SAT math: solving equations with parentheses
Equations with parentheses appear throughout SAT Math, both as plain “solve for x” questions and inside word problems. They are also where many careless errors happen: one skipped sign, or one term left undistributed, changes the answer. This guide shows two ways to clear the parentheses, worked examples with a check each time, and the two special cases where the variable disappears.
What the parentheses mean
A number written directly outside parentheses multiplies everything inside. So 3(x + 2) means 3 times the whole quantity x + 2, which is 3x + 6, not 3x + 2. You can see the difference with a quick test: at x = 2, 3(x + 2) is 12 but 3x + 2 is 8. At x = 0 they are 6 and 2. Two expressions that give different values for the same x are not equal. Testing one or two numbers is the fastest way to catch a distribution mistake.
Method 1: distribute, then collect
- Multiply the outside number by every term inside, keeping each sign.
- Combine like terms on each side.
- Move the x terms to one side and the plain numbers to the other.
- Divide to get x, then check in the original equation.
Example 1
Solve 3(x + 2) = 2x + 11.
Distribute: 3x + 6 = 2x + 11. Subtract 2x from both sides: x + 6 = 11. Subtract 6: x = 5.
Check: the left side is 3(5 + 2) = 21 and the right side is 2(5) + 11 = 21.
Watch the minus sign
When the number outside is negative, or the equation subtracts a parenthesis, the sign goes into every term. Subtracting a negative adds.
Example 2
Solve 5 − 2(x − 3) = 4x − 1.
The −2 multiplies both terms: −2 times x is −2x, and −2 times −3 is +6. So 5 − 2x + 6 = 4x − 1, which combines to 11 − 2x = 4x − 1. Add 2x: 11 = 6x − 1. Add 1: 12 = 6x, so x = 2.
Check: the left side is 5 − 2(2 − 3) = 5 − 2(−1) = 7 and the right side is 4(2) − 1 = 7.
A common slip is writing 5 − 2x − 6, treating −2 times −3 as −6. That path gives x = 0. The check exposes it: at x = 0 the original left side is 5 − 2(−3) = 11 and the right side is −1.
Method 2: divide first when it is clean
If the whole left side is one number times one parenthesis, and that number divides the right side evenly, you can divide first and skip the distribution.
Example 3
Solve 4(2x − 3) = 20.
Divide both sides by 4: 2x − 3 = 5. Add 3: 2x = 8, so x = 4. Check: 4(2 · 4 − 3) = 4 · 5 = 20.
If the right side does not divide evenly, or there are other terms outside the parentheses, distribute instead. Both methods give the same answer; pick the one with fewer steps.
When the variable disappears
Sometimes the x terms cancel after you distribute. Then the equation is either never true or always true.
- 2(x + 3) = 2x + 5 becomes 2x + 6 = 2x + 5, then 6 = 5. That is false for every x, so there is no solution.
- 2(x + 3) = 2x + 6 becomes 6 = 6. That is true for every x, so there are infinitely many solutions.
SAT questions often hide this behind a constant: “For what value of k does 3(2x + 4) = 6x + k have infinitely many solutions?” Distribute to get 6x + 12 = 6x + k. The x terms match, so the equation is always true exactly when the constants match: k = 12. For every other k there is no solution.
A routine for test day
- Circle the number outside each parenthesis, with its sign.
- Write each product on its own line instead of in your head.
- After you solve, substitute your answer into the original equation. The Math section allows a calculator, so entering each side at your value is a fast check.
Try these
Solve 2(x − 4) = 14. Solve 7 − 3(x + 1) = 2x − 6. For what k does 5(x − 2) = 5x + k have infinitely many solutions?
Show answers
First: x = 11 (divide by 2 to get x − 4 = 7). Second: x = 2 (4 − 3x = 2x − 6, so 10 = 5x; both sides equal −2 at 2). Third: k = −10, because 5x − 10 = 5x + k is always true only when k = −10.