Learn on PengiReveal Math, Course 3Module 3: Solve Equations with Variables on Each Side

Lesson 3-1: Solve Equations with Variables on Each Side

In this Grade 8 lesson from Reveal Math, Course 3, students learn to solve equations with variables on each side by applying the properties of equality — including the Addition, Subtraction, Division, and Multiplication Properties — to isolate the variable. The lesson also covers equations with rational coefficients, teaching students two methods: solving directly with fractions or multiplying by the LCD to eliminate fractions first. Students practice verifying solutions by substituting values back into the original equation.

Section 1

Concept: Equivalent Equations

Property

Two equations are equivalent if they have the same solution set. When we apply the properties of equality to an equation, we create equivalent equations: if a=ba = b, then a+c=b+ca + c = b + c, a−c=b−ca - c = b - c, a×c=b×ca \times c = b \times c, and a÷c=b÷ca \div c = b \div c (where c≠0c \neq 0).

Examples

Section 2

Solving Equations with Variables on Both Sides

Property

Step 1. Simplify each side of the equation as much as possible.

  • Use the Distributive Property to remove any parentheses.
  • Combine like terms.

Step 2. Collect all the variable terms on one side of the equation.

  • Use the Addition or Subtraction Property of Equality.

Step 3. Collect all the constant terms on the other side of the equation.

  • Use the Addition or Subtraction Property of Equality.

Step 4. Make the coefficient of the variable term to equal to 1.

  • Use the Multiplication or Division Property of Equality.
  • State the solution to the equation.

Step 5. Check the solution.

  • Substitute the solution into the original equation to make sure the result is a true statement.

Examples

Section 3

Solving with Fraction Coefficients

Property

When a variable has a fraction coefficient, such as in the equation abx=c\frac{a}{b}x = c, you can isolate xx by multiplying both sides of the equation by the reciprocal of the coefficient, ba\frac{b}{a}. The product of a number and its reciprocal is 1.

Examples

  • Solve 23x=10\frac{2}{3}x = 10. Multiply both sides by the reciprocal, 32\frac{3}{2}: 32⋅23x=32⋅10\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 10, which simplifies to x=15x = 15.
  • Solve −35w=9-\frac{3}{5}w = 9. Multiply both sides by the reciprocal, −53-\frac{5}{3}: −53⋅−35w=−53⋅9-\frac{5}{3} \cdot -\frac{3}{5}w = -\frac{5}{3} \cdot 9, which simplifies to w=−15w = -15.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Concept: Equivalent Equations

Property

Two equations are equivalent if they have the same solution set. When we apply the properties of equality to an equation, we create equivalent equations: if a=ba = b, then a+c=b+ca + c = b + c, a−c=b−ca - c = b - c, a×c=b×ca \times c = b \times c, and a÷c=b÷ca \div c = b \div c (where c≠0c \neq 0).

Examples

Section 2

Solving Equations with Variables on Both Sides

Property

Step 1. Simplify each side of the equation as much as possible.

  • Use the Distributive Property to remove any parentheses.
  • Combine like terms.

Step 2. Collect all the variable terms on one side of the equation.

  • Use the Addition or Subtraction Property of Equality.

Step 3. Collect all the constant terms on the other side of the equation.

  • Use the Addition or Subtraction Property of Equality.

Step 4. Make the coefficient of the variable term to equal to 1.

  • Use the Multiplication or Division Property of Equality.
  • State the solution to the equation.

Step 5. Check the solution.

  • Substitute the solution into the original equation to make sure the result is a true statement.

Examples

Section 3

Solving with Fraction Coefficients

Property

When a variable has a fraction coefficient, such as in the equation abx=c\frac{a}{b}x = c, you can isolate xx by multiplying both sides of the equation by the reciprocal of the coefficient, ba\frac{b}{a}. The product of a number and its reciprocal is 1.

Examples

  • Solve 23x=10\frac{2}{3}x = 10. Multiply both sides by the reciprocal, 32\frac{3}{2}: 32⋅23x=32⋅10\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 10, which simplifies to x=15x = 15.
  • Solve −35w=9-\frac{3}{5}w = 9. Multiply both sides by the reciprocal, −53-\frac{5}{3}: −53⋅−35w=−53⋅9-\frac{5}{3} \cdot -\frac{3}{5}w = -\frac{5}{3} \cdot 9, which simplifies to w=−15w = -15.