SAT Linear Equations: Variables on Both Sides

To solve a linear equation with variables on both sides, simplify each side, subtract the same variable term from both sides, and isolate the remaining variable. Check your answer in the original equation. If all variable terms cancel, the remaining statement tells you whether there are no solutions or infinitely many.

College Board includes linear equations in one variable in SAT Algebra. This guide practices that skill with original Kibo examples. These are teaching exercises, not official SAT questions.

Keep the two sides equal

Think of the equals sign as a claim that two expressions have the same value. Adding or subtracting the same quantity on both sides preserves that equality. You can also multiply or divide both sides by the same nonzero number. OpenStax explains these properties and the possible solution cases.

A useful order is: expand parentheses, combine like terms, collect the variable terms, collect the constants, then divide. Write the operation you apply to both sides whenever a sign feels uncertain.

Worked example: collect the variable terms

Solve 5x + 7 = 2x + 22.

1. Subtract 2x from both sides: 3x + 7 = 22.
2. Subtract 7 from both sides: 3x = 15.
3. Divide both sides by 3: x = 5.

Check the original equation: 5(5) + 7 = 32 and 2(5) + 22 = 32. Both sides agree.

You could instead subtract 5x first and get 7 = −3x + 22. Subtracting 22 gives −15 = −3x, so x = 5 again. Choosing the smaller variable coefficient often keeps the arithmetic positive, but either route works.

Worked example: distribute a negative sign

Solve 4 − 3(2x − 5) = x − 2.

The factor −3 multiplies both terms in the parentheses: −3(2x − 5) = −6x + 15. The constant becomes positive because (−3)(−5) = 15.

4 − 6x + 15 = x − 2
19 − 6x = x − 2
19 = 7x − 2
21 = 7x
x = 3

Those last three steps add 6x, add 2, then divide by 7 on both sides. Check: the left side is 4 − 3(6 − 5) = 1; the right side is 3 − 2 = 1.

If you wrote −6x − 15 when expanding, return to the multiplication step. Fixing that sign matters more than doing the later arithmetic faster.

Worked example: clear constant denominators

Solve (x + 2)/3 = (2x − 1)/5. The parentheses mean the entire expression above each slash is divided by the denominator.

Multiply both sides by 15, a common multiple of 3 and 5:

5(x + 2) = 3(2x − 1)
5x + 10 = 6x − 3
10 = x − 3
x = 13

The original fractions both equal 5: (13 + 2)/3 = 15/3 and (26 − 1)/5 = 25/5. Multiplying by 15 applies to the whole side of the equation, including every term if a side has several terms.

Keep fractions exact while solving. If you want a refresher on exact fractions and decimals, see our guide to rational numbers and number types; those definitions also apply to SAT algebra.

What if the variable disappears?

Compare these two original equations after expanding and subtracting 4x from both sides:

• 4(x + 3) = 4x + 9 becomes 12 = 9. That is false, so there is no solution.
• 4(x + 3) = 4x + 12 becomes 12 = 12. That is true for every real x, so there are infinitely many solutions.

Neither result means x = 0. Try zero in the first equation: the sides are 12 and 9, so it fails. In the second, zero works, but so does every other real number.

More generally, ax + b = cx + d becomes (a − c)x = d − b. If a and c differ, division gives one solution. If a = c, compare b and d: equal constants give infinitely many solutions, and different constants give none. Do not divide by a − c when it is zero.

Answer the expression the question asks for

Suppose 3(2x − 1) = 4x + 13, and the question asks for 2x + 5.

Expanding gives 6x − 3 = 4x + 13, so 2x = 16. You can immediately find 2x + 5 = 21. Solving x = 8 first also works: 2(8) + 5 = 21. Reporting 8 alone would answer a different question.

Try six original practice questions

Write an operation beside each step. For a numerical solution, substitute into the original equation before reading the answers.

1. Solve 7x − 4 = 3x + 20.
2. Solve 6 − 2(x + 3) = 3x + 10.
3. Solve (x − 1)/4 = (x + 5)/6.
4. How many real solutions does 5(x − 2) = 5x − 7 have?
5. How many real solutions does 2(3x + 4) = 6x + 8 have?
6. If 4(x + 1) = 2x + 18, what is 3x − 2?

Answers with checks

1. x = 6. Subtracting 3x and adding 4 gives 4x = 24. Check: 7(6) − 4 = 38 and 3(6) + 20 = 38.
2. x = −2. The left side simplifies to 6 − 2x − 6 = −2x. Then −5x = 10. Check: 6 − 2(−2 + 3) = 4 and 3(−2) + 10 = 4.
3. x = 13. Multiply both sides by 12: 3(x − 1) = 2(x + 5). Expanding and collecting gives x = 13. Check: 12/4 = 3 and 18/6 = 3.
4. No solutions. Expanding gives 5x − 10 = 5x − 7. Subtracting 5x leaves the false statement −10 = −7.
5. Infinitely many solutions. Expanding gives 6x + 8 = 6x + 8, which is true for every real x.
6. 19. Expanding gives 4x + 4 = 2x + 18, so x = 7. Both original sides equal 32. The requested value is 3(7) − 2 = 19.

Choose a useful next step

If question 1 was difficult, repeat the collect-and-isolate steps. For question 2, write both products before combining terms. For question 3, show the common multiplier acting on each whole fraction. For questions 4 and 5, compare what remains after the variable terms cancel. For question 6, underline the requested expression before calculating.

Rework one missed problem later with the answer covered, then try a new problem using the same skill. You can try a free SAT practice question with Kibo to continue practicing.

Prepared with AI assistance. These original examples were checked algebraically and by calculation; no human review is claimed. Sources checked October 4, 2026. SAT is a registered trademark of College Board, which is not affiliated with or endorsing this guide.