← IGCSE / O Level Mathematics

IGCSE / O Level Mathematics · Topic 2 Algebra and graphs · Chapter 13 · pp. 126–146

Equations and inequalities

An equation is a balance: whatever you do to one side you do to the other. This chapter solves equations with one unknown, two unknowns at once, and an x² in them, then switches the = to < or > to describe a whole range of answers.

Learning objectives

  1. Solve linear equations, including brackets, fractions and the unknown on both sides.
  2. Construct an equation from a problem or diagram, solve it and check the answer.
  3. Solve a pair of simultaneous equations by elimination or substitution.
  4. Solve quadratic equations by factorising, by the quadratic formula and by completing the square (Extended).
  5. Solve a linear equation and a non-linear equation together (Extended).
  6. Solve linear and double inequalities and show them on a number line (Extended).

Book exercises: 13.1 to 13.13 and Student assessments 1 to 3 (pp. 126–146). Exercises 13.7 to 13.13 are Extended content.

Solving linear equations

🔑 Key ideas

  • Do the same thing to both sides: the equation stays true.
  • Undo the last operation first. Add undoes subtract; multiply undoes divide.
  • Unknown on both sides: collect the x terms on one side and the numbers on the other.
  • Fractions: multiply both sides by the denominator (or by the common multiple of both).
Brackets and fractions x terms to one side Numbers to the other side Divide by the x number Check in the original
The routine for every linear equation. Step 5 is the part students skip, and the part that catches mistakes.

📘 Example 1 · Solve 3x + 8 = 14

  1. Subtract 8 from both sides: 3x = 6.
  2. Divide both sides by 3: x = 2.
  3. Check: 3 × 2 + 8 = 14. ✔

📘 Example 2 · Solve 12 = 20 + 2x

  1. Subtract 20 from both sides: −8 = 2x.
  2. Divide both sides by 2: x = −4.
  3. Check: 20 + 2 × (−4) = 12. ✔

📘 Example 3 · Solve 4(x − 5) = 7(2x − 5)

  1. Expand both brackets: 4x − 20 = 14x − 35.
  2. Add 35 to both sides: 4x + 15 = 14x.
  3. Subtract 4x from both sides: 15 = 10x.
  4. Divide by 10: x = 1.5.
  5. Check: 4(1.5 − 5) = −14 and 7(3 − 5) = −14. ✔

📘 Example 4 · Solve = 3

  1. Multiply both sides by 2: x + 1 = 6.
  2. Subtract 1: x = 5.
  3. Check: (5 + 1) ÷ 2 = 3. ✔

📘 Example 5 · Solve =

  1. Cross-multiply (multiply by 4 and by 6): 6(2x + 3) = 4(4x − 2).
  2. Expand: 12x + 18 = 16x − 8.
  3. Subtract 12x and add 8: 26 = 4x.
  4. Divide by 4: x = 6.5.
  5. Check: (13 + 3) ÷ 4 = 4 and (26 − 2) ÷ 6 = 4. ✔

✏️ Try it

  1. Solve 5x − 7 = 18.
  2. Solve 6 − 2x = 3x − 9.
  3. Solve = 4.
Answers

1. 5x = 25, so x = 5 · 2. 15 = 5x, so x = 3 (check: 6 − 6 = 0 and 9 − 9 = 0) · 3. x − 1 = 12, so x = 13

Next: constructing equations →

Constructing equations

🔑 Key ideas

  • Pick a letter for the unknown, write what is equal to what, then solve.
  • Diagrams hide the equation in a fact. Use it:
FactGives the equation
Angles in a triangleadd up to 180°
Angles in a quadrilateraladd up to 360°
Angles in a pentagonadd up to 540°
Angles at a pointadd up to 360°
Vertically opposite anglesare equal
Isosceles triangletwo sides equal, two base angles equal
Rectanglearea = length × width, perimeter = 2 × (length + width)
90° (x − 30)° (x + 30)° The three angles must add to 180°. (x + 30) + (x − 30) + 90 = 180
Example 1 as a picture: the angle sum becomes the equation. The diagram is not drawn to scale.

📘 Example 1 · Find the angles of the triangle (x + 30)°, (x − 30)° and 90°

  1. Angles in a triangle add to 180°: (x + 30) + (x − 30) + 90 = 180.
  2. Collect: 2x + 90 = 180.
  3. Subtract 90: 2x = 90, so x = 45.
  4. The angles are 90°, 45 + 30 = 75° and 45 − 30 = 15°.
  5. Check: 90 + 75 + 15 = 180. ✔

📘 Example 2 · A quadrilateral has angles (4x + 30)°, (3x + 10)°, 3x° and (2x + 20)°

  1. Angles in a quadrilateral add to 360°: 4x + 30 + 3x + 10 + 3x + 2x + 20 = 360.
  2. Collect: 12x + 60 = 360.
  3. Subtract 60: 12x = 300, so x = 25.
  4. The angles are 130°, 85°, 75° and 70°.
  5. Check: 130 + 85 + 75 + 70 = 360. ✔

📘 Example 3 · A rectangle is 4 by (x + 1) and has area 24. Find x

  1. Area = length × width: 4(x + 1) = 24.
  2. Divide by 4: x + 1 = 6.
  3. Subtract 1: x = 5.
  4. Check: 4 × (5 + 1) = 24. ✔

📘 Example 4 · Add 3 to a number, then quadruple the result. The answer is 36. Find the number

  1. Call the number n. Add 3: n + 3. Quadruple it: 4(n + 3).
  2. The equation is 4(n + 3) = 36.
  3. Divide by 4: n + 3 = 9.
  4. Subtract 3: n = 6. Check: (6 + 3) × 4 = 36. ✔

✏️ Try it

  1. A triangle has angles x°, 2x° and 3x°. Find x and the three angles.
  2. Double a number, then subtract 9. The answer is 11. Find the number.
  3. A rectangle is 6 by (x + 2) and has area 54. Find x.
Answers

1. 6x = 180, so x = 30 and the angles are 30°, 60°, 90° · 2. 2n − 9 = 11, so n = 10 · 3. 6(x + 2) = 54, x + 2 = 9, x = 7

Next: simultaneous equations →

Simultaneous equations

🔑 Key ideas

  • Two unknowns need two equations. The answer is a pair (x, y) that makes both true.
  • Elimination: make the number in front of one letter the same size in both equations, then add or subtract to make that letter disappear.
  • Same signs: subtract. Different signs: add.
  • Put the answer back into an equation to find the other letter, then check in the other equation.
  • Substitution: if one equation says x = … or y = …, put it straight into the other.
Match one letter Add or subtract Solve for one letter Substitute back Check in both equations
The elimination method in five moves.
−112345678 −4−22468 (4, 2) x + y = 6 x − y = 2
Each equation is a straight line. The solution of the pair is where the lines cross: x = 4, y = 2.

📘 Example 1 · Solve x + y = 6 and x − y = 2

  1. The y terms are +y and −y: different signs, so add. (x + y) + (x − y) = 6 + 2.
  2. This gives 2x = 8, so x = 4.
  3. Put x = 4 into x + y = 6: 4 + y = 6, so y = 2.
  4. Check in x − y = 2: 4 − 2 = 2. ✔

📘 Example 2 · Solve 2x + y = 14 and x + y = 9

  1. Both y terms are +y: same sign, so subtract. (2x + y) − (x + y) = 14 − 9.
  2. This gives x = 5.
  3. Put x = 5 into x + y = 9: y = 4.
  4. Check in 2x + y = 14: 10 + 4 = 14. ✔

📘 Example 3 · Solve 2x + y = 7 and 3x + 2y = 12

  1. Label them (1) and (2). Multiply (1) by 2 so both have 2y: 4x + 2y = 14 (call it 3).
  2. Same signs, so subtract: (3) − (2) gives x = 2.
  3. Put x = 2 into (1): 4 + y = 7, so y = 3. The answer is x = 2, y = 3.
  4. Check in (2): 3 × 2 + 2 × 3 = 12. ✔

📘 Example 4 · Solve 2x − 3y = −3 and 3x + 2y = 15

  1. Neither number matches, so multiply both equations. To match y: (1) × 2 gives 4x − 6y = −6, and (2) × 3 gives 9x + 6y = 45.
  2. −6y and +6y have different signs, so add: 13x = 39, so x = 3.
  3. Put x = 3 into (1): 6 − 3y = −3, so −3y = −9 and y = 3.
  4. Check in (2): 3 × 3 + 2 × 3 = 15. ✔

📘 Example 5 · A man is three times as old as his son. Ten years ago he was five times as old. Find both ages

  1. Let the man be m and the son s. Now: m = 3s.
  2. Ten years ago: m − 10 = 5(s − 10), which is m − 10 = 5s − 50.
  3. Substitute m = 3s: 3s − 10 = 5s − 50.
  4. Collect: 40 = 2s, so s = 20 and m = 60.
  5. Check: ten years ago they were 50 and 10, and 50 = 5 × 10. ✔

🧪 Try it: elimination stepper

Type whole numbers for a x + b y = c in both equations. The tool picks the letter that needs the smaller multipliers (or choose one).

    ✏️ Try it

    1. Solve x + y = 10 and x − y = 4.
    2. Solve 3x + y = 11 and x + y = 5.
    3. Solve 3x + 2y = 16 and x = y + 2 by substitution.
    Answers

    1. Add: 2x = 14, so x = 7, y = 3 · 2. Subtract: 2x = 6, so x = 3, y = 2 · 3. 3(y + 2) + 2y = 16, so 5y = 10, y = 2 and x = 4 (check: 12 + 4 = 16)

    Next: factorising quadratics →

    Quadratic equations by factorising Extended

    🔑 Key ideas

    • A quadratic equation has an x² term: ax² + bx + c = 0.
    • If two numbers multiply to give 0, one of them is 0. So (x − 4)(x + 3) = 0 means x − 4 = 0 or x + 3 = 0.
    • To factorise x² + bx + c, find two numbers that multiply to c and add to b.
    • Always make one side 0 first. Never divide by x: you lose the answer x = 0.
    • A quadratic usually has two solutions: the places its graph crosses the x-axis.
    Rearrange to = 0 Factorise into brackets Set each bracket = 0 Solve both Check in the original
    Solving by factorising.
    −5−4−3−2−1123456 −12−8−448 x = −3 x = 4 y = x² − x − 12
    x² − x − 12 = (x − 4)(x + 3). The graph crosses the x-axis at the two solutions of x² − x − 12 = 0.

    📘 Example 1 · Solve x² + 7x + 12 = 0

    1. Find two numbers that multiply to 12 and add to 7: 3 and 4.
    2. Factorise: (x + 3)(x + 4) = 0.
    3. Either x + 3 = 0 or x + 4 = 0.
    4. x = −3 or x = −4.
    5. Check x = −3: 9 − 21 + 12 = 0. ✔

    📘 Example 2 · Solve x² + 5x = −6

    1. Make one side 0: add 6 to both sides to get x² + 5x + 6 = 0.
    2. Two numbers that multiply to 6 and add to 5: 2 and 3.
    3. Factorise: (x + 2)(x + 3) = 0.
    4. x = −2 or x = −3.

    📘 Example 3 · Solve x² − 6x = 0 and x² − 4 = 0

    1. x² − 6x = 0: take out the common factor x to get x(x − 6) = 0.
    2. Either x = 0 or x − 6 = 0, so x = 0 or x = 6.
    3. x² − 4 = 0 is a difference of two squares: (x − 2)(x + 2) = 0.
    4. So x = 2 or x = −2.

    📘 Example 4 · Solve 2x² + 5x + 3 = 0

    1. Multiply the first and last numbers: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 5: 2 and 3.
    2. Split the middle term: 2x² + 2x + 3x + 3 = 0.
    3. Factorise in pairs: 2x(x + 1) + 3(x + 1) = 0.
    4. Take out the common bracket: (2x + 3)(x + 1) = 0.
    5. x = −3/2 or x = −1.
    6. Check x = −1: 2 − 5 + 3 = 0. ✔

    📘 Example 5 · A rectangle is (x + 3) cm by x cm with area 10 cm². Find x

    1. Area = length × width: x(x + 3) = 10.
    2. Expand and make one side 0: x² + 3x − 10 = 0.
    3. Multiply to −10, add to 3: 5 and −2. So (x + 5)(x − 2) = 0.
    4. x = −5 or x = 2. A length cannot be negative, so x = 2.
    5. Check: 5 × 2 = 10. ✔

    ✏️ Try it

    1. Solve x² − 8x + 15 = 0.
    2. Solve x² + 3x = 28.
    3. Solve 2x² + 7x + 3 = 0.
    Answers

    1. (x − 3)(x − 5) = 0, so x = 3 or x = 5 · 2. x² + 3x − 28 = 0, (x + 7)(x − 4) = 0, so x = −7 or x = 4 · 3. (2x + 1)(x + 3) = 0, so x = −1/2 or x = −3

    Next: the quadratic formula →

    The quadratic formula Extended

    🔑 Key ideas

    • For ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) ÷ 2a.
    • The part under the root, b² − 4ac, is the discriminant. It tells you how many answers to expect.
    • Use it when the equation will not factorise. Answers are then usually given to 2 or 3 significant figures.
    • Put negative numbers in brackets: if b = −4, then b² = (−4)² = 16 and −b = 4.
    Rearrange to = 0 Read off a, b, c Work out b² − 4ac Put into the formula Two answers x = … or x = …
    Using the quadratic formula.
    1234 b² − 4ac = 4 > 0 two roots 1234 b² − 4ac = 0 one repeated root 1234 b² − 4ac = −4 < 0 no real roots
    The discriminant tells you what the graph does at the x-axis: crosses twice, just touches, or misses it.

    📘 Example 1 · Solve x² + 7x + 3 = 0 (3 s.f.)

    1. Read off a = 1, b = 7, c = 3.
    2. Discriminant: b² − 4ac = 49 − 12 = 37.
    3. x = (−7 ± √37) ÷ 2, and √37 = 6.083.
    4. x = (−7 + 6.083) ÷ 2 = −0.459.
    5. x = (−7 − 6.083) ÷ 2 = −6.54.

    📘 Example 2 · Solve x² − 4x − 2 = 0 (3 s.f.)

    1. Read off a = 1, b = −4, c = −2.
    2. Discriminant: (−4)² − 4 × 1 × (−2) = 16 + 8 = 24.
    3. x = (4 ± √24) ÷ 2, and √24 = 4.899.
    4. x = (4 + 4.899) ÷ 2 = 4.45.
    5. x = (4 − 4.899) ÷ 2 = −0.449.

    📘 Example 3 · Completing the square: solve x² − 4x − 2 = 0

    1. Half the x number is −2, and (x − 2)² = x² − 4x + 4.
    2. To get −2 instead of +4 take 6 away: x² − 4x − 2 = (x − 2)² − 6.
    3. So (x − 2)² − 6 = 0, which gives (x − 2)² = 6.
    4. Square root both sides: x − 2 = ±√6.
    5. x = 2 ± √6, that is x = 4.45 or x = −0.449 (3 s.f.). Same as Example 2. ✔

    🧪 Try it: quadratic formula solver

      One line, one curve Extended

      🔑 Key ideas

      • Where a line meets a curve, both equations give the same y. So set the two expressions for y equal.
      • That makes a quadratic. Solve it for x.
      • Put each x into the linear equation to get its y.
      • Two x values give two points (the line crosses the curve twice).
      −5−4−3−2−1123456 −8−4481216 (4, 11) (−3, −3) y = x² + x − 9 y = 2x + 3
      The line y = 2x + 3 cuts the curve y = x² + x − 9 at two points, (4, 11) and (−3, −3).

      📘 Example 4 · Solve y = 2x + 3 and y = x² + x − 9

      1. Both equal y, so set them equal: 2x + 3 = x² + x − 9.
      2. Collect on one side: x² − x − 12 = 0.
      3. Factorise: (x − 4)(x + 3) = 0, so x = 4 or x = −3.
      4. Use the line to find y. If x = 4, y = 2 × 4 + 3 = 11. If x = −3, y = 2 × (−3) + 3 = −3.
      5. Solutions: (4, 11) and (−3, −3).
      6. Check (4, 11) in the curve: 16 + 4 − 9 = 11. ✔

      📘 Example 5 · Solve y = 11 − x and y = 30/x

      1. Set equal: 11 − x = 30/x.
      2. Multiply both sides by x: 11x − x² = 30.
      3. Rearrange: x² − 11x + 30 = 0.
      4. Factorise: (x − 5)(x − 6) = 0, so x = 5 or x = 6.
      5. Use y = 11 − x. If x = 5, y = 6. If x = 6, y = 5.
      6. Solutions: (5, 6) and (6, 5). Check: 5 × 6 = 30. ✔

      ✏️ Try it

      1. Use the formula to solve x² + 3x − 3 = 0 to 2 d.p.
      2. Use the formula to solve 2x² − 3x − 4 = 0 to 2 d.p.
      3. How many real solutions does x² + 2x + 5 = 0 have?
      Answers

      1. b² − 4ac = 9 + 12 = 21, √21 = 4.583, so x = 0.79 or x = −3.79 · 2. b² − 4ac = 9 + 32 = 41, √41 = 6.403, so x = (3 + 6.403) ÷ 4 = 2.35 or x = (3 − 6.403) ÷ 4 = −0.85 · 3. 4 − 20 = −16 is negative, so there are none

      Next: inequalities →

      Linear inequalities Extended

      🔑 Key ideas

      • < less than, > greater than, ≤ less than or equal to, ≥ greater than or equal to.
      • Solve exactly like an equation, with one new rule: multiply or divide by a negative number and you must flip the sign.
      • On a number line: open circle for < and > (the number is not included); filled circle for ≤ and ≥ (included).
      • A double inequality such as 7 < 3x + 1 ≤ 13 is solved in one go, doing the same to all three parts.
      x > 2 −3 −2 −1 0 1 2 3 4 5 x ≥ 2 −3 −2 −1 0 1 2 3 4 5 x < −1 −3 −2 −1 0 1 2 3 4 5 x ≤ −1 −3 −2 −1 0 1 2 3 4 5 2 < x ≤ 4 −3 −2 −1 0 1 2 3 4 5
      Open circle: the end value is not a solution. Filled circle: it is. The thick part is every value that works.

      📘 Example 1 · Solve 15 + 3x < 6

      1. Subtract 15 from both sides: 3x < −9.
      2. Divide by 3 (positive, so the sign stays): x < −3.
      3. Number line: open circle at −3, thick line to the left.
      4. Check with x = −4: 15 − 12 = 3, and 3 < 6. ✔

      📘 Example 2 · Solve 17 ≥ 7x + 3

      1. Subtract 3 from both sides: 14 ≥ 7x.
      2. Divide by 7: 2 ≥ x, which is the same as x ≤ 2.
      3. Number line: filled circle at 2, thick line to the left.

      📘 Example 3 · Solve 9 − 4x ≥ 17

      1. Subtract 9 from both sides: −4x ≥ 8.
      2. Divide by −4. This is a negative number, so flip the sign: x ≤ −2.
      3. Number line: filled circle at −2, thick line to the left.
      4. Check with x = −3: 9 + 12 = 21, and 21 ≥ 17. ✔ (Check x = 0: 9 ≥ 17 is false, so 0 is not a solution. ✔)

      📘 Example 4 · Solve 7 < 3x + 1 ≤ 13

      1. Subtract 1 from all three parts: 6 < 3x ≤ 12.
      2. Divide all three parts by 3: 2 < x ≤ 4.
      3. Number line: open circle at 2, filled circle at 4, thick line between (the last row of the figure above).
      4. Integers that work: 3 and 4.

      ✏️ Try it

      1. Solve 2x + 5 > 13.
      2. Solve 7 − 3x ≤ 1.
      3. Solve 3 ≤ 2x − 1 < 9.
      Answers

      1. 2x > 8, so x > 4 · 2. −3x ≤ −6, divide by −3 and flip: x ≥ 2 · 3. Add 1: 4 ≤ 2x < 10, divide by 2: 2 ≤ x < 5

      Next: summary and quiz →

      Chapter summary

      • Linear equations: do the same to both sides. Brackets and fractions first, then x terms one side, numbers the other. Always check.
      • Constructing: use the fact (angle sums, area, ages) to write an equation, solve it, then answer in the units asked.
      • Simultaneous equations: match one letter, add if the signs differ and subtract if they match, solve, substitute back, check.
      • Quadratics by factorising: make it = 0, find two numbers that multiply to c and add to b, set each bracket to 0.
      • Quadratic formula: x = (−b ± √(b² − 4ac)) ÷ 2a. A negative discriminant means no real solutions.
      • Completing the square: x² − 4x − 2 = (x − 2)² − 6.
      • Line and curve: set the two y expressions equal, solve the quadratic, then find each y from the line.
      • Inequalities: solve like equations, but flip the sign when you multiply or divide by a negative. Open circle for < >, filled for ≤ ≥.

      📄 Practice questions

      The book's Exercises 13.1 to 13.13 and Student assessments 1 to 3 are in the printable assignment for this chapter.

      Mini-quiz

      1. Solve 3x + 8 = 14.
      2. Solve 4(x − 5) = 7(2x − 5).
      3. A triangle has angles (x + 30)°, (x − 30)° and 90°. What is x?
      4. Solve x + y = 8 and 4x + y = 23.
      5. To eliminate y from 5x − 3y = 1 and 3x + 4y = 18, multiply the equations by…
      6. Solve x² − x − 12 = 0.
      7. Solve x² − 6x = 0.
      8. For x² − 4x − 2 = 0, what is the discriminant b² − 4ac?
      9. x² + 2x + 5 = 0 has discriminant −16. How many real solutions does it have?
      10. Which is x² − 4x − 2 written by completing the square?
      11. Solve 9 − 4x ≥ 17.
      12. Which number line shows x > 2?

      Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 13 (Equations and inequalities). Explanations, the Try it answers, the tools and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.