Math Core

Real equations rarely arrive in the neat form $ax + b = c$ . They come with parentheses, several $x$ -terms, fractions and decimals. The good news: you already have every tool you need. You simplify each side using Unit 1 skills until the equation becomes a two-step equation, then finish the way you already know.

The plan

Solving a multi-step equation

Distribute to remove any parentheses.
Combine like terms on each side.
Undo addition or subtraction to isolate the variable term.
Undo multiplication or division to isolate the variable.
Check in the original equation.

Steps 1 and 2 simplify each side separately. They don't move anything across the equal sign. Steps 3 and 4 use the properties of equality on both sides.

Combining like terms first

Worked example: Several variable terms

Solve $3x + 5 + 2x - 7 = 23$ .

The left side has two $x$ -terms and two constants. Combine them before doing anything else.

\begin{aligned} 3x + 5 + 2x - 7 &= 23 \\ 5x - 2 &= 23 && \text{combine like terms} \\ 5x &= 25 && \text{add } 2 \\ x &= 5 && \text{divide by } 5 \end{aligned}

Check: $3(5) + 5 + 2(5) - 7 = 15 + 5 + 10 - 7 = 23$ . ✓

Distributing first

When the variable is stuck inside parentheses, distribute to release it.

Worked example: Distribute, combine, solve

Solve $4(2y - 3) + 6 = 26$ .

\begin{aligned} 4(2y - 3) + 6 &= 26 \\ 8y - 12 + 6 &= 26 && \text{distribute the } 4 \\ 8y - 6 &= 26 && \text{combine constants} \\ 8y &= 32 && \text{add } 6 \\ y &= 4 && \text{divide by } 8 \end{aligned}

Check: $4(2 \cdot 4 - 3) + 6 = 4(5) + 6 = 26$ . ✓

Sometimes you can skip distributing. In $5(x - 4) = 35$ , the only thing on the left is a product, so you can divide both sides by $5$ right away: $x - 4 = 7$ , so $x = 11$ . Either route gives the same answer; use whichever is cleaner.

Worked example: Subtracting a group

Solve $20 - 3(n + 2) = 5$ .

The $-3$ multiplies both terms in the parentheses.

\begin{aligned} 20 - 3(n + 2) &= 5 \\ 20 - 3n - 6 &= 5 && \text{distribute } -3 \\ 14 - 3n &= 5 && \text{combine constants} \\ -3n &= -9 && \text{subtract } 14 \\ n &= 3 && \text{divide by } -3 \end{aligned}

Check: $20 - 3(3 + 2) = 20 - 15 = 5$ . ✓

Common mistake

Two classic errors with $20 - 3(n + 2)$ :

Subtracting first. $20 - 3$ is not $17$ here. The order of operations says multiply before subtracting, so the $3$ belongs with the parentheses.
Dropping a sign. $-3(n + 2)$ is $-3n - 6$ , and $-2(x - 3)$ is $-2x + 6$ . The negative multiplies every term inside.

Clearing fractions

Fractions make arithmetic messy, but you can get rid of all of them in one move. Multiply every term on both sides by the least common denominator (LCD). This is the multiplication property of equality, so the new equation has the same solution.

Worked example: Multiply by the LCD

Solve $\dfrac{x}{2} + \dfrac{x}{3} = 10$ .

The LCD of $2$ and $3$ is $6$ . Multiply each term by $6$ :

\begin{aligned} 6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} &= 6 \cdot 10 \\ 3x + 2x &= 60 \\ 5x &= 60 \\ x &= 12 \end{aligned}

Check: $\dfrac{12}{2} + \dfrac{12}{3} = 6 + 4 = 10$ . ✓

Decimals work the same way: multiply by a power of $10$ . In $0.3x + 1.5 = 4.2$ , multiplying every term by $10$ gives $3x + 15 = 42$ , so $x = 9$ .

Common mistake

When you clear fractions, multiply every term, including terms that have no fraction. In $\dfrac{x}{4} + 2 = 5$ , multiplying by $4$ gives $x + 8 = 20$ , not $x + 2 = 20$ .

Setting up multi-step equations

Word problems often lead to multi-step equations because several quantities are described in terms of the same unknown.

Worked example: Consecutive integers

The sum of three consecutive integers is $72$ . What are the integers?

Consecutive integers go up by $1$ . Let $n$ be the smallest; then the others are $n + 1$ and $n + 2$ .

\begin{aligned} n + (n + 1) + (n + 2) &= 72 \\ 3n + 3 &= 72 \\ 3n &= 69 \\ n &= 23 \end{aligned}

The integers are $23$ , $24$ and $25$ . Check: $23 + 24 + 25 = 72$ . ✓

Tip

Write one step per line and keep the equal signs lined up. Most errors in multi-step equations are copying errors: a sign that disappears between lines, or a term that gets left behind. Neat work makes them easy to spot.

Practice

Practice 1

Solve $6a - 2a + 3 = 19$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Solve $5(x - 4) = 35$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Solve $-2(3k + 1) + 4k = 10$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Solve $3(2m - 5) - (m + 1) = 9$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Solve $\dfrac{2x}{3} - \dfrac{x}{4} = 10$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Solve $0.4(x + 5) + 0.2x = 5$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Jordan solved $7 - 2(x - 3) = 15$ like this:

Step 1: $7 - 2x - 6 = 15$
Step 2: $1 - 2x = 15$
Step 3: $-2x = 14$
Step 4: $x = -7$

In which step did Jordan make the first error?

Practice 8

A rectangle's length is $3$ cm more than twice its width. Its perimeter is $54$ cm. Find the width, in centimeters.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.