Lesson 3: Solve Multistep Equations

Property

When solving equations with variables on both sides that contain parentheses, apply the distributive property first: $a(b + c) = ab + ac$.
This eliminates parentheses before collecting like terms and variables.

Examples

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 to one side of the equation. Use the Addition or Subtraction Property of Equality.
Step 3. Collect all the constant terms to 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

• To solve $5(y - 4) = 15$, first distribute to get $5y - 20 = 15$. Add 20 to both sides for $5y = 35$, so $y = 7$.

• For $2(a - 3) + 7 = 9$, first distribute and combine like terms: $2a - 6 + 7 = 9$, which simplifies to $2a + 1 = 9$. Then $2a = 8$, so $a = 4$.

Property

When an equation involves a fraction multiplied by a quantity in parentheses, you can either distribute the fraction first and then clear all resulting fractions with the LCD, or you can clear the initial fraction by multiplying by its denominator first.

Examples

• Solve $-5 = \frac{1}{4}(8x+4)$. Distribute first: $-5 = \frac{1}{4}(8x) + \frac{1}{4}(4)$, which simplifies to $-5 = 2x+1$. Then solve for $x$: $-6 = 2x$, so $x=-3$.

• Solve $\frac{1}{5}(q+3) = \frac{1}{2}(q-3)$. The LCD is 10. Multiply by 10: $10 \cdot \frac{1}{5}(q+3) = 10 \cdot \frac{1}{2}(q-3)$ becomes $2(q+3) = 5(q-3)$. Then $2q+6 = 5q-15$, so $21 = 3q$ and $q=7$.