Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 3: One-Variable Linear Equations

Lesson 3: Word Problems

In this Grade 4 AMC math lesson from AoPS: Introduction to Algebra, students learn how to translate word problems into one-variable linear equations by assigning variables to unknown quantities and building equations from verbal phrases. Using problems drawn from Chapter 3, students practice solving equations through algebraic manipulation and verifying solutions against the original problem conditions. The lesson emphasizes a systematic approach — define the variable, write the equation, solve, and check — as the core strategy for tackling real-world problem contexts.

Section 1

Translate to an equation and solve

Property

To solve word problems, first translate the word sentences into equations with variables, and then solve the equations using the properties of equality. Identify keywords to determine the correct operation.

Examples

  • 'Seven more than nn is equal to 2.' translates to n+7=2n + 7 = 2. Subtracting 7 from both sides gives n=5n = -5.
  • 'The difference of pp and 9 is 15-15.' translates to p9=15p - 9 = -15. Adding 9 to both sides gives p=6p = -6.

Section 2

Use a Problem-Solving Strategy

Property

Use a Problem-Solving Strategy to Solve Word Problems.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Examples

  • A shirt is on sale for 25 dollars, which is one-third of its original price. What was the original price? Let pp be the original price. The equation is 25=13p25 = \frac{1}{3}p. Solving for pp gives p=75p=75. The original price was 75 dollars.
  • A book club has 17 women. The number of women is two more than three times the number of men. How many men are in the club? Let mm be the number of men. The equation is 17=3m+217 = 3m + 2. Solving for mm gives m=5m=5. There are 5 men in the club.

Book overview

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Chapter 3: One-Variable Linear Equations

  1. Lesson 1

    Lesson 1: Solving Linear Equations I

  2. Lesson 2

    Lesson 2: Solving Linear Equations II

  3. Lesson 3Current

    Lesson 3: Word Problems

  4. Lesson 4

    Lesson 4: Linear Equations in Disguise

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Translate to an equation and solve

Property

To solve word problems, first translate the word sentences into equations with variables, and then solve the equations using the properties of equality. Identify keywords to determine the correct operation.

Examples

  • 'Seven more than nn is equal to 2.' translates to n+7=2n + 7 = 2. Subtracting 7 from both sides gives n=5n = -5.
  • 'The difference of pp and 9 is 15-15.' translates to p9=15p - 9 = -15. Adding 9 to both sides gives p=6p = -6.

Section 2

Use a Problem-Solving Strategy

Property

Use a Problem-Solving Strategy to Solve Word Problems.

  1. Read the problem. Make sure all the words and ideas are understood.
  2. Identify what we are looking for.
  3. Name what we are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
  5. Solve the equation using good algebra techniques.
  6. Check the answer in the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Examples

  • A shirt is on sale for 25 dollars, which is one-third of its original price. What was the original price? Let pp be the original price. The equation is 25=13p25 = \frac{1}{3}p. Solving for pp gives p=75p=75. The original price was 75 dollars.
  • A book club has 17 women. The number of women is two more than three times the number of men. How many men are in the club? Let mm be the number of men. The equation is 17=3m+217 = 3m + 2. Solving for mm gives m=5m=5. There are 5 men in the club.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: One-Variable Linear Equations

  1. Lesson 1

    Lesson 1: Solving Linear Equations I

  2. Lesson 2

    Lesson 2: Solving Linear Equations II

  3. Lesson 3Current

    Lesson 3: Word Problems

  4. Lesson 4

    Lesson 4: Linear Equations in Disguise