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§ Algebra

Linear Equations

CCSS.8.EECCSS.HSA.REI3 min read

Linear equations form the backbone of GCSE algebra, yet many Year 11 students still struggle with the systematic approach needed to isolate variables. Whether solving x + 6 = 7 or tackling 8x - 32 = 4x, the key lies in performing identical operations to both sides of the equation.

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§ 01

Why it matters

Linear equations appear everywhere in real-world problem solving, from calculating mobile phone contracts to determining break-even points in business studies. A student working out how many £3 cinema tickets they can buy with £45 is solving 3x = 45. Engineers use linear equations to calculate load distributions, whilst economists model supply and demand curves. In GCSE Foundation papers, linear equations typically account for 15-20 marks across multiple questions. Students who master these skills find simultaneous equations, quadratic formula derivations, and A-level calculus significantly more accessible. The systematic thinking required—collecting like terms, maintaining equation balance, checking solutions—develops logical reasoning that transfers to physics, chemistry, and computer science coursework.

§ 02

How to solve linear equations

Linear equations — how to

  • Collect x-terms on one side, constants on the other.
  • Do the same operation to both sides (add, subtract, multiply, divide).
  • Divide by the coefficient of x to isolate x.

Example: 3x + 7 = 22 → 3x = 15 → x = 5.

§ 03

Worked examples

Beginner§ 01

x + 6 = 7

Answer: x = 1

  1. Subtract 6 from both sides x = 7 − 6 To isolate x, undo the addition.
  2. Calculate x = 1 7 − 6 = 1.
  3. Verify 1 + 6 = 7 ✓ Substitution confirms the solution.
Easy§ 02

7x + 1 = -6

Answer: x = -1

  1. Subtract 1 from both sides 7x = -7 Isolate the x term by removing the constant.
  2. Divide both sides by 7 x = -1 -7 ÷ 7 = -1.
  3. Verify 7(-1) + 1 = -6 ✓ Substitution confirms the solution.
Medium§ 03

8x − 32 = 4x + 0

Answer: x = 8

  1. Subtract 4x from both sides 4x − 32 = 0 Collect all x terms on one side.
  2. Add 32 to both sides 4x = 32 Move constants to the other side.
  3. Divide both sides by 4 x = 8 32 ÷ 4 = 8.
  4. Verify LHS = RHS = 32 ✓ Both sides equal the same value.
§ 04

Common mistakes

  • Adding or subtracting incorrectly when moving terms across the equals sign. Students write 3x + 7 = 22 becomes 3x = 22 + 7 = 29 instead of 3x = 22 - 7 = 15, forgetting to change the sign.
  • Dividing only one side by the coefficient. For 4x = 12, students write x = 12 ÷ 4 = 3 but forget to show 4x ÷ 4 = 12 ÷ 4, missing the balanced operation principle.
  • Incorrectly combining like terms when variables appear on both sides. In 5x - 3 = 2x + 9, students calculate 5x - 2x = 7x instead of 3x, then solve 7x = 12 getting x = 12/7 rather than x = 4.
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§ 05

Frequently asked questions

How do I know which side to collect the x terms on?
Choose the side with the larger coefficient to avoid negative coefficients. In 3x + 5 = 7x - 11, subtract 3x from both sides to get 5 = 4x - 11, keeping the positive 4x. This reduces arithmetic errors and makes the solution cleaner.
What's the difference between one-step and two-step equations?
One-step equations like x + 6 = 7 require just one operation to isolate x. Two-step equations like 3x + 7 = 22 need two operations: first subtract 7, then divide by 3. The coefficient creates the extra step.
Why do we check our answers by substitution?
Substitution catches arithmetic errors and confirms the solution satisfies the original equation. If x = 5 solves 3x + 7 = 22, then 3(5) + 7 should equal 22. This verification step prevents marking errors in GCSE exams.
How do I handle fractional coefficients like ½x + 3 = 8?
Multiply everything by the denominator first to clear fractions. For ½x + 3 = 8, multiply by 2: x + 6 = 16, then solve normally to get x = 10. Always clear fractions early to simplify calculations.
What if my answer is negative or a fraction?
Negative and fractional solutions are perfectly valid in linear equations. For 2x + 10 = 4, you get x = -3. For 3x = 2, you get x = ⅔. Always check these solutions work when substituted back into the original equation.
§ 06

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