Equations with Variables on Both Sides

To solve equations with variables on both sides, get all the variable terms on one side and all the numbers on the other. Add or subtract a variable term from both sides first. Then solve the two-step equation that is left. For example, $5x + 2 = 3x + 10$ becomes $2x + 2 = 10$ , so $x = 4$ .

Before this lesson: You should know how to solve multi-step equations, combine like terms and use the distributive property.

What Does “Variables on Both Sides” Mean?

In the equations you have solved so far, the variable was on one side only. Now it shows up on both sides, like $5x + 2 = 3x + 10$ . You can’t divide yet, because there are two $x$ -terms in different places.

Think of an equation as a balance scale. You can take the same thing off both sides and it stays balanced. That includes variable terms. Subtract $3x$ from both sides, and only one $x$ -term is left.

Subtract 3x from both sides. Now the variable is on one side only.

How to Solve Equations with Variables on Both Sides

Simplify each side. Distribute and combine like terms.
Move the variable terms to one side. Add or subtract a variable term on both sides.
Move the numbers to the other side.
Divide by the coefficient (the number in front of the variable).
Check your answer in the original equation.

Tip: Move the smaller variable term. In $5x + 2 = 3x + 10$ , $3x$ is smaller than $5x$ . Subtracting $3x$ keeps the coefficient positive, so there are fewer negative signs to track.

\begin{aligned} 5x + 2 &= 3x + 10 \\ 2x + 2 &= 10 & &\text{subtract } 3x \\ 2x &= 8 & &\text{subtract 2} \\ x &= 4 & &\text{divide by 2} \end{aligned}

Check: $5(4) + 2 = 22$ and $3(4) + 10 = 22$ . Both sides match.

Worked Examples

Example 1: Basic equation

Solve $6x - 5 = 2x + 11$ .

Step 1. Subtract $2x$ from both sides.

Step 2. Add 5 to both sides, then divide by 4.

\begin{aligned} 6x - 5 &= 2x + 11 \\ 4x - 5 &= 11 \\ 4x &= 16 \\ x &= 4 \end{aligned}

Check: $6(4) - 5 = 19$ and $2(4) + 11 = 19$ . It works.

Answer: $x = 4$

Example 2: A negative variable term

Solve $4n - 9 = -2n + 21$ .

Step 1. The right side has $-2n$ . Add $2n$ to both sides to cancel it.

Step 2. Add 9 to both sides, then divide by 6.

\begin{aligned} 4n - 9 &= -2n + 21 \\ 6n - 9 &= 21 \\ 6n &= 30 \\ n &= 5 \end{aligned}

Check: $4(5) - 9 = 11$ and $-2(5) + 21 = 11$ . It works.

Answer: $n = 5$

Example 3: Parentheses on both sides

Solve $3(a + 4) = 5(a - 2)$ .

Step 1. Distribute on each side.

Step 2. Subtract $3a$ from both sides. The variable ends up on the right, and that’s fine.

Step 3. Add 10 to both sides, then divide by 2.

\begin{aligned} 3(a + 4) &= 5(a - 2) \\ 3a + 12 &= 5a - 10 \\ 12 &= 2a - 10 \\ 22 &= 2a \\ 11 &= a \end{aligned}

Check: $3(11 + 4) = 45$ and $5(11 - 2) = 45$ . It works.

Answer: $a = 11$

Example 4: Fractions

Solve $\frac{1}{2}x + 3 = \frac{1}{4}x + 5$ .

Step 1. Clear the fractions. The denominators are 2 and 4, so multiply every term by 4.

Step 2. Subtract $x$ from both sides, then subtract 12.

\begin{aligned} 4 \cdot \tfrac{1}{2}x + 4 \cdot 3 &= 4 \cdot \tfrac{1}{4}x + 4 \cdot 5 \\ 2x + 12 &= x + 20 \\ x + 12 &= 20 \\ x &= 8 \end{aligned}

Check: $\frac{1}{2}(8) + 3 = 7$ and $\frac{1}{4}(8) + 5 = 7$ . It works.

Answer: $x = 8$

Example 5: Word problem

Bike Shop A charges $12 plus $3 per hour. Bike Shop B charges $4 plus $5 per hour. After how many hours do the two shops cost the same?

Step 1. Let $h$ = the number of hours. Shop A costs $12 + 3h$ . Shop B costs $4 + 5h$ .

Step 2. Set the costs equal and solve.

\begin{aligned} 12 + 3h &= 4 + 5h \\ 12 &= 4 + 2h \\ 8 &= 2h \\ 4 &= h \end{aligned}

Step 3. Check the cost. Shop A: $12 + 3(4) = 24$ . Shop B: $4 + 5(4) = 24$ . Both cost $24.

Answer: The shops cost the same, $24, after 4 hours.

Common Mistakes

Mistake	What it looks like	How to fix it
Changing only one side	$5x + 2 = 3x + 10$ becomes $2x + 2 = 3x + 10$	Subtract $3x$ from both sides: $2x + 2 = 10$ .
Adding when you should subtract	$5x + 2 = 3x + 10$ becomes $8x + 2 = 10$	To remove $+3x$ , subtract $3x$ . Do the opposite operation.
Losing the sign of a negative term	$4n - 9 = -2n + 21$ becomes $2n - 9 = 21$	To remove $-2n$ , add $2n$ : $6n - 9 = 21$ .
Distributing to only one term	$5(a - 2)$ becomes $5a - 2$	Multiply both terms: $5a - 10$ .
Reading a true statement as “x = 0”	$3x + 6 = 3x + 6$ gives $6 = 6$ , so “ $x = 0$ ”	A true statement means every number works. The answer is all real numbers.

When the Variables Cancel Out

Sometimes the variable terms cancel on both sides. Then you are left with only numbers. Look at that statement to decide the answer.

What is left	Example	Answer
A false statement, like $5 = 1$	$2x + 5 = 2x + 1$ gives $5 = 1$	No solution. No number makes it true.
A true statement, like $6 = 6$	$3(x + 2) = 3x + 6$ gives $6 = 6$	Infinitely many solutions. Every number works.

An equation that is true for every number is called an identity. Both sides are the same expression, written two ways.

You will see the same idea later with lines. Two parallel lines never meet, which matches “no solution.” See number of solutions for systems of equations.

Practice Problems

Solve each equation. If there is no solution or every number works, say so.

$6x + 1 = 4x + 9$
$9k - 5 = 4k + 20$
$2m + 7 = 5m - 14$
$8 - 2y = 3y + 23$
$4(b - 1) = 2(b + 5)$
$0.5x + 6 = 0.2x + 9$
$\frac{x}{3} + 2 = \frac{x}{6} + 5$
Maya has $50 and saves $8 a week. Leo has $90 and saves $3 a week. After how many weeks will they have the same amount? How much will each have?
$5(t + 2) - t = 4t + 10$
$3(2w - 1) = 6w + 4$

Answers

Show answers

Subtract $4x$ : $2x + 1 = 9$ , so $2x = 8$ and $x = 4$
Subtract $4k$ : $5k - 5 = 20$ , so $5k = 25$ and $k = 5$
Subtract $2m$ : $7 = 3m - 14$ , so $21 = 3m$ and $m = 7$
Add $2y$ : $8 = 5y + 23$ , so $-15 = 5y$ and $y = -3$
$4b - 4 = 2b + 10$ , so $2b - 4 = 10$ , $2b = 14$ and $b = 7$
Subtract $0.2x$ : $0.3x + 6 = 9$ , so $0.3x = 3$ and $x = 10$
Multiply by 6: $2x + 12 = x + 30$ , so $x = 18$
Let $w$ = weeks. $50 + 8w = 90 + 3w$ , so $5w = 40$ and $w = 8$ . After 8 weeks, each has $114.
$5t + 10 - t = 4t + 10$ , so $4t + 10 = 4t + 10$ and $10 = 10$ . Infinitely many solutions (all real numbers).
$6w - 3 = 6w + 4$ , so $-3 = 4$ , which is false. No solution.