Learn on PengienVision, Mathematics, Grade 7Chapter 5: Solve Problems Using Equations and Inequalities

Lesson 2: Solve Two-Step Equations

In this Grade 7 enVision Mathematics lesson from Chapter 5, students learn how to solve two-step equations by applying the properties of equality — including the Subtraction, Addition, and Division Properties — to isolate a variable in two sequential steps. The lesson connects algebraic and arithmetic solution methods and uses real-world contexts such as movie ticket costs and gift card balances to build understanding. Students also compare one-step and two-step equation strategies to see how inverse operations work consistently across both.

Section 1

Strategy 3, Part C: Solving Two-Step Equations with Inverse Operations

Property

To solve an equation, the goal is to determine the value of the unknown.
This is done by using inverse operations to isolate the variable.
For an equation like ax+b=cax + b = c, first use the addition or subtraction property of equality to isolate the term with the variable (axax).
Then, use the multiplication or division property of equality to find the value of xx.

Examples

  • To solve 4x+5=254x + 5 = 25, first subtract 5 from both sides to get 4x=204x = 20. Then, divide both sides by 4 to find x=5x = 5.
  • To solve 3p8=133p - 8 = 13, first add 8 to both sides to get 3p=213p = 21. Then, divide both sides by 3 to find p=7p = 7.
  • To solve k3+2=6\frac{k}{3} + 2 = 6, first subtract 2 from both sides to get k3=4\frac{k}{3} = 4. Then, multiply both sides by 3 to find k=12k = 12.

Explanation

Think of solving an equation as unwrapping a present. You must undo the operations in reverse order to find the variable. First, undo any addition or subtraction, then undo any multiplication or division to get the variable by itself.

Section 2

Application: Modeling Equations with Tape Diagrams

Property

A tape diagram models a word problem by representing a total quantity as a whole bar and its parts as sections of the bar.
The visual relationship between the parts and the whole helps determine the sequence of operations (addition, subtraction, multiplication, division) needed to find the unknown value, xx.

Examples

Section 3

Comparing Algebraic and Arithmetic Solutions

Property

A problem can be solved arithmetically by working backward or algebraically by writing and solving an equation. Both methods use inverse operations to find the unknown value. For a problem modeled by ax+b=cax + b = c, the arithmetic approach undoes the addition/subtraction first, then the multiplication/division, just as the algebraic approach does.

Examples

  • Problem: A taxi ride costs a flat fee of 3 dollars plus 2 dollars per mile. If a ride costs 19 dollars, how many miles was it?

Arithmetic Solution: Start with the total cost of 19 dollars. Subtract the flat fee: 193=1619 - 3 = 16 dollars. Divide by the cost per mile: 16÷2=816 \div 2 = 8 miles.
Algebraic Solution: Let mm be the number of miles. The equation is 2m+3=192m + 3 = 19. Subtract 3 from both sides: 2m=162m = 16. Divide by 2: m=8m = 8.

  • Problem: Sarah sold 4 equal-sized boxes of cookies, but first she ate 5 cookies. In total, she sold 43 cookies. How many cookies were in each full box?

Arithmetic Solution: Start with the 43 cookies sold. Add back the 5 cookies she ate: 43+5=4843 + 5 = 48. Divide by the 4 boxes: 48÷4=1248 \div 4 = 12 cookies per box.
Algebraic Solution: Let cc be the cookies in a box. The equation is 4c5=434c - 5 = 43. Add 5 to both sides: 4c=484c = 48. Divide by 4: c=12c = 12.

Explanation

The arithmetic method involves "working backward" from the final result, step-by-step, using inverse operations. The algebraic method involves representing the unknown with a variable, writing an equation, and then using properties of equality to solve. Both approaches rely on the same logic of undoing operations to isolate the unknown quantity. Understanding both methods helps connect your intuitive problem-solving skills to the formal process of algebra.

Book overview

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

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Chapter 5: Solve Problems Using Equations and Inequalities

  1. Lesson 1

    Lesson 1: Write Two-Step Equations

  2. Lesson 2Current

    Lesson 2: Solve Two-Step Equations

  3. Lesson 3

    Lesson 3: Solve Equations Using the Distributive Property

  4. Lesson 4

    Lesson 4: Solve Inequalities Using Addition or Subtraction

  5. Lesson 5

    Lesson 5: Solve Inequalities Using Multiplication or Division

  6. Lesson 6

    Lesson 6: Solve Two-Step Inequalities

  7. Lesson 7

    Lesson 7: Solve Multi-Step Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Strategy 3, Part C: Solving Two-Step Equations with Inverse Operations

Property

To solve an equation, the goal is to determine the value of the unknown.
This is done by using inverse operations to isolate the variable.
For an equation like ax+b=cax + b = c, first use the addition or subtraction property of equality to isolate the term with the variable (axax).
Then, use the multiplication or division property of equality to find the value of xx.

Examples

  • To solve 4x+5=254x + 5 = 25, first subtract 5 from both sides to get 4x=204x = 20. Then, divide both sides by 4 to find x=5x = 5.
  • To solve 3p8=133p - 8 = 13, first add 8 to both sides to get 3p=213p = 21. Then, divide both sides by 3 to find p=7p = 7.
  • To solve k3+2=6\frac{k}{3} + 2 = 6, first subtract 2 from both sides to get k3=4\frac{k}{3} = 4. Then, multiply both sides by 3 to find k=12k = 12.

Explanation

Think of solving an equation as unwrapping a present. You must undo the operations in reverse order to find the variable. First, undo any addition or subtraction, then undo any multiplication or division to get the variable by itself.

Section 2

Application: Modeling Equations with Tape Diagrams

Property

A tape diagram models a word problem by representing a total quantity as a whole bar and its parts as sections of the bar.
The visual relationship between the parts and the whole helps determine the sequence of operations (addition, subtraction, multiplication, division) needed to find the unknown value, xx.

Examples

Section 3

Comparing Algebraic and Arithmetic Solutions

Property

A problem can be solved arithmetically by working backward or algebraically by writing and solving an equation. Both methods use inverse operations to find the unknown value. For a problem modeled by ax+b=cax + b = c, the arithmetic approach undoes the addition/subtraction first, then the multiplication/division, just as the algebraic approach does.

Examples

  • Problem: A taxi ride costs a flat fee of 3 dollars plus 2 dollars per mile. If a ride costs 19 dollars, how many miles was it?

Arithmetic Solution: Start with the total cost of 19 dollars. Subtract the flat fee: 193=1619 - 3 = 16 dollars. Divide by the cost per mile: 16÷2=816 \div 2 = 8 miles.
Algebraic Solution: Let mm be the number of miles. The equation is 2m+3=192m + 3 = 19. Subtract 3 from both sides: 2m=162m = 16. Divide by 2: m=8m = 8.

  • Problem: Sarah sold 4 equal-sized boxes of cookies, but first she ate 5 cookies. In total, she sold 43 cookies. How many cookies were in each full box?

Arithmetic Solution: Start with the 43 cookies sold. Add back the 5 cookies she ate: 43+5=4843 + 5 = 48. Divide by the 4 boxes: 48÷4=1248 \div 4 = 12 cookies per box.
Algebraic Solution: Let cc be the cookies in a box. The equation is 4c5=434c - 5 = 43. Add 5 to both sides: 4c=484c = 48. Divide by 4: c=12c = 12.

Explanation

The arithmetic method involves "working backward" from the final result, step-by-step, using inverse operations. The algebraic method involves representing the unknown with a variable, writing an equation, and then using properties of equality to solve. Both approaches rely on the same logic of undoing operations to isolate the unknown quantity. Understanding both methods helps connect your intuitive problem-solving skills to the formal process of algebra.

Book overview

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

Continue this chapter

Chapter 5: Solve Problems Using Equations and Inequalities

  1. Lesson 1

    Lesson 1: Write Two-Step Equations

  2. Lesson 2Current

    Lesson 2: Solve Two-Step Equations

  3. Lesson 3

    Lesson 3: Solve Equations Using the Distributive Property

  4. Lesson 4

    Lesson 4: Solve Inequalities Using Addition or Subtraction

  5. Lesson 5

    Lesson 5: Solve Inequalities Using Multiplication or Division

  6. Lesson 6

    Lesson 6: Solve Two-Step Inequalities

  7. Lesson 7

    Lesson 7: Solve Multi-Step Inequalities