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HomeLessonsGrade 9
Grade 9 · Equations · 15 min read

Solving Linear Equations: Every Type in the Exam

One balancing method solves every linear equation. Here it is applied to all the exam types — one-step, two-step, variables on both sides, brackets, negatives and fractions — plus word problems and a full worksheet.

An equation says two things are equal. Think of it as a balance scale: whatever you do to one side, you must do to the other. Your goal is always the same — get the variable (usually x) alone on one side. This guide works through every type of linear equation an exam can set, from the simplest to fractions and word problems.

1The order that always works

2x + 3 11 = whatever you do to one side, do to the other
An equation is a balance. Whatever you do to one side you must do to the other, or it tips over.
  • If there are fractions, multiply every term by the common denominator to clear them.
  • Remove brackets by expanding.
  • Get all the x-terms on one side and all the plain numbers on the other.
  • Combine like terms on each side.
  • Divide both sides by the number in front of x.
  • Check by substituting your answer back into the original equation.

2Type 1: One-step equations

Worked ExampleExample 1a: solve x + 7 = 12
  1. 1
    Subtract 7 from both sides: x = 12 − 7.
    Undo the +7 to isolate x.
  2. 2
    x = 5.
    Check: 5 + 7 = 12 ✓
Worked ExampleExample 1b: solve 3x = 21
  1. 1
    Divide both sides by 3: x = 21 ÷ 3.
    Undo the ×3.
  2. 2
    x = 7.
    Check: 3 × 7 = 21 ✓

3Type 2: Two-step equations

Worked ExampleExample 2: solve 2x + 5 = 17
  1. 1
    Subtract 5 from both sides: 2x = 12.
    Deal with the +5 first (undo addition before multiplication).
  2. 2
    Divide both sides by 2: x = 6.
    Undo the ×2 last.
  3. 3
    Check: 2(6) + 5 = 17 ✓
    Substitute back.

4Type 3: Variable on both sides

Worked ExampleExample 3: solve 5x − 3 = 2x + 9
  1. 1
    Subtract 2x from both sides: 3x − 3 = 9.
    Gather the x-terms on the side with more x's.
  2. 2
    Add 3 to both sides: 3x = 12.
    Move the numbers to the other side.
  3. 3
    Divide by 3: x = 4.
    Check: 5(4) − 3 = 17 and 2(4) + 9 = 17 ✓

5Type 4: Equations with brackets

Worked ExampleExample 4: solve 3(x − 2) = x + 8
  1. 1
    Expand the bracket: 3x − 6 = x + 8.
    Multiply the 3 into both terms inside.
  2. 2
    Subtract x: 2x − 6 = 8.
    Gather x-terms on the left.
  3. 3
    Add 6: 2x = 14, then divide by 2: x = 7.
    Check: 3(7 − 2) = 15 and 7 + 8 = 15 ✓

6Type 5: Negative coefficient of x

Worked ExampleExample 5: solve 8 − 2x = 14
  1. 1
    Subtract 8 from both sides: −2x = 6.
    Move the number away from the x-term.
  2. 2
    Divide both sides by −2: x = −3.
    Dividing by a negative flips the sign of the answer.
  3. 3
    Check: 8 − 2(−3) = 8 + 6 = 14 ✓
    Two negatives make a plus.

7Type 6: Equations with fractions

Clear the fractions first by multiplying every term by the lowest common denominator (LCD). Then solve as normal.

Worked ExampleExample 6: solve x⁄2 + x⁄3 = 5
  1. 1
    LCD of 2 and 3 is 6. Multiply every term by 6.
    This removes the denominators entirely.
  2. 2
    6 · x⁄2 + 6 · x⁄3 = 6 · 5 → 3x + 2x = 30.
    6÷2 = 3 and 6÷3 = 2.
  3. 3
    Combine: 5x = 30, so x = 6.
    Check: 6⁄2 + 6⁄3 = 3 + 2 = 5 ✓

8Type 7: Word problems (set up the equation)

Worked ExampleExample 7: 'Three times a number, increased by 4, is 19. Find the number.'
  1. 1
    Let the number be x. Translate the words: 3x + 4 = 19.
    'Three times a number' = 3x; 'increased by 4' = + 4; 'is' = equals.
  2. 2
    Subtract 4: 3x = 15.
    Two-step method from here.
  3. 3
    Divide by 3: x = 5.
    The number is 5. Check: 3(5) + 4 = 19 ✓
💡

Show every line. Markers award method marks for the balancing steps, not just the final answer — and a neat column of steps makes your own slips easy to spot.

⚠️

When you move a term across the equals sign, its sign changes. Moving +6 to the other side makes it −6. Picturing 'subtract 6 from both sides' keeps the sign automatic.

This same balancing idea scales up to quadratic, simultaneous and trigonometric equations in the senior grades — so mastering these seven types now pays off for years. Try the worksheet below.

Practice exercises

Work each one out, then click to reveal the answer.

  1. 1
    Solve x + 9 = 15 (one-step)
    Show answer ▾
    x = 15 − 9 = 6
  2. 2
    Solve 4x = 32 (one-step)
    Show answer ▾
    x = 32 ÷ 4 = 8
  3. 3
    Solve 3x + 7 = 22 (two-step)
    Show answer ▾
    3x = 15 → x = 5
  4. 4
    Solve 5x − 4 = 21 (two-step)
    Show answer ▾
    5x = 25 → x = 5
  5. 5
    Solve 6x − 5 = 2x + 11 (both sides)
    Show answer ▾
    4x = 16 → x = 4
  6. 6
    Solve 7x + 3 = 4x + 18 (both sides)
    Show answer ▾
    3x = 15 → x = 5
  7. 7
    Solve 2(x + 4) = 18 (brackets)
    Show answer ▾
    2x + 8 = 18 → 2x = 10 → x = 5
  8. 8
    Solve 5(x − 3) = 2(x + 3) (brackets)
    Show answer ▾
    5x − 15 = 2x + 6 → 3x = 21 → x = 7
  9. 9
    Solve 10 − 3x = 1 (negative x)
    Show answer ▾
    −3x = −9 → x = 3
  10. 10
    Solve 12 − 5x = 2x − 2 (negative + both sides)
    Show answer ▾
    12 + 2 = 2x + 5x → 14 = 7x → x = 2
  11. 11
    Solve x⁄3 + x⁄6 = 3 (fractions)
    Show answer ▾
    ×6: 2x + x = 18 → 3x = 18 → x = 6
  12. 12
    Solve x⁄4 − 1 = 2 (fractions)
    Show answer ▾
    x⁄4 = 3 → x = 12
  13. 13
    A number doubled and then increased by 5 gives 23. Find it. (word problem)
    Show answer ▾
    2x + 5 = 23 → 2x = 18 → x = 9
  14. 14
    Five less than four times a number is 27. Find it. (word problem)
    Show answer ▾
    4x − 5 = 27 → 4x = 32 → x = 8
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