Linear equations practice: brackets, both sides and worked answers
Use these eight original questions to find the step that needs practice. You need a pen and paper; a calculator is optional for checking arithmetic. The set builds from one-step equations to equations with brackets and variables on both sides. It is a focused algebra exercise, not an exam paper or a complete GCSE topic checklist.
Keep the equation balanced
An equation states that two expressions have the same value. Keep that equality by applying the same operation to both sides. In 2x + 5 = 17, subtract 5 from both sides to get 2x = 12, then divide both sides by 2. The answer is x = 6. Check it in the original equation: 2 × 6 + 5 = 17.
“Move it across and change the sign” can hide the operation. Instead, write what you added, subtracted, multiplied or divided on both sides. When brackets appear, expand every term accurately before collecting the variable terms. A negative solution is allowed unless the question’s context rules it out.
Try all eight questions before opening the answers
Write at least one line of working for each question. If you get stuck, mark the exact line where you stopped, then continue with another question. Return to the worked answer only after a genuine attempt.
- x + 9 = 23
- 4x = 28
- 3x − 5 = 16
- x ÷ 4 + 3 = 8
- 2(x + 3) = 18
- 5x + 2 = 3x + 14
- 3(x − 2) = x + 8
- 4 − 2x = 10
Questions 1–4 isolate inverse operations. Questions 5–7 add brackets or a variable on both sides. Question 8 checks whether you can divide by a negative coefficient without changing the equation’s meaning.
Worked answers and substitution checks
Show all eight worked solutions
- x = 14. Subtract 9 from both sides: x = 23 − 9. Check: 14 + 9 = 23.
- x = 7. Divide both sides by 4. Check: 4 × 7 = 28.
- x = 7. Add 5 to get 3x = 21, then divide by 3. Check: 21 − 5 = 16.
- x = 20. Subtract 3 to get x ÷ 4 = 5, then multiply by 4. Check: 20 ÷ 4 + 3 = 8.
- x = 6. Divide both sides by 2 to get x + 3 = 9, then subtract 3. Check: 2(6 + 3) = 18. Expanding first also works.
- x = 6. Subtract 3x to get 2x + 2 = 14; subtract 2 and divide by 2. Check: both original sides equal 32.
- x = 7. Expand to 3x − 6 = x + 8. Subtract x, add 6 and divide by 2: 2x = 14. Check: both sides equal 15.
- x = −3. Subtract 4 to get −2x = 6, then divide by −2. Check: 4 − 2(−3) = 4 + 6 = 10.
A substitution check tests your answer in the original equation, not just the last line of your working. If the two sides differ, trace your operations backwards to the first line where equality was lost.
Use the result to choose your next task
If questions 1–4 were difficult, practise identifying the final operation applied to x and undoing it on both sides. If you solved them but missed question 7, compare 3(x − 2) with 3x − 6. If only question 8 caused trouble, focus on signed multiplication and division.
For a transfer challenge, invent an equation with solution x = 4, then make a second version containing a bracket. Substitute 4 to confirm that each equation really works. One example is 2(x + 1) = 10. There are many valid answers; explain why yours is balanced.
Add the specific error and a fresh example to the revision error log. Return to the question in a later session before deciding that the method is secure. For help diagnosing a persistent gap, see our maths tuition information.
Common questions
How do you solve linear equations with brackets?
Expand each bracket by multiplying every term inside it, then collect like terms and apply the same operation to both sides. For 3(x − 2) = x + 8, expand to 3x − 6 = x + 8. Subtract x and add 6 to get 2x = 14, so x = 7.
How do you solve equations with variables on both sides?
Add or subtract the same variable term on both sides so that the variable remains on one side. For 5x + 2 = 3x + 14, subtract 3x to get 2x + 2 = 14. Subtract 2, then divide by 2: x = 6. Substitute 6 into the original equation to check.
Can a linear equation have a negative answer?
Yes. In 4 − 2x = 10, subtract 4 to get −2x = 6, then divide both sides by −2: x = −3. Checking gives 4 − 2(−3) = 10. A negative answer is valid unless a word problem imposes a restriction, such as a positive length.
How do I check an answer to a linear equation?
Substitute your answer into the original equation and calculate both sides separately. For x = 7 in 3(x − 2) = x + 8, the left side is 3 × 5 = 15 and the right side is 7 + 8 = 15. Matching values confirm the solution; checking only your final rearranged line can miss an earlier mistake.