = −8y³ − 6y² + 10y.

📍 Order does not matter

• The monomial can be written first or second: (x + 3)p = px + 3p.
• Distribute in the same way regardless of position.

🔗 Multiplying two binomials

🔗 Using the Distributive Property

• Treat the second binomial as a single quantity and distribute it over the first binomial.
• Then distribute again within each term.
• Example: (x + 3)(x + 7)
  • Distribute (x + 7): x(x + 7) + 3(x + 7)
  • Distribute again: x² + 7x + 3x + 21
  • Combine like terms: x² + 10x + 21
• This method always works but requires careful bookkeeping.

🎯 The FOIL method

FOIL: an acronym for First, Outer, Inner, Last—the four products obtained when multiplying two binomials.

• First: multiply the first term of each binomial.
• Outer: multiply the outer terms (first of the first binomial, second of the second).
• Inner: multiply the inner terms (second of the first binomial, first of the second).
• Last: multiply the last term of each binomial.
• Example: (x + 5)(x + 9)
  • First: x · x = x²
  • Outer: x · 9 = 9x
  • Inner: 5 · x = 5x
  • Last: 5 · 9 = 45
  • Combine: x² + 9x + 5x + 45 = x² + 14x + 45
• FOIL only applies to binomials—it is a shortcut for the Distributive Property applied twice.

📐 The Vertical Method

• Write one binomial above the other, similar to multiplying two-digit numbers.
• Multiply the top binomial by each term of the bottom binomial, writing each result on a separate line (partial products).
• Add the partial products, aligning like terms.
• Example: (3y − 1)(2y − 6)
  • Multiply (3y − 1) by −6: −18y + 6
  • Multiply (3y − 1) by 2y: 6y² − 2y
  • Add: 6y² − 20y + 6
• The partial products match the terms from FOIL.

🧩 When like terms do not combine

• Sometimes the four products have no like terms, so the result has four terms.
• Example: (x − 2)(x − y) = x² − xy − 2x + 2y.
• Do not force combination; leave all four terms.

⚠️ Watch exponents and coefficients

• Example: (n² + 4)(n − 1) = n³ − n² + 4n − 4.
• Example: (3pq + 5)(6pq − 11) = 18p²q² − 33pq + 30pq − 55 = 18p²q² − 3pq − 55.

🏗️ Multiplying a trinomial by a binomial

🏗️ Using the Distributive Property

• Distribute each term of the binomial over the trinomial.
• Multiply each term of the trinomial by each term of the binomial.
• Combine like terms.
• Example: (b + 3)(2b² − 5b + 8)
  • Distribute b: b(2b² − 5b + 8) = 2b³ − 5b² + 8b
  • Distribute 3: 3(2b² − 5b + 8) = 6b² − 15b + 24
  • Add: 2b³ − 5b² + 8b + 6b² − 15b + 24 = 2b³ + b² − 7b + 24

📐 Using the Vertical Method

• Write the trinomial on top and the binomial below (fewer terms on the bottom reduces the number of partial products).
• Multiply the trinomial by each term of the binomial, writing each result on a separate line.
• Add the partial products, aligning like terms.
• Example: (b + 3)(2b² − 5b + 8)
  • Multiply by 3: 6b² − 15b + 24
  • Multiply by b: 2b³ − 5b² + 8b
  • Add: 2b³ + b² − 7b + 24

🚫 FOIL does not apply here

• FOIL is only for multiplying two binomials.
• For a trinomial × binomial, use the Distributive Property or Vertical Method.

🔄 Choosing a method

🔄 Summary of methods

Multiplication type	Recommended methods
Monomial × polynomial	Distributive Property
Binomial × binomial	Distributive Property, FOIL, or Vertical Method
Trinomial × binomial	Distributive Property or Vertical Method

• Distributive Property: works for all cases; requires careful distribution and combining like terms.
• FOIL: fast and systematic for binomials only; easy to remember with the acronym.
• Vertical Method: mirrors arithmetic multiplication; good for larger polynomials and visual learners.

🧠 Practice and preference

• The excerpt emphasizes practicing each method and choosing the one you prefer.
• All methods produce the same result when applied correctly.
• Example: Mental math applications—multiply 13 × 15 by thinking of it as (10 + 3)(10 + 5) and using FOIL to get 100 + 50 + 30 + 15 = 195.

⚠️ Common mistake: misapplying FOIL

• Do not use FOIL for anything other than binomial × binomial.
• For larger polynomials, revert to the Distributive Property or Vertical Method.

🧭 Overview

🧠 One-sentence thesis

Special product patterns—binomial squares and products of conjugates—provide shortcuts for multiplying specific types of binomials without using the full FOIL method.

📌 Key points (3–5)

• Binomial Squares Pattern: squaring a binomial (a + b)² or (a − b)² produces a trinomial with three predictable terms.
• Product of Conjugates Pattern: multiplying conjugates (a − b)(a + b) produces a binomial called a difference of squares.
• Common confusion: binomial squares vs. conjugates—squares produce trinomials with a middle term; conjugates produce binomials with no middle term because the middle terms cancel.
• Recognition skill: identifying which pattern applies (same binomial squared vs. sum times difference) determines which shortcut to use.
• Why it matters: these patterns reduce calculation steps and help with mental math.

🟦 Binomial Squares Pattern

🔲 What the pattern is

Binomial Squares Pattern: If a and b are real numbers,
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²

• Squaring a binomial means multiplying it by itself: (x + 9)² means (x + 9)(x + 9).
• The result is always a trinomial (three terms).

🧩 How the pattern works

To square a binomial:

1. Square the first term → a²
2. Square the last term → b²
3. Double their product → 2ab (positive for sum, negative for difference)

Example: (x + 5)²

• Square first term: x²
• Square last term: 25
• Double the product: 2 · x · 5 = 10x
• Result: x² + 10x + 25

Example: (y − 3)²

• Square first term: y²
• Square last term: 9
• Double the product: −2 · y · 3 = −6y
• Result: y² − 6y + 9

🔍 Why the middle term appears

• When you use FOIL on (a + b)(a + b), the outer and inner terms are both ab.
• Adding them gives 2ab, which becomes the middle term.
• This is why the pattern says "double their product."

Don't confuse: The middle term is not just a · b; it is 2ab (doubled).

📐 Examples with coefficients and variables

Example: (4x + 6)²

• First term squared: (4x)² = 16x²
• Last term squared: 6² = 36
• Double the product: 2 · 4x · 6 = 48x
• Result: 16x² + 48x + 36

Example: (2x − 3y)²

• First term squared: (2x)² = 4x²
• Last term squared: (3y)² = 9y²
• Double the product: −2 · 2x · 3y = −12xy
• Result: 4x² − 12xy + 9y²

🔗 Product of Conjugates Pattern

🔗 What conjugates are

Conjugate pair: two binomials of the form (a − b) and (a + b).

• Same first term, same last term.
• One is a sum, the other is a difference.
• Examples: (x − 9)(x + 9), (2x − 5)(2x + 5), (y − 8)(y + 8).

⚡ The pattern for conjugates

Product of Conjugates Pattern: If a and b are real numbers,
(a − b)(a + b) = a² − b²

• The result is a binomial (two terms only), called a difference of squares.
• There is no middle term.

To multiply conjugates:

1. Square the first term → a²
2. Square the last term → b²
3. Write as a difference: a² − b²

Example: (x − 8)(x + 8)

• Square first term: x²
• Square last term: 64
• Result: x² − 64

🧠 Why the middle term disappears

• When you use FOIL on (a − b)(a + b), you get: a² + ab − ab − b².
• The middle terms +ab and −ab are opposites and cancel to zero.
• Only the first and last terms remain: a² − b².

Don't confuse: Conjugates produce a binomial (no middle term); binomial squares produce a trinomial (with middle term).

🔢 Examples with various forms

Example: (2x + 5)(2x − 5)

• Square first term: (2x)² = 4x²
• Square last term: 5² = 25
• Result: 4x² − 25

Example: (3 + 5x)(3 − 5x)

• The variable can be in the second position; still conjugates.
• Square first term: 3² = 9
• Square last term: (5x)² = 25x²
• Result: 9 − 25x²

Example: (5m − 9n)(5m + 9n)

• Square first term: (5m)² = 25m²
• Square last term: (9n)² = 81n²
• Result: 25m² − 81n²

Example: (6u² − 11v⁵)(6u² + 11v⁵)

• Square first term: (6u²)² = 36u⁴
• Square last term: (11v⁵)² = 121v¹⁰
• Result: 36u⁴ − 121v¹⁰

🎯 Recognizing Which Pattern to Use

🎯 Comparison table

Feature	Binomial Squares	Product of Conjugates
Form	(a + b)² or (a − b)²	(a − b)(a + b)
Operation	Squaring one binomial	Multiplying two different binomials
Result type	Trinomial (3 terms)	Binomial (2 terms)
Middle term	Yes: ±2ab (doubled product)	No (cancels to zero)
FOIL middle terms	Same (both ab)	Opposites (+ab and −ab)
Name of result	Perfect square trinomial	Difference of squares

🔎 How to decide

Ask yourself:

1. Is it the same binomial twice (squared)? → Use binomial squares pattern.
2. Are there two binomials, one sum and one difference, with the same terms? → Use product of conjugates pattern.
3. Does it fit neither pattern? → Use FOIL or the distributive property.

Example decision process:

• (2x − 3)(2x + 3): Conjugates (sum and difference) → use conjugates pattern → result: 4x² − 9.
• (8x − 5)²: Same binomial squared → use binomial squares pattern → result: 64x² − 80x + 25.
• (5x − 6)(6x + 5): Different binomials, not conjugates → use FOIL → result: 30x² − 11x − 30.

⚠️ Common mistakes

• Mistake: Treating (a + b)² as a² + b² (forgetting the middle term).
  • Correct: (a + b)² = a² + 2ab + b².
• Mistake: Thinking conjugates produce a trinomial.
  • Correct: Conjugates produce a binomial (difference of squares).
• Mistake: Confusing which pattern to use when both binomials look similar.
  • Check: Are they identical (square) or one sum and one difference (conjugates)?

🧮 Mental Math Applications

🧮 Using conjugates for mental multiplication

The excerpt shows how to multiply numbers like 47 × 53 without a calculator:

• Think of 47 as (50 − 3) and 53 as (50 + 3).
• Recognize this as conjugates: (50 − 3)(50 + 3).
• Apply the pattern: 50² − 3² = 2500 − 9 = 2491.

Why it works: The conjugates pattern turns multiplication into simpler squaring and subtraction.

🧮 Using binomial squares for mental squaring

To square 65 mentally:

• Think of 65 as (60 + 5).
• Apply the binomial squares pattern: (60 + 5)² = 60² + 2·60·5 + 5².
• Calculate: 3600 + 600 + 25 = 4225.

Why it works: Breaking a number into tens and ones makes squaring easier using the pattern.

🧭 Overview

🧠 One-sentence thesis

The Quotient Property for Exponents allows us to divide monomials by subtracting exponents when bases are the same, and combining this with other exponent properties enables simplification of complex algebraic fractions.

📌 Key points (3–5)

• Quotient Property core rule: When dividing powers with the same base, subtract the exponents (smaller from larger) and place the result in the numerator if the numerator exponent is larger, or in the denominator if the denominator exponent is larger.
• Zero exponent definition: Any nonzero number or expression raised to the zero power equals 1.
• Quotient to a Power Property: To raise a fraction to a power, raise both the numerator and denominator to that power separately.
• Common confusion: Direction matters—compare exponents first to determine whether the result will have factors in the numerator (when numerator exponent is larger) or denominator (when denominator exponent is larger).
• Combining properties: Simplifying complex expressions requires applying multiple properties in the correct order (parentheses first, then exponents, then multiplication/division).

📐 The Quotient Property for Exponents

📐 What the property states

Quotient Property for Exponents: If a is a real number (a ≠ 0) and m and n are whole numbers, then:

• a to the m divided by a to the n equals a to the (m minus n), when m > n
• a to the m divided by a to the n equals 1 divided by a to the (n minus m), when n > m

• This property works because division cancels common factors.
• Example: x to the 5 divided by x to the 2 means (x·x·x·x·x) divided by (x·x), which simplifies to x·x·x or x to the 3.

🔍 Comparing exponents first

Key decision: Before applying the property, determine which exponent is larger.

• If numerator exponent is larger: Result has factors in the numerator.
  • Example: x to the 9 divided by x to the 7 equals x to the (9 minus 7) equals x squared.
• If denominator exponent is larger: Result has factors in the denominator.
  • Example: b to the 8 divided by b to the 12 equals 1 divided by b to the (12 minus 8) equals 1 divided by b to the 4.

Don't confuse: The location of the larger exponent determines where the result appears, not the order you subtract.

🎯 Zero Exponent Property

🎯 Why zero exponents equal one

The excerpt shows that when exponents are equal, two approaches give the same result:

• Any number divided by itself equals 1 (e.g., 8 divided by 8 equals 1).
• Using the Quotient Property: a to the m divided by a to the m equals a to the (m minus m) equals a to the 0.
• Therefore, a to the 0 must equal 1.

Zero Exponent: If a is a nonzero number, then a to the 0 equals 1.

🎯 Applying to expressions

• A single number or variable: 9 to the 0 equals 1; n to the 0 equals 1.
• An entire expression in parentheses: (5b) to the 0 equals 1; (negative 4 times a squared times b) to the 0 equals 1.
• The zero exponent applies to the entire grouped expression.

🔺 Quotient to a Power Property

🔺 The property definition

Quotient to a Power Property: If a and b are real numbers (b ≠ 0) and m is a counting number, then (a divided by b) to the m equals a to the m divided by b to the m.

• To raise a fraction to a power, raise both the numerator and denominator to that power.
• Example: (3 divided by 7) squared equals 3 squared divided by 7 squared equals 9 divided by 49.

🔺 With variables

• (b divided by 3) to the 4 equals b to the 4 divided by 3 to the 4 equals b to the 4 divided by 81.
• (k divided by j) to the 3 equals k to the 3 divided by j to the 3.
• Apply the power to each part of the fraction separately.

🧮 Combining Multiple Properties

🧮 Order of operations matters

When simplifying complex expressions, follow this sequence:

1. Parentheses first: Simplify inside grouping symbols.
2. Apply Power Property: Multiply exponents when raising a power to a power.
3. Apply Quotient Property: Subtract exponents when dividing like bases.

🧮 Common patterns

Pattern	Steps	Result
(y to the 4) squared divided by y to the 6	Multiply exponents in numerator: y to the 8 divided by y to the 6	y squared
b to the 12 divided by (b squared) to the 6	Multiply exponents in denominator: b to the 12 divided by b to the 12	1
((y to the 9 divided by y to the 4)) squared	Simplify inside first: (y to the 5) squared	y to the 10

🧮 With coefficients and multiple variables

Example: (2m squared divided by 5n) to the 4

• Raise numerator and denominator to the fourth power separately.
• Apply power to each factor: (2 to the 4)(m squared to the 4) divided by (5 to the 4)(n to the 4).
• Simplify: 16m to the 8 divided by 625n to the 4.

Don't confuse: When you cannot simplify inside parentheses (different bases), apply the Quotient to a Power Property first.

➗ Dividing Monomials

➗ Basic division process

To divide monomials:

1. Rewrite as a fraction if given with division symbol.
2. Separate into coefficient divided by coefficient and variables divided by variables.
3. Simplify coefficients by regular division.
4. Apply Quotient Property to each variable.

Example: 56x to the 7 divided by 8x to the 3

• Rewrite: (56 divided by 8) times (x to the 7 divided by x to the 3)
• Simplify: 7 times x to the 4 equals 7x to the 4

➗ Multiple variables

Example: 45a squared b to the 3 divided by negative 5ab to the 5

• Separate: (45 divided by negative 5) times (a squared divided by a) times (b to the 3 divided by b to the 5)
• Simplify: negative 9 times a times (1 divided by b squared)
• Result: negative 9a divided by b squared

➗ When numerator has products

If the numerator contains a product of monomials, multiply them first (following order of operations—fraction bar is a grouping symbol).

Example: (6x squared y to the 3)(5x to the 3 y squared) divided by (3x to the 4 y to the 5)

• Multiply numerator first: 30x to the 5 y to the 5 divided by 3x to the 4 y to the 5
• Simplify: 10x

Don't confuse: Always complete operations in the numerator or denominator before simplifying the fraction itself.

🧭 Overview

🧠 One-sentence thesis

Dividing polynomials—whether by monomials or binomials—extends fraction arithmetic to algebraic expressions and relies on splitting each term separately or using a long-division procedure similar to numerical division.

📌 Key points (3–5)

• Dividing by a monomial: split the polynomial into separate fractions, one for each term, then simplify each fraction independently.
• Dividing by a binomial: use long division (analogous to numeric long division) by repeatedly dividing, multiplying, subtracting, and bringing down terms.
• Handling remainders: when division does not come out evenly, write the remainder as a fraction with the divisor as the denominator.
• Common confusion: forgetting to include placeholder terms (e.g., 0x³) when a polynomial has missing degrees can lead to alignment errors in long division.
• Sign care: when the divisor or any term is negative, change signs carefully during subtraction steps to avoid errors.

🧩 Dividing a polynomial by a monomial

🧩 The core rule

Division of a Polynomial by a Monomial: To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.

• This rule is based on the reverse of fraction addition: (a + b)/c = a/c + b/c.
• You split one big fraction into separate fractions, one per term in the numerator.
• Each fraction simplifies independently using the rules for dividing monomials (subtract exponents, divide coefficients).

Example: (7y² + 21)/7 becomes (7y²)/7 + 21/7, which simplifies to y² + 3.

➗ Writing division as a fraction

• If the problem is written with a division symbol (÷), rewrite it as a fraction first.
• Put the polynomial in the numerator and the monomial in the denominator.
• Then apply the term-by-term splitting rule.

Example: (18x³ − 36x²) ÷ 6x becomes (18x³ − 36x²)/(6x), then split into (18x³)/(6x) − (36x²)/(6x) = 3x² − 6x.

⚠️ Watch out for negatives

• When dividing by a negative monomial, each term's sign may flip.
• Remember that subtracting a negative is the same as adding a positive.

Example: (12d² − 16d)/(−4) splits into (12d²)/(−4) − (16d)/(−4) = −3d² + 4d (the second term becomes positive).

🔢 Multiple variables and non-integer results

• When the polynomial has multiple variables, divide coefficients and subtract exponents for each variable separately.
• If a term in the numerator has a lower degree than the denominator, the result will include a fraction or negative exponent.

Example: (10x² + 5x − 20)/(5x) becomes (10x²)/(5x) + (5x)/(5x) − 20/(5x) = 2x + 1 − 4/x.

Don't confuse: The term 4/x is not a polynomial term (it has a variable in the denominator), but it is a valid algebraic expression resulting from division.

📐 Dividing a polynomial by a binomial

📐 Long division setup

• Write the division as a long-division bracket, with the dividend (polynomial being divided) inside and the divisor (binomial) outside.
• Ensure the dividend is in standard form (descending order of degrees) and include placeholder terms (e.g., 0x²) for any missing degrees.

🔁 The long-division steps

1. Divide the leading term of the current dividend by the leading term of the divisor; write the result in the quotient.
2. Multiply that quotient term by the entire divisor and write the product below the dividend, aligning like terms.
3. Subtract the product from the dividend (change signs and add to avoid errors).
4. Bring down the next term from the original dividend.
5. Repeat steps 1–4 until no terms remain to bring down.

Example: Dividing (x² + 9x + 20) by (x + 5):

• Divide x² by x → x; multiply x(x + 5) = x² + 5x; subtract to get 4x; bring down 20.
• Divide 4x by x → 4; multiply 4(x + 5) = 4x + 20; subtract to get 0.
• Quotient: x + 4.

📝 Handling remainders

• If after the final subtraction you have a nonzero constant or term, that is the remainder.
• Write the final answer as: quotient + (remainder)/(divisor).

Example: Dividing (x³ − x² + x + 4) by (x + 1) leaves a remainder of 1, so the answer is x² − 2x + 3 + 1/(x + 1).

🕳️ Missing terms and placeholders

• If the dividend is missing a degree (e.g., x⁴ − x² + 5x − 2 has no x³ term), insert 0x³ as a placeholder.
• This keeps terms aligned during subtraction and prevents mistakes.

Example: (x⁴ − x² + 5x − 2) ÷ (x + 2) should be written with 0x³ included: x⁴ + 0x³ − x² + 5x − 2.

✅ Checking your work

• Multiply the quotient by the divisor (and add any remainder).
• The result should equal the original dividend.
• This verifies that the division was performed correctly.

🔍 Common pitfalls and tips

🔍 Sign errors in subtraction

• When subtracting polynomials in long division, it is safer to change all signs in the subtracted row and then add.
• This reduces mistakes, especially when the divisor contains a subtraction (e.g., x − 3).

Example: Subtracting (2x² − 6x) from (2x² − 5x) is easier if you change signs: 2x² − 5x − (2x² − 6x) becomes 2x² − 5x − 2x² + 6x = x.

🔍 Forgetting to bring down terms

• After each subtraction, bring down the next term from the dividend before dividing again.
• Skipping this step will misalign your work and produce an incorrect quotient.

🔍 Dividing only the first term

• A common error is to divide only the leading term and stop, rather than continuing through all terms.
• Always repeat the divide-multiply-subtract-bring down cycle until no terms remain.

🔍 Placeholder terms

• If you skip placeholder terms for missing degrees, your like terms will not line up during subtraction.
• Always write the dividend in complete standard form, inserting 0 coefficients where needed.

Don't confuse: A missing term (e.g., no x² in x³ + x + 1) is not the same as a zero result—it means you must write 0x² explicitly during long division.

🧭 Overview

🧠 One-sentence thesis

Negative exponents and scientific notation provide systematic ways to represent very large and very small numbers efficiently, using the reciprocal property and powers of ten.

📌 Key points (3–5)

• Negative exponents mean "take the reciprocal": a to the power of negative n equals 1 divided by a to the power of n.
• Scientific notation expresses numbers as a product of a number between 1 and 10 and a power of 10.
• Common confusion: The base matters—parentheses determine whether the negative sign is part of the base or separate from it.
• All exponent properties (Product, Power, Quotient) extend from whole-number exponents to integer exponents.
• Why it matters: Scientists use scientific notation to handle extremely large (astronomical distances) or extremely small (atomic sizes) measurements without writing many zeros.

🔄 Negative exponents

🔄 Definition and meaning

Negative exponent: If n is an integer and a ≠ 0, then a to the power of negative n equals 1 divided by a to the power of n.

• A negative exponent does not make the number negative.
• It tells you to "flip" the base to the reciprocal position and change the sign of the exponent.
• Example: 4 to the power of negative 2 equals 1 divided by 4 squared, which equals 1/16.

🔄 Reciprocal property

Property of Negative Exponents: If n is an integer and a ≠ 0, then 1 divided by (a to the power of negative n) equals a to the power of n.

• When you have 1 in the numerator and a negative exponent in the denominator, move the base to the numerator and make the exponent positive.
• Example: 1 divided by (y to the power of negative 4) equals y to the power of 4.

🔄 Quotient to a negative power

Quotient to a Negative Exponent Property: If a and b are real numbers (a ≠ 0, b ≠ 0) and n is an integer, then (a/b) to the power of negative n equals (b/a) to the power of n.

• Take the reciprocal of the entire fraction and change the sign of the exponent.
• Example: (5/7) to the power of negative 2 equals (7/5) squared, which equals 49/25.

⚠️ Base identification matters

The excerpt emphasizes careful identification of the base:

Expression	Base	Result
(negative 3) to the power of negative 2	negative 3	1/9
negative (3 to the power of negative 2)	3 (with a separate negative sign)	negative 1/9
(negative 1/3) to the power of negative 2	negative 1/3	9
negative (1/3) to the power of negative 2	1/3 (with a separate negative sign)	negative 9

• Don't confuse: Parentheses determine whether the negative sign is part of the base or applied afterward.
• The Order of Operations applies: exponents before multiplication by negative 1.

🧮 Simplifying with integer exponents

🧮 All properties still apply

The excerpt lists all exponent properties that now work with integer (not just whole-number) exponents:

• Product Property: a to the power of m times a to the power of n equals a to the power of (m plus n).
• Power Property: (a to the power of m) to the power of n equals a to the power of (m times n).
• Product to a Power: (ab) to the power of m equals (a to the power of m) times (b to the power of m).
• Quotient Property: a to the power of m divided by a to the power of n equals a to the power of (m minus n), where a ≠ 0.
• Zero Exponent: a to the power of 0 equals 1, where a ≠ 0.
• Quotient to a Power: (a/b) to the power of m equals (a to the power of m) divided by (b to the power of m), where b ≠ 0.

🧮 Simplest form requirement

• An expression is not in simplest form if it contains negative exponents.
• After applying properties, rewrite any negative exponents using the definition (move to denominator or numerator and make positive).
• Example: x to the power of negative 3 simplifies to 1 divided by (x cubed).

🧮 Working with variables

• The same rules apply to variables as to numbers.
• Example: (x to the power of negative 4) times (x to the power of 6) equals x squared (add exponents: negative 4 plus 6 equals 2).
• Example: (y to the power of negative 6) times (y to the power of 4) equals y to the power of negative 2, which then becomes 1 divided by (y squared).

🧮 Combining monomials

When multiplying monomials with negative exponents:

1. Group like bases together (Commutative Property).
2. Multiply numerical coefficients.
3. Add exponents for each variable base.
4. Rewrite any remaining negative exponents in reciprocal form.

Example: (2 times x to the power of negative 6 times y to the power of 8) times (negative 5 times x to the power of 5 times y to the power of negative 3) equals negative 10 times x to the power of negative 1 times y to the power of 5, which simplifies to (negative 10 times y to the power of 5) divided by x.

📐 Scientific notation

📐 What it is

Scientific notation: A number is expressed in scientific notation when it is of the form a times 10 to the power of n, where 1 ≤ a < 10 and n is an integer.

• The first factor must be greater than or equal to 1 but less than 10.
• The second factor is a power of 10.
• Example: 4,000 equals 4 times 10 cubed; 0.004 equals 4 times 10 to the power of negative 3.

📐 Converting from decimal to scientific notation

Steps:

1. Move the decimal point so the first factor is between 1 and 10.
2. Count how many places (n) the decimal moved.
3. Write as a product with a power of 10:
   • If the original number is greater than 1, the power is 10 to the power of n (positive).
   • If the original number is between 0 and 1, the power is 10 to the power of negative n (negative).
4. Check your work.

Example: 37,000 becomes 3.7 times 10 to the power of 4 (decimal moved 4 places left, original number greater than 1).

Example: 0.0052 becomes 5.2 times 10 to the power of negative 3 (decimal moved 3 places right, original number between 0 and 1).

📐 Converting from scientific notation to decimal

Steps:

1. Identify the exponent n on the factor 10.
2. Move the decimal n places:
   • If the exponent is positive, move the decimal right.
   • If the exponent is negative, move the decimal left (use absolute value of n).
3. Add zeros as placeholders if needed.
4. Check.

Example: 6.2 times 10 cubed becomes 6,200 (move decimal 3 places right).

Example: 8.9 times 10 to the power of negative 2 becomes 0.089 (move decimal 2 places left).

📐 Multiplying and dividing in scientific notation

Multiplication:

• Use the Commutative Property to group numerical factors and powers of 10 separately.
• Multiply the numerical parts.
• Add the exponents on the powers of 10 (Product Property).
• Convert the result to decimal form if requested.

Example: (4 times 10 to the power of 5) times (2 times 10 to the power of negative 7) equals 8 times 10 to the power of negative 2, which equals 0.08.

Division:

• Separate into two fractions: numerical parts and powers of 10.
• Divide the numerical parts.
• Subtract the exponents on the powers of 10 (Quotient Property).
• Convert to decimal form if requested.

Example: (9 times 10 cubed) divided by (3 times 10 to the power of negative 2) equals 3 times 10 to the power of 5, which equals 300,000.

🌍 Why scientists use it

• Astronomers work with very large numbers (distances in the universe, ages of stars).
• Chemists work with very small numbers (size of atoms, charge on electrons).
• Scientific notation avoids writing many zeros and makes calculations with extreme values manageable.
• The Properties of Exponents make multiplication and division straightforward.

🧭 Overview

🧠 One-sentence thesis

Factoring reverses multiplication by splitting a polynomial into simpler factors, starting with the greatest common factor (GCF) and using grouping for polynomials with four or more terms.

📌 Key points (3–5)

• What factoring is: the reverse process of multiplying—splitting a product into factors.
• Greatest Common Factor (GCF): the largest expression that is a factor of two or more expressions; always check for it first.
• How to find the GCF: factor coefficients into primes, expand variables, circle common factors in each column, then multiply them.
• Factor by grouping: used for polynomials with four or more terms—group terms with common factors, factor each group, then factor out the common binomial.
• Common confusion: factoring is not simplifying or combining like terms; it is rewriting a sum as a product.

🔍 What factoring means

🔍 Factoring as reverse multiplication

Factoring is splitting a product into factors; in other words, it is the reverse process of multiplying.

• Multiplication takes factors and produces a product.
• Factoring takes a product (a polynomial) and breaks it back into factors.
• Example: If multiplying gives you 6x + 9, factoring finds 3(2x + 3).

🎯 Why start with the GCF

• The GCF is the largest expression that divides all terms in the polynomial.
• Always factor out the GCF first before attempting other factoring methods.
• This simplifies the remaining polynomial and makes further factoring easier.

🧮 Finding the Greatest Common Factor

🧮 Step-by-step GCF process

The excerpt provides a detailed procedure:

1. Factor each coefficient into primes: break down numbers into their prime factors.
2. Write all variables with exponents in expanded form: for example, x³ becomes x · x · x.
3. List all factors—matching common factors in a column: align common factors vertically.
4. Circle the common factors in each column: identify what appears in every term.
5. Bring down the common factors that all expressions share: collect only the circled factors.
6. Multiply the factors: the product is the GCF.

• Don't confuse: the GCF is not just the largest number; it includes both numerical and variable factors.
• Example reference: the excerpt mentions "as in Example 7.2" for the complete procedure.

🔧 Factoring out the GCF from a polynomial

Once you have the GCF:

1. Find the GCF of all the terms (using the process above).
2. Rewrite each term as a product using the GCF: express each term as GCF times something.
3. Use the 'reverse' Distributive Property: factor out the GCF, leaving the remaining factors in parentheses.
4. Check by multiplying the factors: distribute to verify you get the original polynomial.

• The "reverse" Distributive Property means undoing distribution: ab + ac becomes a(b + c).
• Example reference: "as in Example 7.5."

🧩 Factor by Grouping

🧩 When to use grouping

To factor a polynomial with four or more terms.

• Grouping is a technique specifically for polynomials that have four or more terms.
• It works when terms can be paired so that each pair shares a common factor.

🧩 Grouping procedure

The excerpt outlines four steps:

1. Group terms with common factors: pair terms that share factors (often the first two and last two).
2. Factor out the common factor in each group: apply GCF factoring to each pair separately.
3. Factor the common factor from the expression: after step 2, a common binomial should appear; factor it out.
4. Check by multiplying the factors: distribute to confirm the original polynomial.

• Example reference: "as in Example 7.15."

⚠️ Key insight for grouping

• After factoring each group, you should see the same binomial in both groups.
• If the binomials don't match, try regrouping the terms differently.
• Don't confuse: grouping is not the same as combining like terms; you are creating a common binomial factor.

📋 Summary of factoring strategies

📋 General approach

Step	Action	Purpose
1	Check for GCF first	Simplify the polynomial
2	If 4+ terms, try grouping	Create common binomial factors
3	Check by multiplying	Verify the factoring is correct

• The excerpt emphasizes checking by multiplying in every method.
• Factoring is complete when the polynomial is written as a product of simpler factors.

📋 Related concepts mentioned

The excerpt lists related terms in the chapter review:

• Prime polynomials: polynomials that cannot be factored.

• Perfect square trinomials pattern: specific factoring patterns (a² + 2ab + b² and a² − 2ab + b²).

• Difference of squares pattern: another special factoring case.

• Zero Product Property: if a product equals zero, at least one factor is zero (used for solving quadratic equations after factoring).

• Don't confuse: factoring is a tool for simplifying and solving; the Zero Product Property is what makes factoring useful for solving equations.

🧭 Overview

🧠 One-sentence thesis

Factoring trinomials of the form x²+bx+c requires finding two numbers that multiply to c and add to b, then writing the expression as two binomials.

📌 Key points (3–5)

• What the method does: converts a trinomial x²+bx+c into a product of two binomials (x+m)(x+n).
• The key step: find two numbers m and n where m·n = c and m+n = b.
• How to structure the factors: write them as (x )(x ) first, then fill in the last terms with m and n.
• Common confusion: the two numbers must satisfy both conditions—multiplying to c and adding to b—not just one.
• Always verify: check by multiplying the factors back together to confirm the original trinomial.

🔢 The factoring process

🔢 Starting structure

• Begin by writing the factors as two binomials with first terms x: (x )(x ).
• This structure comes from the fact that the original trinomial has x² as its first term.
• The blanks will be filled with the two numbers you find in the next step.

🎯 Finding the two numbers

The two numbers m and n must:

• Multiply to c: m · n = c
• Add to b: m + n = b

• These are the only two conditions the numbers must satisfy.
• Both conditions must be met simultaneously—one alone is not enough.
• Example: If you need to factor x² + 7x + 12, you need two numbers that multiply to 12 and add to 7 (which are 3 and 4).

✍️ Completing the factors

• Once you find m and n, use them as the last terms of the binomials: (x + m)(x + n).
• The order of m and n doesn't matter because multiplication is commutative.
• Example: (x + 3)(x + 4) and (x + 4)(x + 3) are equivalent.

✅ Verification step

✅ Why checking matters

• The excerpt emphasizes: Check by multiplying the factors.
• Multiplying the binomials back together should give you the original trinomial x² + bx + c.
• This step catches errors in finding m and n or in writing the factors.

✅ How to check

• Multiply (x + m)(x + n) using the distributive property (FOIL).
• First terms: x · x = x²
• Outer and inner terms: x · n + m · x = (m + n)x = bx (because m + n = b)
• Last terms: m · n = c
• Result: x² + bx + c, which matches the original.

🔍 Common pitfalls

🔍 Both conditions required

• Don't confuse: finding two numbers that only multiply to c, or only add to b.
• Both m · n = c and m + n = b must be true.
• Example: For x² + 5x + 6, the numbers 1 and 6 multiply to 6, but they add to 7, not 5—so they don't work. The correct pair is 2 and 3.

🔍 Signs matter

• The excerpt doesn't detail sign rules in this section, but the method applies to all values of b and c.
• When c is positive, m and n have the same sign (both positive or both negative).
• When c is negative, m and n have opposite signs.

📋 Step-by-step summary

Step	Action	Purpose
1	Write (x )(x )	Set up the binomial structure
2	Find m and n where m·n = c and m+n = b	Identify the last terms
3	Write (x + m)(x + n)	Complete the factored form
4	Check by multiplying	Verify the result matches the original trinomial

🧭 Overview

🧠 One-sentence thesis

Factoring trinomials where the leading coefficient is not 1 requires systematically testing factor pairs of both the first and third terms until the correct combination produces the original middle term.

📌 Key points (3–5)

• What makes this harder: when the coefficient of x² (called "a") is not 1, you must find factor pairs for both the first term and the third term.
• Trial and error method: write the trinomial in descending order, list all factor pairs of the first and third terms, then test combinations until you find the one that gives the correct middle term.
• Order matters: always write the trinomial in descending order of degrees before factoring.
• Common confusion: don't forget to check your answer by multiplying the factors back together—this catches errors in sign or combination.
• Key difference from simpler case: unlike x²+bx+c (section 7.2), here you have multiple possible first terms in your binomial factors, not just (x)(x).

🔢 Understanding the form ax²+bx+c

🔢 What the "a" changes

• In the simpler form x²+bx+c, the first term is always x², so factors always start with (x)(x).
• When the first term is ax² (where a is not 1), you must find two numbers that multiply to give a.
• Example: if the trinomial is 6x²+bx+c, the first term factor pairs could be (6x)(x), (3x)(2x), (2x)(3x), or (x)(6x).

📐 Descending order requirement

• The excerpt emphasizes: "Write the trinomial in descending order of degrees."
• This means arrange terms from highest power to lowest: ax² term first, then bx term, then constant c.
• Why it matters: the systematic method depends on identifying which term is first, middle, and last.

🧪 The trial and error method

🧪 Step-by-step process

The excerpt outlines a five-step approach:

1. Write in descending order of degrees
2. Find all factor pairs of the first term (the ax² term)
3. Find all factor pairs of the third term (the constant c)
4. Test all possible combinations of the factors until the correct product is found
5. Check by multiplying the factors

🎯 What "test combinations" means

• You're looking for two binomials whose product gives you back the original trinomial.
• The "correct product" means when you multiply the binomials using FOIL or distribution, the middle term matches the original bx term.
• Don't confuse: you're not just matching the first and last terms—the middle term must also work out correctly.

✅ Why checking matters

• The excerpt lists "Check by multiplying" as the final step in both methods described.
• Multiply your two binomial factors together; if you get the original trinomial, your factorization is correct.
• This catches sign errors, wrong factor pair choices, or incorrect ordering.

🔄 Comparison with simpler trinomials

🔄 Section 7.2 vs 7.3

Aspect	Section 7.2: x²+bx+c	Section 7.3: ax²+bx+c
First term factors	Always (x)(x)	Must find factor pairs of a
Complexity	Find two numbers m and n	Test multiple combinations
What to find	m·n = c and m+n = b	Factor pairs of both first and third terms

🔄 Building on previous knowledge

• Section 7.2 taught factoring when the leading coefficient is 1.
• Section 7.3 extends this to cases where the leading coefficient is any number.
• The core idea (finding binomial factors) remains the same, but the search space is larger.

🧭 Overview

🧠 One-sentence thesis

Certain polynomial patterns—perfect square trinomials and differences of squares—can be factored quickly by recognizing their structure and applying reverse formulas.

📌 Key points (3–5)

• Two special patterns: perfect square trinomials (a² + 2ab + b² or a² − 2ab + b²) and differences of squares (a² − b²).
• Recognition is key: check whether terms are perfect squares and whether the middle term (for trinomials) or sign (for binomials) fits the pattern.
• Common confusion: not all trinomials are perfect squares—the middle term must be exactly 2ab; not all binomials are differences of squares—there must be a subtraction sign and both terms must be perfect squares.
• Factored forms: perfect square trinomials factor into (a + b)² or (a − b)²; differences of squares factor into (a + b)(a − b).
• Always check: multiply the factors back to verify the result.

🎯 Perfect square trinomials

🔍 What the pattern looks like

Perfect square trinomial: a trinomial of the form a² + 2ab + b² or a² − 2ab + b².

• The excerpt emphasizes three checks:
  • First term is a perfect square: (a)²
  • Last term is a perfect square: (b)²
  • Middle term is exactly 2ab (twice the product of a and b)
• If all three conditions hold, the trinomial is a perfect square.

✅ How to recognize it

1. Check the first term: Can it be written as something squared?
2. Check the last term: Can it be written as something squared?
3. Check the middle term: Is it exactly 2 · (first square root) · (second square root)?

Example: For a trinomial like x² + 6x + 9:

• First term x² = (x)²
• Last term 9 = (3)²
• Middle term 6x = 2 · x · 3 ✓
• So it fits the pattern a² + 2ab + b².

🧩 How to factor it

• Once you confirm the pattern, write the factored form as a binomial squared:
  • a² + 2ab + b² becomes (a + b)²
  • a² − 2ab + b² becomes (a − b)²
• The sign in the binomial matches the sign of the middle term in the trinomial.
• Always multiply back to check.

⚠️ Common confusion

• Don't assume every trinomial is a perfect square: the middle term must be exactly 2ab, not just any coefficient.
• Example: x² + 5x + 4 has perfect square first and last terms (x² and 2²), but 5x ≠ 2 · x · 2, so it is not a perfect square trinomial.

🔲 Differences of squares

🔍 What the pattern looks like

Difference of squares: a binomial of the form a² − b².

• The excerpt specifies two checks:
  • It must be a difference (subtraction, not addition)
  • Both the first and last terms must be perfect squares
• If both conditions hold, the binomial is a difference of squares.

✅ How to recognize it

1. Check for subtraction: Is there a minus sign between the two terms?
2. Check both terms: Can each be written as something squared?

Example: For x² − 25:

• It is a difference (−)
• x² = (x)²
• 25 = (5)²
• So it fits the pattern a² − b².

🧩 How to factor it

• Write each term as a square: (a)² − (b)²
• Factor into the product of a sum and a difference: (a + b)(a − b)
• The excerpt instructs "Write the product," meaning the factored form is always (first square root + second square root)(first square root − second square root).
• Always multiply back to check.

⚠️ Common confusion

• Don't confuse with sums of squares: a² + b² does not factor using this method (the excerpt says "Is this a difference?").
• Both terms must be perfect squares: if one term is not a perfect square, the pattern does not apply.

📋 Step-by-step summary

📝 For perfect square trinomials

Step	Action	What to check
1	Does it fit the pattern?	First term = (a)², last term = (b)², middle = 2ab
2	Write the square of the binomial	(a + b)² or (a − b)² depending on the middle sign
3	Check by multiplying	Expand to verify you get the original trinomial

📝 For differences of squares

Step	Action	What to check
1	Does it fit the pattern?	Difference (−) and both terms are perfect squares
2	Write them as squares	(a)² − (b)²
3	Write the product	(a + b)(a − b)
4	Check by multiplying	Expand to verify you get the original binomial

🧭 Overview

🧠 One-sentence thesis

A systematic strategy for factoring polynomials requires first checking for a greatest common factor, then choosing the appropriate method based on whether the polynomial is a binomial, trinomial, or has more terms, and finally verifying the result by multiplying the factors back together.

📌 Key points (3–5)

• Always start with GCF: before applying any other method, factor out the greatest common factor if one exists.
• Structure determines method: binomials, trinomials, and polynomials with more than three terms each require different factoring approaches.
• Binomial patterns matter: sums of squares cannot be factored, but differences of squares, sums of cubes, and differences of cubes each have specific patterns.
• Common confusion: not all binomials factor—sums of squares have no factorization, while differences of squares do.
• Always verify: check by multiplying the factors to confirm they produce the original polynomial.

🔍 First step: greatest common factor

🔍 Why GCF comes first

• The strategy begins with a mandatory question: "Is there a greatest common factor?"
• If a GCF exists, factor it out before attempting any other method.
• This simplifies the remaining polynomial and makes subsequent factoring easier.
• Example: if a polynomial has a common factor of 3 in all terms, remove it first, then work with the simpler expression inside.

🧩 Choosing by structure

🧩 Binomials: sum vs difference

The excerpt distinguishes two main cases for binomials:

Type	Sub-case	Method
Sum	Sum of squares	Does not factor
Sum	Sum of cubes	Use the sum of cubes pattern
Difference	Difference of squares	Factor as product of conjugates
Difference	Difference of cubes	Use the difference of cubes pattern

• Don't confuse: sums of squares have no factorization, but sums of cubes do.
• Example: a squared plus b squared cannot be factored, but a cubed plus b cubed follows a specific pattern.

🧩 Trinomials: two forms

The strategy splits trinomials by their leading coefficient:

Form x² + bx + c: use the "Undo FOIL" method.

Form ax² + bx + c: check if it fits the perfect square trinomial pattern, or use trial and error or the "ac" method.

• If both 'a' and 'c' are perfect squares, check whether the trinomial matches the perfect square pattern before trying other methods.
• The "ac" method involves finding two numbers that multiply to a·c and add to b, then splitting the middle term and factoring by grouping.

🧩 More than three terms: grouping

• When a polynomial has more than three terms, use the grouping method.
• This involves pairing terms and factoring out common factors from each pair.

🎯 Special product patterns

🎯 Perfect square trinomials

The excerpt provides a three-step process:

Step 1: Pattern recognition

• Check if the trinomial fits a² + 2ab + b² or a² - 2ab + b².
• Verify the first term is a perfect square: write it as (a)².
• Verify the last term is a perfect square: write it as (b)².
• Check the middle term: is it exactly 2·a·b?

Step 2: Write as a binomial square

• If the pattern matches, write as (a + b)² or (a - b)².

Step 3: Verify

• Multiply the binomial square to confirm it produces the original trinomial.

🎯 Difference of squares

Pattern: a² - b² factors as (a - b)(a + b), the product of conjugates.

Recognition steps:

• Confirm it is a difference (subtraction), not a sum.
• Verify both terms are perfect squares.
• Write each term as a square: (a)² - (b)².
• Apply the conjugate pattern.
• Check by multiplying.

🎯 Sum and difference of cubes

The excerpt mentions specific patterns for cubes but does not detail them in the strategy section. The key points are:

• Determine whether the binomial is a sum or difference.
• Verify both terms are perfect cubes.
• Write them as cubes.
• Apply the appropriate sum or difference of cubes pattern.
• Simplify inside the parentheses.
• Check by multiplying.

✅ Verification and completeness

✅ Always check by multiplying

• After factoring, multiply the factors back together.
• The product must equal the original polynomial.
• This step catches errors in pattern recognition or arithmetic.

✅ Is it factored completely?

• The final question in the strategy: "Is it factored completely?"
• Ensure no factor can be broken down further.
• Example: if one factor is itself a difference of squares or a trinomial that can be factored, continue factoring until all factors are prime (cannot be factored further).

📋 Complete decision flowchart

The excerpt presents a comprehensive decision tree:

1. GCF check: factor out any greatest common factor first.
2. Count terms: determine if the polynomial is a binomial, trinomial, or has more terms.
3. Apply the appropriate method:
   • Binomial → check sum/difference and squares/cubes patterns
   • Trinomial → check form (x² + bx + c vs ax² + bx + c) and apply corresponding method
   • More than three terms → use grouping
4. Verify: multiply factors to confirm they produce the original polynomial.
5. Completeness check: ensure all factors are fully factored.

This systematic approach ensures no factoring opportunity is missed and reduces errors through structured decision-making.

🧭 Overview

🧠 One-sentence thesis

Quadratic equations can be solved by factoring and applying the Zero Product Property, which states that if a product equals zero then at least one factor must be zero, and this method applies to many real-world problems involving products and areas.

📌 Key points (3–5)

• Zero Product Property: if a times b equals zero, then either a equals zero or b equals zero (or both).
• Solving by factoring: rewrite the quadratic in standard form, factor it completely, set each factor equal to zero, and solve the resulting linear equations.
• Standard form requirement: the equation must be written as a x squared plus b x plus c equals zero before factoring.
• Common confusion: don't forget to check your solutions by substituting back into the original equation.
• Real applications: quadratic equations model problems about consecutive numbers, rectangular areas, and other product relationships.

🔢 The Zero Product Property

🔢 What it says

Zero Product Property: If a · b = 0, then either a = 0 or b = 0 or both.

• This is the foundation for solving quadratic equations by factoring.
• It works because zero is the only number that, when multiplied by something, gives zero.
• Example: if (x minus 3) times (x plus 7) equals zero, then either x minus 3 equals zero or x plus 7 equals zero.

🎯 Why it matters

• Once you factor a quadratic expression into a product of simpler expressions, the Zero Product Property lets you break one equation into multiple simpler equations.
• Each simpler equation is linear (first-degree), which is much easier to solve.

🛠️ Solving quadratic equations by factoring

📝 Step-by-step method

The excerpt gives a five-step process:

1. Write in standard form: rewrite the equation as a x squared plus b x plus c equals zero.
2. Factor the quadratic expression: use any appropriate factoring method (GCF, trinomial patterns, special products, grouping, etc.).
3. Use the Zero Product Property: set each factor equal to zero.
4. Solve the linear equations: solve each resulting equation for the variable.
5. Check: substitute your solutions back into the original equation to verify.

⚠️ Standard form is required

• The equation must equal zero before you can apply the Zero Product Property.
• Don't confuse: if the equation is written as x squared equals x plus 132, you must first rewrite it as x squared minus x minus 132 equals zero.
• Example: to solve 2 p squared minus 11 p equals 40, rewrite as 2 p squared minus 11 p minus 40 equals zero, then factor.

🔍 Checking your work

• Always substitute your solutions back into the original equation.
• This catches arithmetic errors and ensures the solutions make sense in context.

📐 Applications modeled by quadratic equations

📐 Word problem strategy

The excerpt provides a six-step problem-solving framework:

1. Read the problem carefully; make sure all words and ideas are understood.
2. Identify what you are looking for.
3. Name what you are looking for; choose a variable to represent that quantity.
4. Translate into an equation (restate the problem in one sentence with all important information, then translate to algebra).
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem context and make sure it makes sense.
7. Answer the question with a complete sentence.

🧮 Typical scenarios

The excerpt mentions two common application types:

Problem type	What it involves	Example from excerpt
Consecutive numbers	Product of two numbers that differ by 1	The product of two consecutive numbers is 462; find the numbers
Rectangular area	Length and width relationship with a given area	A rectangular patio has area 400 square feet; the length is 9 feet more than the width; find length and width

💡 Why these lead to quadratics

• Consecutive numbers: if one number is n, the next is n plus 1, so their product is n times (n plus 1), which expands to a quadratic.
• Rectangular area: area equals length times width; if width is w and length is w plus 9, then area equals w times (w plus 9), again a quadratic.
• Don't confuse: the problem may not look like a quadratic at first, but translating the words into algebra reveals the quadratic structure.

🔎 Practice and review

🔎 Types of exercises

The excerpt lists several exercise categories:

• Using the Zero Product Property directly: solve equations already factored, like (a minus 3) times (a plus 7) equals zero.
• Solving by factoring: equations in various forms (standard trinomials, equations with GCF, perfect squares, differences of squares).
• Applications: word problems about consecutive integers and rectangular dimensions.

🧪 Example problems from the excerpt

• Solve x squared plus 9 x plus 20 equals zero (standard trinomial).
• Solve y squared minus y minus 72 equals zero (trinomial with negative middle term).
• Solve 2 p squared minus 11 p equals 40 (not in standard form initially).
• Solve 144 m squared minus 25 equals zero (difference of squares).
• The product of two consecutive integers is 156; find the integers.
• A rectangular place mat has area 168 square inches; its length is two inches longer than the width; find length and width.

✅ Key reminders

• Always check whether the equation is in standard form before factoring.
• Factor completely (look for GCF first, then apply other methods).
• Set each factor equal to zero separately.
• Verify solutions by substitution.
• In word problems, make sure your answer makes sense in the real-world context (e.g., dimensions must be positive).

🧭 Overview

🧠 One-sentence thesis

Rational expressions—fractions with polynomial numerators and denominators—are simplified by factoring completely and removing common factors, while always excluding values that make the denominator zero.

📌 Key points (3–5)

• What a rational expression is: a fraction of the form p(x)/q(x) where p and q are polynomials and q ≠ 0.
• When expressions are undefined: any value that makes the denominator zero must be excluded.
• How to simplify: factor the numerator and denominator completely, then divide out common factors (not terms).
• Common confusion: factors vs. terms—you can only remove common factors (things multiplied together), not common terms (things added together).
• Opposite factors: expressions like (a − b)/(b − a) simplify to −1 because the numerator and denominator are opposites.

🔍 What rational expressions are and when they're undefined

🔍 Definition of a rational expression

Rational expression: an expression of the form p(x)/q(x), where p and q are polynomials and q ≠ 0.

• This is the polynomial version of a numerical fraction.
• Examples range from simple constants like −13/42 to complex expressions like (4x² + 3x − 1)/(2x − 8).
• The key restriction: the denominator can never be zero.

⚠️ Finding excluded values

How to determine when a rational expression is undefined:

1. Set the denominator equal to zero.
2. Solve the equation for the variable.

• Example: For 9y/x, set x = 0, so the expression is undefined when x = 0.
• Example: For (4b − 3)/(2b + 5), set 2b + 5 = 0, solve to get b = −5/2.
• Example: For (x + 4)/(x² + 5x + 6), factor the denominator as (x + 2)(x + 3), so the expression is undefined when x = −2 or x = −3.

Why this matters: These excluded values are like "void where prohibited"—they must be kept in mind throughout all work with the expression.

🧮 Evaluating rational expressions

🧮 Substitution and simplification

To evaluate a rational expression:

• Substitute the given values for the variables.
• Simplify the resulting numerical fraction.

Example: Evaluate (2x + 3)/(3x − 5) when x = 0:

• Substitute: (2·0 + 3)/(3·0 − 5) = 3/(−5) = −3/5.

🚫 Recognizing undefined cases

When evaluating, you may encounter a value that makes the denominator zero—the expression is undefined for that input.

Example: Evaluate (x² + 8x + 7)/(x² − 4) when x = 2:

• Substitute: (4 + 16 + 7)/(4 − 4) = 27/0.
• Result: undefined (division by zero).

Don't confuse: A zero numerator (which gives 0) vs. a zero denominator (which makes the expression undefined).

✂️ Simplifying rational expressions

✂️ The simplification process

Simplified rational expression: one with no common factors (other than 1) in the numerator and denominator.

Steps to simplify:

1. Factor the numerator and denominator completely.
2. Divide out (remove) common factors using the Equivalent Fractions Property.

Equivalent Fractions Property: If a, b, and c are numbers where b ≠ 0 and c ≠ 0, then a/b = (a·c)/(b·c) and (a·c)/(b·c) = a/b.

🔢 Numerical example

Simplify −36/63:

• Rewrite showing common factors: (−4·9)/(7·9).
• Remove the common factor 9: −4/7.

🔤 Algebraic examples

Example: Simplify 3xy/(18x²y²):

• Rewrite: (3·x·y)/(3·x·6·x·y·y).
• Remove common factors 3, x, and y: 1/(6xy).

Example: Simplify (2x + 8)/(5x + 20):

• Factor numerator and denominator: 2(x + 4)/[5(x + 4)].
• Remove common factor (x + 4): 2/5.

⚠️ Critical warning: factors vs. terms

Can do	Cannot do
Remove common factors (multiplied)	Remove common terms (added)
Example: (2·x)/(3·x) = 2/3	Wrong: (x + 5)/x ≠ 5

• Removing x's from (x + 5)/x would be like canceling 2's in (2 + 5)/2—incorrect!
• You can only remove factors from a product, not terms from a sum.

📐 Factoring techniques

The excerpt shows various factoring methods:

• Simple trinomials: x² + 5x + 6 = (x + 2)(x + 3).
• Grouping: p³ − 2p² + 2p − 4 = p²(p − 2) + 2(p − 2) = (p² + 2)(p − 2).
• GCF first: 2n² − 14n = 2n(n − 7).
• Difference of squares: y² − 36 = (y + 6)(y − 6).
• Sum/difference of cubes: m³ + 8 = (m + 2)(m² − 2m + 4).

Example: Simplify (x² + 5x + 6)/(x² + 8x + 12):

• Factor: [(x + 2)(x + 3)]/[(x + 2)(x + 6)].
• Remove common factor (x + 2): (x + 3)/(x + 6).

Example: Simplify (3b² − 12b + 12)/(6b² − 24):

• Factor out GCF: [3(b² − 4b + 4)]/[6(b² − 4)].
• Factor further: [3(b − 2)(b − 2)]/[6(b + 2)(b − 2)].
• Remove common factors 3 and (b − 2): (b − 2)/[2(b + 2)].

🔄 Opposite factors

🔄 Recognizing opposites

Opposites in rational expressions: The opposite of (a − b) is (b − a), and (a − b)/(b − a) = −1 (when a ≠ b).

• An expression and its opposite divide to −1.
• Example: (x − 8)/(8 − x) = −1 immediately.

🔄 Simplifying with opposites

Example: Simplify (14 − 2x)/(x² − 49):

• Factor: [2(7 − x)]/[(x + 7)(x − 7)].
• Recognize (7 − x) and (x − 7) are opposites: (7 − x)/(x − 7) = −1.
• Result: −2/(x + 7).

Example: Simplify (x² − 4x − 32)/(64 − x²):

• Factor: [(x − 8)(x + 4)]/[(8 + x)(8 − x)].
• Recognize (x − 8) and (8 − x) are opposites; also (x + 4) and (8 + x) are the same as (x + 4) and (x + 8) [note: the excerpt shows (8 + x) which equals (x + 8)].
• The common factor (x + 4) cancels; (x − 8)/(8 − x) = −1.
• Result: −(x + 4)/(x + 8) or −(x + 4)/(8 + x).

Don't confuse: Opposite factors (which give −1) with unrelated factors (which don't cancel).

🧭 Overview

🧠 One-sentence thesis

Multiplying and dividing rational expressions follows the same rules as numerical fractions—multiply numerators and denominators together, rewrite division as multiplication by the reciprocal, then factor and simplify by removing common factors.

📌 Key points (3–5)

• Multiplication rule: Multiply numerators together and denominators together, then simplify by factoring and canceling common factors.
• Division rule: Rewrite division as multiplication by the reciprocal (flip the second fraction), then follow multiplication steps.
• Factoring is essential: Always factor numerators and denominators completely before simplifying to identify all common factors.
• Common confusion: Division vs. multiplication—remember to flip the second fraction before factoring and multiplying; don't try to divide directly.
• Opposite factors: Expressions like (4 − x) and (x − 4) are opposites; factoring out −1 reveals the common factor.

✖️ Multiplying rational expressions

✖️ The multiplication rule

Multiplication of Rational Expressions: If p, q, r, s are polynomials where q ≠ 0 and s ≠ 0, then (p/q) · (r/s) = (pr)/(qs).

• Multiply the numerators together to get the new numerator.
• Multiply the denominators together to get the new denominator.
• This is exactly the same process as multiplying numerical fractions.

🔧 Step-by-step procedure

The excerpt provides a three-step method:

1. Factor each numerator and denominator completely.
2. Multiply the numerators and denominators.
3. Simplify by dividing out common factors.

Why factor first? Factoring reveals common factors that can be canceled, making the final answer simpler and avoiding large polynomial multiplication.

📝 Example walkthrough

Example: Multiply (2x)/(x² − 7x − 12) · (x² − 9)/(6x²)

• Factor numerators and denominators: 2x / [(x − 3)(x + 4)] · [(x − 3)(x + 3)] / [6x²]
• Multiply: [2x(x − 3)(x + 3)] / [6x²(x − 3)(x + 4)]
• Cancel common factors: x and (x − 3) appear in both numerator and denominator.
• Simplified result: (x + 3) / [3x(x + 4)]

Don't confuse: Multiplying before factoring leads to messier expressions; always factor first to spot cancellations early.

⚠️ Handling opposite factors

Example from the excerpt: (16 − 4x)/(2x − 12) · (x² − 5x − 6)/(x² − 16)

• Factor: [4(4 − x)] / [2(x − 6)] · [(x − 6)(x + 1)] / [(x − 4)(x + 4)]
• Notice (4 − x) and (x − 4) are opposites: 4 − x = −1(x − 4)
• Factor out −1 to reveal the common factor, then cancel.
• Final result includes the −1 factor: −2(x + 1)/(x + 4)

Key insight: When you see terms that differ only in sign order, factor out −1 to expose the common factor.

➗ Dividing rational expressions

➗ The division rule

Division of Rational Expressions: If p, q, r, s are polynomials where q ≠ 0, r ≠ 0, s ≠ 0, then (p/q) ÷ (r/s) = (p/q) · (s/r).

• To divide, multiply the first fraction by the reciprocal of the second.
• The reciprocal of a/b is b/a—flip the numerator and denominator.
• After rewriting as multiplication, follow the multiplication steps.

🔄 Four-step procedure

1. Rewrite the division as multiplication by the reciprocal of the second expression.
2. Factor the numerators and denominators completely.
3. Multiply the numerators and denominators together.
4. Simplify by dividing out common factors.

📝 Example walkthrough

Example: Divide (3n²)/(n² − 4n) ÷ (9n² − 45n)/(n² − 7n + 10)

• Rewrite: (3n²)/(n² − 4n) · (n² − 7n + 10)/(9n² − 45n)
• Factor: [3n²] / [n(n − 4)] · [(n − 2)(n − 5)] / [9n(n − 5)]
• Multiply and cancel common factors: n and (n − 5)
• Simplified result: (n − 2) / [3(n − 4)]

Don't confuse: Forgetting to flip the second fraction is a common error—always rewrite division as multiplication by the reciprocal first.

🔢 Dividing by a whole polynomial

When dividing by a polynomial (not a fraction), first write it as a fraction over 1.

Example: (a² − b²)/(3ab) ÷ (a² + 2ab + b²)

• Write the polynomial as a fraction: (a² − b²)/(3ab) ÷ (a² + 2ab + b²)/1
• Flip to get the reciprocal: (a² − b²)/(3ab) · 1/(a² + 2ab + b²)
• Factor and simplify: [(a − b)(a + b)] / [3ab(a + b)(a + b)] = (a − b) / [3ab(a + b)]

🧮 Special cases and complex fractions

🧮 Complex fractions

A complex fraction is a fraction where the numerator or denominator (or both) contains a fraction—it represents division.

Example: [6x² − 7x + 2] / [4x − 8] over [2x² − 7x + 3] / [x² − 5x + 6]

• Rewrite the fraction bar as a division sign: top ÷ bottom
• Apply the division rule: multiply by the reciprocal of the bottom
• Factor, multiply, and simplify as usual

🔗 Multiple operations

When an expression contains both multiplication and division:

• First, rewrite all divisions as multiplication by reciprocals.
• Then, factor everything and multiply.
• Finally, cancel common factors.

Example: (3x − 6)/(4x − 4) · (x² + 2x − 3)/(x² − 3x − 10) ÷ (2x + 12)/(8x + 16)

• Rewrite division: (3x − 6)/(4x − 4) · (x² + 2x − 3)/(x² − 3x − 10) · (8x + 16)/(2x + 12)
• Factor all numerators and denominators
• Multiply and cancel common factors across all three fractions

Key principle: Handle all operations in order, converting division to multiplication first, then treat the entire expression as one big multiplication problem.

⚠️ Restrictions on variables

The excerpt notes that throughout the chapter, values that make any denominator zero are excluded (though not written explicitly each time).

• Always remember: denominators cannot equal zero.
• Example: In (2x)/(3y²), both x ≠ 0 and y ≠ 0 are implied restrictions.
• These restrictions come from the original expression, not just the simplified form.

🧭 Overview

🧠 One-sentence thesis

Adding and subtracting rational expressions follows the same rules as numerical fractions: combine numerators over the common denominator, then simplify by factoring and removing common factors.

📌 Key points (3–5)

• Common denominator requirement: Rational expressions must share the same denominator before you can add or subtract them.
• Addition rule: Add numerators and place the sum over the common denominator; subtraction works the same way but subtract numerators.
• Always simplify: After combining, factor the numerator and denominator to identify and remove common factors.
• Common confusion—opposite denominators: When denominators are opposites (e.g., x - 3 and 3 - x), multiply one fraction by -1/-1 to make them identical, not a new common denominator.
• Sign distribution matters: When subtracting a binomial or polynomial, distribute the negative sign to every term in the numerator.

➕ Adding rational expressions

➕ The addition rule

Rational Expression Addition: If p, q, and r are polynomials where r ≠ 0, then p/r + q/r = (p + q)/r

• To add rational expressions with a common denominator, add the numerators and place the sum over the common denominator.
• This mirrors how numerical fractions work: 5/18 + 7/18 = (5 + 7)/18 = 12/18 = 2/3.

🔧 Steps for adding

1. Check the denominators are identical.
2. Add the numerators and write the result over the common denominator.
3. Simplify: Factor both numerator and denominator, then remove common factors.

Example: To add (3y)/(4y - 3) + 7/(4y - 3), the denominators match, so combine numerators: (3y + 7)/(4y - 3). This cannot be factored further, so it is simplified.

📐 When the result factors nicely

• After adding numerators, you may get a factorable polynomial.
• Example: Adding (7x + 12)/(x + 3) + x²/(x + 3) gives (x² + 7x + 12)/(x + 3).
• Factor the numerator: (x + 3)(x + 4)/(x + 3).
• Remove the common factor (x + 3) to get x + 4.

➖ Subtracting rational expressions

➖ The subtraction rule

Rational Expression Subtraction: If p, q, and r are polynomials where r ≠ 0, then p/r - q/r = (p - q)/r

• Subtract the numerators and place the difference over the common denominator.
• Always factor after subtracting to identify common factors.

⚠️ Watch the signs

• Distribute the negative sign to every term in the numerator you are subtracting.
• Example: Subtracting y²/(y - 6) - (2y + 24)/(y - 6) becomes [y² - (2y + 24)]/(y - 6).
• Distribute: (y² - 2y - 24)/(y - 6).
• Factor: (y - 6)(y + 4)/(y - 6) = y + 4.

🧮 Subtracting polynomials with multiple terms

• When subtracting a trinomial or larger polynomial, use parentheses and distribute the negative.
• Example: [5x² - 7x + 3]/(x² - 3x - 18) - [4x² + x - 9]/(x² - 3x - 18).
• Subtract: [5x² - 7x + 3 - 4x² - x + 9]/(x² - 3x - 18).
• Combine like terms: (x² - 8x + 12)/(x² - 3x - 18).
• Factor both: (x - 2)(x - 6)/[(x + 3)(x - 6)] = (x - 2)/(x + 3).

🔄 When denominators are opposites

🔄 Recognizing opposites

• Denominators like (3u - 1) and (1 - 3u) are opposites, not different denominators.
• Also: (m² - 1) and (1 - m²) are opposites.
• Key insight: Multiply one fraction by -1/-1 to flip the denominator and make them identical.

🔧 The technique

• Multiply the fraction with the "flipped" denominator by -1/-1.
• This changes both the numerator and denominator signs.
• Example: To add (4u - 1)/(3u - 1) + u/(1 - 3u), multiply the second fraction by -1/-1.
• Result: (4u - 1)/(3u - 1) + (-u)/(3u - 1) = (4u - 1 - u)/(3u - 1) = (3u - 1)/(3u - 1) = 1.

📝 Subtraction with opposites

• The same technique applies when subtracting.
• Example: [m² - 6m]/(m² - 1) - [3m + 2]/(1 - m²).
• Multiply the second by -1/-1: [m² - 6m]/(m² - 1) - [-(3m + 2)]/(m² - 1).
• Simplify: [m² - 6m + 3m + 2]/(m² - 1) = (m² - 3m + 2)/(m² - 1).
• Factor: (m - 1)(m - 2)/[(m - 1)(m + 1)] = (m - 2)/(m + 1).

Don't confuse: Opposite denominators are not "different" denominators requiring a least common denominator—just multiply by -1/-1 to make them match.

🎯 Simplification reminders

🎯 Always factor and simplify

• After every addition or subtraction, factor the numerator and denominator.
• Remove any common factors.
• This is the final step and cannot be skipped.

🚫 Excluded values

• Remember that values making the denominator zero are excluded.
• Example: In expressions with denominator (4y - 3), y cannot equal 3/4.
• These restrictions come from the original denominators, not the simplified form.

🧭 Overview

🧠 One-sentence thesis

Adding and subtracting rational expressions with different denominators requires finding a least common denominator (LCD), rewriting each expression as an equivalent fraction with that LCD, and then combining the numerators while keeping the common denominator.

📌 Key points (3–5)

• Finding the LCD: Factor each denominator completely, line up common factors, bring down one from each column, and leave the LCD in factored form.
• Rewriting as equivalent expressions: Multiply each numerator and denominator by the "missing" factors needed to reach the LCD.
• Adding vs subtracting: The process is identical except you must be extra careful with signs when subtracting numerators.
• Common confusion: Don't simplify too early—you must keep expressions in their LCD form until after combining numerators.
• When denominators are opposites: Multiply one expression by negative one over negative one to create matching denominators.

🔍 Finding the least common denominator

🔍 The LCD procedure for rational expressions

The excerpt extends the numerical fraction LCD method to algebraic expressions:

Least Common Denominator (LCD): the least common multiple of the denominators; the smallest expression we can use as a common denominator.

How to find it:

1. Factor each denominator completely
2. List factors of each expression
3. Match factors vertically when possible (line up common factors in columns)
4. Bring down one factor from each column
5. Multiply the factors together
6. Key difference from numbers: leave the LCD in factored form (don't multiply out)

📐 Example walkthrough

For the denominators x² − 2x − 3 and x² + 4x + 3:

• First factors to (x + 1)(x − 3)
• Second factors to (x + 1)(x + 3)
• Common factor: (x + 1) appears in both
• LCD: (x + 1)(x − 3)(x + 3)

Don't confuse: The LCD includes each unique factor; common factors appear only once, but unique factors from each denominator must all be included.

🔄 Creating equivalent rational expressions

🔄 The "missing factor" method

Once you have the LCD, rewrite each fraction:

1. Identify what factor(s) each denominator is "missing" to become the LCD
2. Multiply both numerator and denominator by those missing factors
3. Simplify the numerators (expand/distribute)
4. Do NOT simplify the denominators—keep them in factored LCD form

⚠️ Why not to simplify too soon

The excerpt emphasizes: "Avoid the temptation to simplify too soon!"

Example: When adding 2a/(2ab + b²) + 3a/(4a² − b²):

• After rewriting with LCD, you get expressions like [2a(2a − b)]/[b(2a + b)(2a − b)]
• You must keep this form to add it to the other fraction
• Only simplify after combining numerators

Don't confuse: Simplifying individual fractions before finding the LCD will prevent you from combining them correctly.

➕ Adding rational expressions

➕ The addition procedure

1. Check if denominators are already common
   • Yes → skip to step 3
   • No → continue to step 2
2. Rewrite with LCD:
   • Find the LCD
   • Rewrite each expression as equivalent fraction with LCD
3. Add the numerators (keep the common denominator)
4. Simplify if possible (factor numerator, look for common factors with denominator)

🔢 From numbers to variables

The excerpt shows the parallel:

• Numerical: 7/12 + 5/18 requires LCD = 36
• Monomials: 5/(12x²y) + 4/(21xy²) requires LCD = 84x²y²
• Polynomials: 3/(x−3) + 2/(x−2) requires LCD = (x−3)(x−2)

📝 Key steps after combining

After adding numerators:

1. Simplify the numerator expression (combine like terms)
2. Factor the numerator completely
3. Look for common factors between numerator and denominator
4. Cancel common factors if they exist
5. State the simplified result

Example: After adding, if numerator becomes 3x² + 11x + 8 and factors to (x + 1)(3x + 8), check if either factor appears in the denominator.

➖ Subtracting rational expressions

➖ The subtraction procedure

Identical to addition except for step 3:

1. Check for common denominators
2. Rewrite with LCD if needed
3. Subtract the numerators (keep common denominator)
4. Simplify if possible

⚠️ Sign management is critical

The excerpt warns: "Be extra careful of the signs when subtracting the numerators."

When subtracting, you must distribute the negative sign across the entire second numerator:

• Write as one fraction: [first numerator − (second numerator)] / LCD
• Expand carefully: if second numerator is (x − 2), subtracting gives −x + 2, not −x − 2

🔄 Handling opposite denominators

When denominators are opposites (like n − 2 and 2 − n):

1. Recognize that (n − 2) = −(2 − n)
2. Multiply the fraction with (2 − n) by (−1)/(−1)
3. This flips the denominator to match: (2 − n) becomes (n − 2)
4. The numerator also changes sign
5. Now denominators match and you can proceed

Example: For expressions with denominators (n² + n − 6) and (2 − n), factor the first and recognize the opposite relationship, then multiply by −1/−1 to align.

🎯 Special cases and multiple expressions

🎯 Non-fraction terms

When one term is not a fraction (like subtracting 3 from a rational expression):

1. Write the whole number as a fraction with denominator 1
2. The LCD becomes the denominator of the rational expression
3. Rewrite the whole number with that LCD
4. Proceed with subtraction

Example: 5c + 4/(c − 2) − 3 becomes [5c + 4/(c − 2)] − 3/1, then rewrite 3/1 as 3(c − 2)/(c − 2).

🎯 Three or more expressions

For expressions like 2u/(u−1) + 1/u − (2u−1)/(u²−u):

1. Factor all denominators
2. Find LCD from all three (or more) denominators
3. Rewrite each expression with the LCD
4. Combine all numerators in order (respecting addition and subtraction)
5. Simplify the combined numerator
6. Factor and reduce if possible

Don't confuse: With multiple terms, track signs carefully—each subtraction applies only to the term immediately following it.

🧭 Overview

🧠 One-sentence thesis

Complex rational expressions—fractions containing other fractions in their numerator or denominator—can be simplified either by rewriting them as division problems or by multiplying both parts by the least common denominator (LCD) of all internal fractions.

📌 Key points (3–5)

• What a complex rational expression is: a rational expression that has rational expressions (fractions) in its numerator or denominator.
• Two methods to simplify: Method 1 rewrites the complex fraction as a division problem; Method 2 multiplies numerator and denominator by the LCD of all internal fractions.
• Both methods yield the same result: choose the method that feels more natural for the given problem.
• Common confusion: forgetting to exclude values that make any denominator zero in the original expression, even if those denominators disappear after simplification.
• Order of operations matters: simplify the numerator and denominator separately before performing the division.

🧩 What is a complex rational expression?

🧩 Definition and structure

Complex Rational Expression: A rational expression in which the numerator or denominator contains a rational expression.

• Not just a simple fraction like 3/4 or x/2.
• Contains "fractions within fractions."
• Examples from the excerpt:
  • (4)/(y - 3) divided by (8)/(y² - 9)
  • (1/x + 1/y) divided by (x/y - y/x)
  • (2)/(x + 6) divided by [4/(x - 6) - 4/(x² - 36)]

⚠️ Excluded values

• Always exclude values that make any denominator in the original expression equal to zero.
• Even after simplification, remember the original restrictions.
• Example: if the original expression had (y - 3) in a denominator, y = 3 must be excluded even if (y - 3) cancels out during simplification.

🔀 Method 1: Rewrite as division

🔀 The core idea

• Fraction bars mean division.
• A complex fraction like (A/B)/(C/D) is the same as (A/B) ÷ (C/D).
• Once rewritten as division, multiply the first expression by the reciprocal of the second.

📝 Step-by-step process

Step 1: Simplify the numerator and denominator separately (if needed).

• Find the LCD and combine fractions in the numerator.
• Find the LCD and combine fractions in the denominator.

Step 2: Rewrite the complex rational expression as a division problem.

• The numerator divided by the denominator.

Step 3: Divide the expressions.

• Multiply the first by the reciprocal of the second.
• Factor all expressions.
• Remove common factors.
• Simplify.

💡 Example walkthrough

For (4)/(y - 3) divided by (8)/(y² - 9):

• Rewrite as: [4/(y - 3)] ÷ [8/(y² - 9)]
• Multiply by reciprocal: [4/(y - 3)] · [(y² - 9)/8]
• Result: 4(y² - 9) / [8(y - 3)]
• Factor: 4(y - 3)(y + 3) / [4 · 2(y - 3)]
• Cancel common factors: (y + 3)/2
• Don't forget: y ≠ 3 and y ≠ -3 from the original denominators.

🔍 When to use this method

• When the numerator and denominator are already single fractions.
• When you're comfortable with the "invert and multiply" rule for division.

🧮 Method 2: Multiply by the LCD

🧮 The core idea

• Find the LCD of all fractions appearing anywhere in the complex expression.
• Multiply both the numerator and denominator by this LCD.
• This "clears" all the internal fractions at once.
• Multiplying by LCD/LCD is the same as multiplying by 1, so the value doesn't change.

📝 Step-by-step process

Step 1: Find the LCD of all fractions in the complex rational expression.

• Look at every denominator in both the numerator and denominator of the complex fraction.
• Factor all denominators first to identify the LCD correctly.

Step 2: Multiply the numerator and denominator by the LCD.

• Distribute the LCD to every term.

Step 3: Simplify the expression.

• Cancel out denominators.
• Combine like terms.
• Factor and reduce if possible.

💡 Example walkthrough

For (1/3 + 1/6) divided by (1/2 - 1/3):

• LCD of all fractions (3, 6, 2, 3) is 6.
• Multiply top and bottom by 6:
  • Numerator: 6(1/3 + 1/6) = 6/3 + 6/6 = 2 + 1 = 3
  • Denominator: 6(1/2 - 1/3) = 6/2 - 6/3 = 3 - 2 = 1
• Result: 3/1 = 3

🔍 When to use this method

• When there are multiple fractions in the numerator or denominator.
• When you want to avoid intermediate division steps.
• When the LCD is easy to identify.

⚖️ Comparing the two methods

Aspect	Method 1: Division	Method 2: LCD
First step	Simplify numerator and denominator separately	Find LCD of all fractions
Key operation	Rewrite as division, then multiply by reciprocal	Multiply both parts by LCD
Best for	Simple cases with single fractions top and bottom	Multiple fractions; complex denominators
Result	Same final answer	Same final answer

🤔 Don't confuse

• Both methods are valid; neither is "more correct."
• The excerpt shows the same problem solved both ways to demonstrate equivalence.
• Choose based on which feels clearer for the specific problem.

🛠️ Common techniques and tips

🛠️ Factoring denominators first

• Always factor denominators before finding the LCD.
• Example: x² - 36 factors to (x + 6)(x - 6).
• The LCD must include all unique factors.

🛠️ Distributing carefully

• When multiplying by the LCD, distribute to every term in the numerator and denominator.
• Don't forget terms that are just constants or single variables.

🛠️ Simplifying step by step

• Combine like terms after distributing.
• Factor the result to look for common factors between numerator and denominator.
• Remove common factors to reach the simplest form.

⚠️ Remembering restrictions

• The simplified expression may look simpler, but the original restrictions still apply.
• If the original had (x - 4) in a denominator, x ≠ 4 even if (x - 4) cancels out.

🧭 Overview

🧠 One-sentence thesis

Rational equations can be solved by multiplying both sides by the least common denominator to clear fractions, but any solution that makes a denominator zero must be discarded as extraneous.

📌 Key points (3–5)

• What a rational equation is: two rational expressions connected by an equal sign (unlike rational expressions, which have no equal sign).
• Core solving strategy: multiply both sides by the LCD to eliminate fractions, then solve the resulting simpler equation.
• Extraneous solutions: algebraic solutions that would make any denominator zero must be identified at the start and discarded at the end.
• Common confusion: rational expressions vs rational equations—expressions are simplified; equations are solved and contain an equal sign.
• Solving for a specific variable: the same LCD-clearing method applies when isolating one variable in a formula.

🔍 Distinguishing expressions from equations

🔍 Rational expression vs rational equation

Type	Structure	What you do
Rational expression	No equal sign, e.g. 1/(8x) + 1/2	Simplify
Rational equation	Has an equal sign, e.g. 1/(8x) + 1/2 = 1/4	Solve

• The excerpt emphasizes: "You must make sure to know the difference."
• Example: (y + 6)/(y² - 36) is an expression; (y + 6)/(y² - 36) = (y + 1) is an equation.

📌 Definition of rational equation

Rational equation: two rational expressions connected by an equal sign.

🧩 Core solving method

🧩 The LCD-clearing strategy

• The excerpt recalls solving linear equations with fractions: multiply both sides by the LCD to "clear" the fractions.
• The same strategy applies to rational equations.
• After clearing fractions, you get an equation without rational expressions, which is easier to solve.

⚠️ Why we must check for extraneous solutions

• Because the original equation may have a variable in a denominator, a solution might make a denominator zero.
• Before solving, identify values that would make any denominator zero and note them (e.g., x ≠ 0).
• After solving, discard any algebraic solution that equals one of those forbidden values.

📖 Definition of extraneous solution

Extraneous solution to a rational equation: an algebraic solution that would cause any of the expressions in the original equation to be undefined.

🛠️ Step-by-step procedure

🛠️ How to solve equations with rational expressions

1. Note any value of the variable that would make any denominator zero.
2. Find the LCD of all denominators in the equation.
3. Clear the fractions by multiplying both sides by the LCD.
4. Solve the resulting equation (which no longer has fractions).
5. Check:
   • If any values from Step 1 appear as algebraic solutions, discard them.
   • Check any remaining solutions in the original equation.

🔢 Example walkthrough (from Example 8.59)

• Equation: 1/x + 1/3 = 5/6
• Step 1: Note x ≠ 0.
• Step 2: LCD is 6x.
• Step 3: Multiply both sides by 6x: 6x(1/x) + 6x(1/3) = 6x(5/6) → 6 + 2x = 5x.
• Step 4: Solve: 6 = 3x → x = 2.
• Step 5: Check: x = 2 does not make any denominator zero, so it is valid.

🚫 Handling extraneous solutions

🚫 When all solutions are extraneous

• Some equations have only extraneous solutions, leaving no solution to the equation.
• Example (from Example 8.64): after solving, the only algebraic solution is m = 4, but m = 4 makes a denominator zero, so the equation has no solution.

🚫 Identifying forbidden values early

• Factor all denominators at the start to see clearly which values are forbidden.
• Example: if a denominator is m² - 5m + 4, factor it to (m - 4)(m - 1) and note m ≠ 4, m ≠ 1.

🚫 Example with no solution (from Example 8.65)

• Equation: n/12 + (n + 3)/(3n) = 1/n
• Note n ≠ 0.
• After clearing fractions and solving, the only solution is n = 0.
• But n = 0 is extraneous, so the equation has no solution.

🔧 Solving for a specific variable

🔧 Applying the method to formulas

• Many formulas in science, business, and economics are rational equations.
• To solve for a specific variable, use the same LCD-clearing strategy.

🔧 Example: distance, rate, time (from Example 8.68)

• Formula: D/T = R
• Solve for T:
  • Note T ≠ 0.
  • Multiply both sides by T: D = RT.
  • Divide both sides by R: T = D/R.

🔧 Example: slope formula (from Example 8.69)

• Formula: m = (x - 2)/(y - 3)
• Solve for y:
  • Note y ≠ 3.
  • Multiply both sides by (y - 3): m(y - 3) = x - 2.
  • Expand: my - 3m = x - 2.
  • Isolate y: my = x - 2 + 3m.
  • Divide by m: y = (x - 2 + 3m)/m.

🔧 Example with multiple fractions (from Example 8.70)

• Equation: 1/c + 1/m = 1
• Solve for c:
  • Note c ≠ 0, m ≠ 0.
  • LCD is cm.
  • Multiply both sides by cm: m + c = cm.
  • Collect terms with c: c - cm = -m.
  • Factor: c(1 - m) = -m.
  • Divide: c = -m/(1 - m) = m/(m - 1).
  • Also note m ≠ 1 (to avoid division by zero in the final answer).

⚠️ Don't confuse: instant solving vs proper steps

• Even a simple-looking formula like 1/c + 1/m = 1 cannot be solved instantly for c.
• You must follow all steps: note forbidden values, clear fractions, isolate the variable, and state any new restrictions.

📝 Summary of key distinctions

📝 Expression vs equation reminder

• Expression: simplify it (no equal sign).
• Equation: solve it (has an equal sign).

📝 Algebraic solution vs valid solution

• Algebraic solution: what you get after solving the equation.
• Valid solution: an algebraic solution that does not make any denominator zero.
• Extraneous solution: an algebraic solution that is not valid because it makes a denominator zero.

📝 When there is no solution

• If all algebraic solutions are extraneous, the equation has no solution.
• Example: the excerpt shows several cases where the only solution found is forbidden, leaving no valid answer.

🧭 Overview

🧠 One-sentence thesis

Proportions provide a powerful tool for solving real-world scaling problems—from medication dosages to currency exchange to finding unmeasurable heights—by setting up equations where two ratios are equal.

📌 Key points (3–5)

• What a proportion is: an equation stating that two rational expressions (fractions) are equal, written as a/b = c/d.
• How to solve proportions: multiply both sides by the LCD (least common denominator) to clear fractions, then solve the resulting equation.
• Units must match correctly: when setting up proportions, numerators must have the same units and denominators must have the same units.
• Common confusion: proportional vs equal—proportional means "in the same ratio," not "the same amount"; also, be careful to match corresponding sides correctly in similar figures.
• Similar figures: two shapes with the same angles and sides in the same ratio; their corresponding sides can be related through proportions.

📐 Understanding proportions

📐 Definition and notation

Proportion: An equation of the form a/b = c/d, where b ≠ 0 and d ≠ 0.

• Read as "a is to b, as c is to d."
• Example: 1/2 = 4/8 is a proportion because both fractions are equal.
• The excerpt emphasizes that proportions are used to "scale up" quantities in many applications.

🔧 Solving method

The excerpt explains that proportions are solved like rational equations:

1. Multiply both sides by the LCD to clear the fractions.
2. Simplify by removing common factors.
3. Solve the resulting equation.
4. Check by substituting back into the original proportion.

Example from the excerpt: To solve x/63 = 4/7, multiply both sides by 63, giving x = (4 × 63)/7 = 36.

⚠️ Excluded values

• Just like all rational expressions, exclude any values that make a denominator zero.
• Example: In 144/a = 9/4, the value a = 0 must be excluded.

🧮 Setting up proportions correctly

🧮 Unit matching is critical

The excerpt repeatedly stresses:

• Numerators must match: if the left numerator is "ml," the right numerator must also be "ml."
• Denominators must match: if the left denominator is "pounds," the right denominator must also be "pounds."

Example structure from the excerpt:

• ml/pounds = ml/pounds
• calories/ounce = calories/ounce
• $/pesos = $/pesos

Don't confuse: mixing units (e.g., ml/pounds = pounds/ml) will give a nonsensical answer.

🏥 Real-world application pattern

The excerpt provides a step-by-step pattern for word problems:

1. Identify what you're looking for and choose a variable.
2. Write a sentence giving the relationship (e.g., "If 5 ml is prescribed for every 25 pounds, how much for 80 pounds?").
3. Translate into a proportion with correct units.
4. Solve by multiplying by the LCD.
5. Check if the answer is reasonable.
6. Answer with a complete sentence.

Example: Pediatricians prescribe 5 ml of acetaminophen for every 25 pounds. For an 80-pound child:

• Set up: 5/25 = a/80
• Solve: a = (5 × 80)/25 = 16 ml

💱 Currency exchange applications

The excerpt shows how proportions work for currency conversion:

• If $1 US = 12.54 Mexican pesos, then $325 = how many pesos?
• Set up: $/pesos = $/pesos → 1/12.54 = 325/p
• Solve: p = 325 × 12.54 = 4075.5 pesos

The excerpt introduces the term proportional: "If two quantities are related by a proportion, we say that they are proportional."

📏 Similar figures

📏 What similar figures are

Similar figures: Two figures with the same shape but different sizes; their corresponding angles are equal and their corresponding sides are in the same ratio.

• One figure is a scale model of the other.
• Example from the excerpt: If triangle ABC is similar to triangle XYZ, and each side of ABC is 4 times the corresponding side of XYZ, then AB/XY = BC/YZ = AC/XZ = 4/3.

🔺 Property of similar triangles

The excerpt states:

If triangle ABC is similar to triangle XYZ, then their corresponding angle measures are equal and their corresponding sides are in the same ratio.

This means you can set up proportions between corresponding sides.

🗺️ Applications of similar triangles

Map distances: The excerpt shows how similar triangles on a map relate to actual distances.

Example: On a map, if the triangle formed by three cities has sides in a certain ratio, and you know one actual distance is 270 miles, you can find other actual distances by setting up proportions with the map measurements.

Shadow problems: The excerpt demonstrates finding heights using shadows.

Example: Tyler is 6 feet tall with an 8-foot shadow. A tree has a 24-foot shadow. How tall is the tree?

• Set up: height/shadow = height/shadow → 6/8 = h/24
• Solve: h = (6 × 24)/8 = 18 feet
• The key: at the same time of day, the ratio of height to shadow is the same for all objects.

Don't confuse: You must match corresponding sides correctly. The excerpt warns to "be careful to match up corresponding sides correctly" when writing AB/XY = BC/YZ = AC/XZ.

🎯 Problem-solving strategy

🎯 Seven-step geometry application method

The excerpt provides a formal "How To" strategy:

Step	Action
1. Read	Understand all words and ideas; draw and label the figure
2. Identify	What are we looking for?
3. Name	Choose a variable to represent it
4. Translate	Write the appropriate formula or proportion; substitute given information
5. Solve	Use algebra techniques
6. Check	Does the answer make sense in the problem?
7. Answer	Write a complete sentence

✅ Reasonableness checks

The excerpt emphasizes checking if answers are reasonable:

• For the medication problem: "Is the answer reasonable? Yes, since 80 is about 3 times 25, the medicine should be about 3 times 5. So 16 ml makes sense."
• For the calorie problem: "Yes, 345 calories for 24 ounces is more than 290 calories for 16 ounces, but not too much more."
• For the shadow problem: "Tyler's height is less than his shadow's length so it makes sense that the tree's height is less than the length of its shadow."

This sanity-check step helps catch setup errors (like inverted fractions or mismatched units).

🧭 Overview

🧠 One-sentence thesis

Rational equations can model real-world situations where objects move at different speeds or people work at different rates, allowing us to find unknown speeds or times by setting up equations based on distance-rate-time or work-rate relationships.

📌 Key points (3–5)

• Uniform motion problems use the formula D = r·t, which can be rearranged to t = D/r when solving for time.
• Wind/current effects: tailwinds add to speed (r + wind), headwinds subtract from speed (r - wind).
• Work problems use the principle that (part of job done per hour by person A) + (part done per hour by person B) = (part done per hour together).
• Common confusion: In motion problems, don't confuse the vehicle's own speed with its effective speed when wind or current is present; in work problems, don't confuse hours-to-complete-job with fraction-of-job-per-hour.
• Key strategy: organize information in a chart, identify what times or rates are equal or related, then translate to a rational equation.

🛫 Uniform motion with wind or current

✈️ The airplane with headwind/tailwind problem

The excerpt shows an airplane flying 200 miles into a 30 mph headwind and 300 miles with a 30 mph tailwind in the same time.

Setup:

• Let r = airplane's speed (no wind).
• With tailwind: effective speed = r + 30.
• Against headwind: effective speed = r - 30.
• Since D = r·t, rearrange to get t = D/r.

Chart structure:

Direction	Rate	Distance	Time
Into headwind	r - 30	200	200/(r - 30)
With tailwind	r + 30	300	300/(r + 30)

Equation:

• The times are equal: 200/(r - 30) = 300/(r + 30).
• Multiply both sides by the LCD (r + 30)(r - 30).
• Simplify: 200(r + 30) = 300(r - 30).
• Solve: 200r + 6000 = 300r - 9000 → 15000 = 100r → r = 150 mph.

Check: At 150 mph, with 30 mph wind, tailwind gives 180 mph and 300/180 = 5/3 hours; headwind gives 120 mph and 200/120 = 5/3 hours. Times match.

🚴 When total time is known

Example: Jazmine trained for 3 hours total. She ran 8 miles then biked 24 miles. Her biking speed is 4 mph faster than running speed.

Setup:

• Let r = running speed.
• Biking speed = r + 4.
• Time running = 8/r.
• Time biking = 24/(r + 4).
• Total time = 3 hours.

Equation:

• 8/r + 24/(r + 4) = 3.
• Multiply by LCD r(r + 4).
• Simplify: 8(r + 4) + 24r = 3r(r + 4).
• Expand: 8r + 32 + 24r = 3r² + 12r.
• Rearrange: 0 = 3r² - 20r - 32.
• Factor: 0 = (3r + 4)(r - 8).
• Solutions: r = -4/3 (reject, negative speed) or r = 8 mph.

Don't confuse: The problem gives total time, not equal times for each leg.

🚲 When time difference is known

Example: Hamilton rode 12 miles downhill to the ocean, then 12 miles uphill home. Uphill speed was 8 mph slower than downhill. Uphill took 2 hours longer.

Setup:

• Let r = downhill speed.
• Uphill speed = r - 8.
• Time downhill = 12/r.
• Time uphill = 12/(r - 8).
• Uphill time is 2 hours more than downhill time.

Equation:

• 12/(r - 8) = 12/r + 2.
• Multiply by LCD r(r - 8).
• Simplify: 12r = 12(r - 8) + 2r(r - 8).
• Expand: 12r = 12r - 96 + 2r² - 16r.
• Rearrange: 0 = 2r² - 16r - 96.
• Factor: 0 = 2(r - 12)(r + 4).
• Solutions: r = 12 mph (accept) or r = -4 (reject).

Check: Downhill at 12 mph takes 1 hour; uphill at 4 mph takes 3 hours. Difference is 2 hours. ✓

🛠️ Work application problems

🖨️ The core work principle

If one person can complete a job in h hours, then in 1 hour they complete 1/h of the job.

Example: Pete paints a room in 10 hours, so in 1 hour he paints 1/10 of the room. Alicia paints the same room in 8 hours, so in 1 hour she paints 1/8 of the room.

Working together:

• Let t = hours to complete the job together.
• In 1 hour together they complete 1/t of the job.
• Equation: 1/10 + 1/8 = 1/t.
• Multiply by LCD 40t: 4t + 5t = 40.
• Solve: 9t = 40 → t = 40/9 ≈ 4.44 hours (about 4 hours 27 minutes).

Key insight: Working together takes less time than either person alone.

📰 Two machines working together

Example: Press #1 takes 6 hours to print a magazine; Press #2 takes 12 hours. How long together?

Chart:

Press	Hours per job	Part completed per hour
#1	6	1/6
#2	12	1/12
Together	t	1/t

Equation:

• 1/6 + 1/12 = 1/t.
• Multiply by LCD 12t: 2t + t = 12.
• Solve: 3t = 12 → t = 4 hours.

⛄ Finding one person's time when given combined time

Example: Corey can shovel snow in 4 hours. Together with Casey they finish in 2 hours. How long would Casey take alone?

Setup:

• Let t = hours for Casey alone.
• Corey's rate: 1/4 per hour.
• Casey's rate: 1/t per hour.
• Together: 1/2 per hour.

Equation:

• 1/4 + 1/t = 1/2.
• Multiply by LCD 4t: t + 4 = 2t.
• Solve: 4 = t → Casey takes 4 hours alone.

Don't confuse: The "together time" (2 hours) is not the same as either individual time; you must use the rate equation, not a simple average.

📋 Problem-solving strategy

📊 Organize with a chart

For motion problems:

Leg/Direction	Rate	Distance	Time (= D/r)
...	...	...	...

For work problems:

Worker	Hours per job	Part per hour (= 1/hours)
...	...	...

🔍 Identify the relationship

• Equal times: set time expressions equal.
• Time sum: add time expressions to equal total.
• Time difference: set one time equal to another plus/minus a constant.
• Work rates: sum of individual rates equals combined rate.

⚙️ Solve the rational equation

1. Write the equation from the relationship.
2. Multiply both sides by the LCD to clear denominators.
3. Simplify and solve (may result in linear or quadratic equation).
4. Check solutions for reasonableness (reject negative speeds/times, verify the original condition).

✅ Always check

• Does the answer make physical sense?
• Plug back into the original word problem (not just the equation) to verify the condition is satisfied.