Learn on PengiYoshiwara Core MathChapter 5: Using Variables

Lesson 4: More Equations

In this Grade 8 lesson from Yoshiwara Core Math, Chapter 5, students learn how to solve two-step equations by undoing operations in reverse order using inverse operations such as addition, subtraction, multiplication, and division. The lesson introduces a systematic strategy for isolating variables in equations like 5n + 4 = 34 and 2x βˆ’ 5 = 13, and extends the skill to working with formulas and real-world problem solving. Students practice writing and solving algebraic equations from word problems, then verify solutions by substituting back into the original equation.

Section 1

πŸ“˜ More Equations

New Concept

This lesson introduces a powerful strategy for solving equations with multiple steps. We'll learn to isolate the variable by "unwrapping" itβ€”undoing each operation in reverse order to find the solution.

What’s next

You'll soon tackle interactive examples and practice cards that apply this strategy to solve equations, formulas, and real-world problems.

Section 2

Equations with Two Operations

Property

To solve an equation with two or more operations, we must isolate the variable on one side of the equation. We undo the operations in reverse order. Typically, we undo addition or subtraction first, before undoing multiplication or division.

Examples

  • To solve 4x+5=294x + 5 = 29, first subtract 5 from both sides to get 4x=244x = 24. Then, divide both sides by 4 to find x=6x = 6.
  • To solve y3βˆ’2=7\frac{y}{3} - 2 = 7, first add 2 to both sides to get y3=9\frac{y}{3} = 9. Then, multiply both sides by 3 to find y=27y = 27.
  • To solve 18=6+2z18 = 6 + 2z, first subtract 6 from both sides to get 12=2z12 = 2z. Then, divide both sides by 2 to find z=6z = 6.

Explanation

Think of it as reversing your morning routine. To get back to the start, you undo the last thing you did first. In equations, this means handling addition or subtraction before dealing with multiplication or division to isolate the variable.

Section 3

A Strategy for Solving Equations

Property

To solve an equation:

  1. List the operations performed on the variable in order.
  2. Undo those operations in reverse order.

Examples

  • Solve 4aβˆ’53=9\frac{4a - 5}{3} = 9. The operations on aa are: multiply by 4, subtract 5, divide by 3. Undo in reverse: multiply by 3 to get 4aβˆ’5=274a - 5 = 27, add 5 to get 4a=324a = 32, and divide by 4 to get a=8a = 8.
  • Solve b+72βˆ’3=8\frac{b + 7}{2} - 3 = 8. Undo in reverse: add 3 to get b+72=11\frac{b+7}{2} = 11, multiply by 2 to get b+7=22b+7 = 22, and subtract 7 to get b=15b = 15.
  • Solve 10+c5=1410 + \frac{c}{5} = 14. The operations on cc are: divide by 5, then add 10. Undo in reverse: subtract 10 to get c5=4\frac{c}{5} = 4, then multiply by 5 to get c=20c = 20.

Explanation

Solving an equation is like unwrapping a gift. The variable is the gift inside, and the operations are the wrapping paper and box. To get to the gift, you must undo each layer in the reverse order it was applied.

Section 4

Using Formulas to Solve Equations

Property

A formula is an equation that describes a relationship between variables. We use formulas to solve for an unknown value by substituting the known values into the formula and then solving the resulting equation.

Examples

  • The perimeter of a rectangle is P=2l+2wP = 2l + 2w. If P=60P=60 cm and w=12w=12 cm, find the length ll. Substitute to get 60=2l+2(12)60 = 2l + 2(12), which is 60=2l+2460 = 2l + 24. Solving gives 36=2l36 = 2l, so l=18l=18 cm.
  • Using the temperature formula F=95C+32F = \frac{9}{5}C + 32, find the Celsius temperature for 59Β°F. Substitute to get 59=95C+3259 = \frac{9}{5}C + 32. Subtract 32 to get 27=95C27 = \frac{9}{5}C. Solving for CC gives C=15C = 15 degrees.
  • The area of a triangle is A=12bhA = \frac{1}{2}bh. If A=50A=50 square inches and b=10b=10 inches, find the height hh. Substitute to get 50=12(10)h50 = \frac{1}{2}(10)h, which is 50=5h50 = 5h. Solving gives h=10h=10 inches.

Explanation

Formulas are like pre-built equations for real-world situations. Just plug in the numbers you know, and the formula turns into a simple equation puzzle. Solve it to find the missing piece of information you need.

Section 5

Using Algebra to Solve Problems

Property

Steps for Modeling a Problem:

  1. Identify the unknown quantity and choose a variable to represent it.
  2. Find some quantity that can be expressed in two different ways, and write an equation.
  3. Solve the equation, and answer the question in the problem.

Examples

  • A taxi ride costs 3 dollars plus 2 dollars per mile. If the total fare was 21 dollars, how many miles was the ride? Let mm be the miles. The equation is 3+2m=213 + 2m = 21. Solving gives 2m=182m = 18, so the ride was m=9m=9 miles.
  • Jamal is saving for a 150 dollars bike. He has 30 dollars and earns 15 dollars per week. How many weeks will it take? Let ww be the weeks. The equation is 30+15w=15030 + 15w = 150. Solving gives 15w=12015w = 120, so it will take w=8w=8 weeks.
  • A phone company charges 20 dollars a month plus 0.05 dollars per text. If a monthly bill is 24 dollars, how many texts were sent? Let tt be the number of texts. The equation is 20+0.05t=2420 + 0.05t = 24. Solving gives 0.05t=40.05t = 4, so t=80t=80 texts were sent.

Explanation

Algebra is a powerful tool that turns word problems into solvable puzzles. First, you pick a letter for the unknown value. Then, you translate the problem's words into a math sentence (an equation) and solve it to find the answer.

Book overview

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

Continue this chapter

Chapter 5: Using Variables

  1. Lesson 1

    Lesson 1: Working with Variables

  2. Lesson 2

    Lesson 2: More Algebraic Expressions

  3. Lesson 3

    Lesson 3: Problem Solving

  4. Lesson 4Current

    Lesson 4: More Equations

  5. Lesson 5

    Lesson 5: Graphs

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ More Equations

New Concept

This lesson introduces a powerful strategy for solving equations with multiple steps. We'll learn to isolate the variable by "unwrapping" itβ€”undoing each operation in reverse order to find the solution.

What’s next

You'll soon tackle interactive examples and practice cards that apply this strategy to solve equations, formulas, and real-world problems.

Section 2

Equations with Two Operations

Property

To solve an equation with two or more operations, we must isolate the variable on one side of the equation. We undo the operations in reverse order. Typically, we undo addition or subtraction first, before undoing multiplication or division.

Examples

  • To solve 4x+5=294x + 5 = 29, first subtract 5 from both sides to get 4x=244x = 24. Then, divide both sides by 4 to find x=6x = 6.
  • To solve y3βˆ’2=7\frac{y}{3} - 2 = 7, first add 2 to both sides to get y3=9\frac{y}{3} = 9. Then, multiply both sides by 3 to find y=27y = 27.
  • To solve 18=6+2z18 = 6 + 2z, first subtract 6 from both sides to get 12=2z12 = 2z. Then, divide both sides by 2 to find z=6z = 6.

Explanation

Think of it as reversing your morning routine. To get back to the start, you undo the last thing you did first. In equations, this means handling addition or subtraction before dealing with multiplication or division to isolate the variable.

Section 3

A Strategy for Solving Equations

Property

To solve an equation:

  1. List the operations performed on the variable in order.
  2. Undo those operations in reverse order.

Examples

  • Solve 4aβˆ’53=9\frac{4a - 5}{3} = 9. The operations on aa are: multiply by 4, subtract 5, divide by 3. Undo in reverse: multiply by 3 to get 4aβˆ’5=274a - 5 = 27, add 5 to get 4a=324a = 32, and divide by 4 to get a=8a = 8.
  • Solve b+72βˆ’3=8\frac{b + 7}{2} - 3 = 8. Undo in reverse: add 3 to get b+72=11\frac{b+7}{2} = 11, multiply by 2 to get b+7=22b+7 = 22, and subtract 7 to get b=15b = 15.
  • Solve 10+c5=1410 + \frac{c}{5} = 14. The operations on cc are: divide by 5, then add 10. Undo in reverse: subtract 10 to get c5=4\frac{c}{5} = 4, then multiply by 5 to get c=20c = 20.

Explanation

Solving an equation is like unwrapping a gift. The variable is the gift inside, and the operations are the wrapping paper and box. To get to the gift, you must undo each layer in the reverse order it was applied.

Section 4

Using Formulas to Solve Equations

Property

A formula is an equation that describes a relationship between variables. We use formulas to solve for an unknown value by substituting the known values into the formula and then solving the resulting equation.

Examples

  • The perimeter of a rectangle is P=2l+2wP = 2l + 2w. If P=60P=60 cm and w=12w=12 cm, find the length ll. Substitute to get 60=2l+2(12)60 = 2l + 2(12), which is 60=2l+2460 = 2l + 24. Solving gives 36=2l36 = 2l, so l=18l=18 cm.
  • Using the temperature formula F=95C+32F = \frac{9}{5}C + 32, find the Celsius temperature for 59Β°F. Substitute to get 59=95C+3259 = \frac{9}{5}C + 32. Subtract 32 to get 27=95C27 = \frac{9}{5}C. Solving for CC gives C=15C = 15 degrees.
  • The area of a triangle is A=12bhA = \frac{1}{2}bh. If A=50A=50 square inches and b=10b=10 inches, find the height hh. Substitute to get 50=12(10)h50 = \frac{1}{2}(10)h, which is 50=5h50 = 5h. Solving gives h=10h=10 inches.

Explanation

Formulas are like pre-built equations for real-world situations. Just plug in the numbers you know, and the formula turns into a simple equation puzzle. Solve it to find the missing piece of information you need.

Section 5

Using Algebra to Solve Problems

Property

Steps for Modeling a Problem:

  1. Identify the unknown quantity and choose a variable to represent it.
  2. Find some quantity that can be expressed in two different ways, and write an equation.
  3. Solve the equation, and answer the question in the problem.

Examples

  • A taxi ride costs 3 dollars plus 2 dollars per mile. If the total fare was 21 dollars, how many miles was the ride? Let mm be the miles. The equation is 3+2m=213 + 2m = 21. Solving gives 2m=182m = 18, so the ride was m=9m=9 miles.
  • Jamal is saving for a 150 dollars bike. He has 30 dollars and earns 15 dollars per week. How many weeks will it take? Let ww be the weeks. The equation is 30+15w=15030 + 15w = 150. Solving gives 15w=12015w = 120, so it will take w=8w=8 weeks.
  • A phone company charges 20 dollars a month plus 0.05 dollars per text. If a monthly bill is 24 dollars, how many texts were sent? Let tt be the number of texts. The equation is 20+0.05t=2420 + 0.05t = 24. Solving gives 0.05t=40.05t = 4, so t=80t=80 texts were sent.

Explanation

Algebra is a powerful tool that turns word problems into solvable puzzles. First, you pick a letter for the unknown value. Then, you translate the problem's words into a math sentence (an equation) and solve it to find the answer.

Book overview

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

Continue this chapter

Chapter 5: Using Variables

  1. Lesson 1

    Lesson 1: Working with Variables

  2. Lesson 2

    Lesson 2: More Algebraic Expressions

  3. Lesson 3

    Lesson 3: Problem Solving

  4. Lesson 4Current

    Lesson 4: More Equations

  5. Lesson 5

    Lesson 5: Graphs