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IGCSE Mathematics 0580 · Chapter 4 notes · ~21 min read

Equations & inequalities

Solve linear, simultaneous and quadratic equations and inequalities with confidence. Balance diagrams, graphs and step-by-step methods show you what each answer really means.

Linear equationsSimultaneous equationsQuadratic equationsQuadratic formulaCompleting the squareInequalities
RSby Riyaz Sir
mentorriyaz.in
No tag = Core (everyone)EXTENDED Extended onlyhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • An equation is a balance: whatever you do to one side you must do to the other. To solve, undo the operations in reverse order and check the answer.
  • Simultaneous equations are solved where two lines cross. Use elimination (make a letter match, then add or subtract) or substitution.
  • Quadratics are solved by factorising, the formula or completing the square. Inequalities are solved like equations, but flip the sign when multiplying or dividing by a negative.

By the end you can

  • Solve linear equations, including brackets, fractions and unknowns on both sides.
  • Form and solve equations from words.
  • Solve two simultaneous linear equations by graph, elimination and substitution.
  • Solve linear inequalities and show them on a number line.
  • Solve quadratic equations by factorising, using the formula and by completing the square. EXT
  • Solve simultaneous equations where one is linear and one is quadratic. EXT
  • Show inequalities as regions on a graph, and solve quadratic inequalities. EXT

1Linear equations

Solving step by step

An equation is a balance: do the same to both sidesxxx203x + 5 = 20startxxx153x = 15− 5 from both sidesx5x = 5÷ 3 on both sides
Keep the balance level: do exactly the same to both sides until x is on its own.

Equation — a statement that two expressions are equal. To solve it means to find the value(s) of the letter that make it true.

  • Order: (1) expand any brackets; (2) collect the x-terms on one side and the numbers on the other; (3) divide to get x.
  • Undo + and − first, then × and ÷. Reverse BIDMAS.
  • Check: put your answer back into the original equation. Both sides must match.
Worked example

Solve 3x + 5 = 20.

  1. subtract 5 from both sides: 3x = 15
  2. divide both sides by 3: x = 5
  3. check: 3(5) + 5 = 20 ✓
Worked example

Solve 4(2x − 3) = 20.

  1. expand: 8x − 12 = 20
  2. add 12: 8x = 32
  3. divide by 8: x = 4
  4. check: 4(8 − 3) = 20 ✓
Worked example

Solve 7x − 4 = 3x + 20.

  1. subtract 3x from both sides: 4x − 4 = 20
  2. add 4: 4x = 24
  3. divide by 4: x = 6
Worked example

Solve x3 + 2 = 7.

  1. subtract 2: x3 = 5
  2. multiply both sides by 3: x = 15

Unknowns on both sides and fractions

Unknown on both sides5x − 3 = 2x + 9− 2x3x − 3 = 9+ 33x = 12÷ 3x = 4check: 5(4) − 3 = 17 and 2(4) + 9 = 17 ✓With fractions(x + 1) ÷ 2 = (x − 2) ÷ 3× 6 (the LCM)3(x + 1) = 2(x − 2)expand3x + 3 = 2x − 4− 2x and − 3x = −7check: −6 ÷ 2 = −3 and −9 ÷ 3 = −3 ✓
With fractions, multiply every term by the lowest common multiple of the denominators to clear them.
  • Fractions: multiply both sides by the LCM of the denominators. Every term must be multiplied, not just one.
  • Keep the x-terms positive when you can: move the smaller x-term across. This avoids sign mistakes.
Worked example

Solve x + 32 − x − 14 = 3.

  1. multiply every term by 4: 2(x + 3) − (x − 1) = 12
  2. expand (watch the minus): 2x + 6 − x + 1 = 12
  3. collect: x + 7 = 12
  4. x = 5
  5. check: 8 ÷ 2 − 4 ÷ 4 = 4 − 1 = 3 ✓
Common mistakeExpanding −(x − 1) as −x − 1. The minus changes both signs: −(x − 1) = −x + 1.

Forming equations from words

  • Let a letter stand for the unknown, write what you know as an expression, set up an equation, solve it and answer the question.
Worked example

The sides of a triangle are x, x + 3 and 2x − 1 cm. The perimeter is 30 cm. Find the sides.

  1. perimeter: x + (x + 3) + (2x − 1) = 30
  2. 4x + 2 = 30
  3. 4x = 28, so x = 7
  4. sides: 7 cm, 10 cm, 13 cm
  5. (7 + 10 + 13 = 30 ✓)
Worked example

Sam is 3 times as old as his daughter. In 4 years the sum of their ages will be 52. How old is Sam now?

  1. daughter = d, Sam = 3d
  2. in 4 years: (d + 4) + (3d + 4) = 52
  3. 4d + 8 = 52, so d = 11
  4. Sam = 3 × 11 = 33 years old
Exam tipAlways answer the question asked: x is often not the final answer. Check by putting the values back into the words of the question.

2Simultaneous equations

Solving by graph

xy-2-1123456-11234560x + y = 5y = 2x − 1(2, 3)Each line is all thepoints that fit oneequation.Where the lines cross,both equations are true:x = 2 and y = 3check: 2 + 3 = 5 ✓2(2) − 1 = 3 ✓
Every point on a line fits that line's equation. The crossing point fits both.
  • Simultaneous means "at the same time": find the values of x and y that make both equations true.
  • Graph method: draw both lines. The coordinates of the crossing point are the solution.
  • Parallel lines never cross, so there is no solution.

Elimination and substitution

EliminationSubstitution(1) 3x + 2y = 12(2) x + y = 5(2) × 2: 2x + 2y = 10 (3)(1) − (3): x = 2put x = 2 in (2): 2 + y = 5y = 3(1) y = 2x − 1(2) x + y = 5replace y in (2):x + (2x − 1) = 53x = 6, so x = 2y = 2(2) − 1 = 3same signs: subtract, different: addone equation must give x or yBoth methods give x = 2 and y = 3.Check in the equation you did not use.
Elimination removes one letter by adding or subtracting. Substitution replaces one letter using the other equation.
  • Elimination: (1) make the coefficients of one letter the same size (multiply an equation); (2) same signs: subtract; different signs: add; (3) solve; (4) substitute back to find the other letter.
  • Substitution: use when one equation already gives y = … or x = …. Put it into the other equation.
  • Always check in the equation you did not use to find the second letter.
Worked example

Solve 2x + y = 7 and x − y = 2.

  1. the y terms have different signs: add the equations
  2. 3x = 9, so x = 3
  3. put in (1): 6 + y = 7, so y = 1
  4. check in (2): 3 − 1 = 2 ✓
Worked example

Solve 2x + 3y = 13 and 5x − 2y = 4.

  1. (1) × 2: 4x + 6y = 26
  2. (2) × 3: 15x − 6y = 12
  3. different signs: add: 19x = 38, so x = 2
  4. put in (1): 4 + 3y = 13, so y = 3
  5. check in (2): 10 − 6 = 4 ✓
Worked example

Solve x = y + 2 and 3x + y = 14.

  1. substitute x: 3(y + 2) + y = 14
  2. 4y + 6 = 14, so y = 2
  3. x = 2 + 2 = 4
  4. check: 12 + 2 = 14 ✓
Worked example

2 pens and 3 notebooks cost £9. 3 pens and 1 notebook cost £6.50. Find the cost of each.

  1. let a pen cost p and a notebook cost n: 2p + 3n = 9 and 3p + n = 6.5
  2. from (2): n = 6.5 − 3p
  3. substitute: 2p + 3(6.5 − 3p) = 9
  4. 2p + 19.5 − 9p = 9, so −7p = −10.5
  5. p = £1.50, n = £2.00

One linear and one quadraticEXTENDED

  • EXT Substitute the linear equation into the quadratic, solve the quadratic, then find y for each x. You normally get two pairs of answers.
Worked example

EXT Solve y = x + 1 and y = x² − 5.

  1. set equal: x² − 5 = x + 1
  2. x² − x − 6 = 0
  3. (x − 3)(x + 2) = 0
  4. x = 3 or x = −2
  5. y = x + 1: (3, 4) and (−2, −1)
Worked example

EXT Solve x² + y² = 25 and y = x + 1.

  1. x² + (x + 1)² = 25
  2. 2x² + 2x + 1 = 25
  3. x² + x − 12 = 0, so (x + 4)(x − 3) = 0
  4. x = −4 gives y = −3; x = 3 gives y = 4
  5. (−4, −3) and (3, 4)

3Quadratic equations EXT

Solving by factorising

Solving by factorising: it must equal 0 first1Rearrange so one side is 0x² − 2x = 15 → x² − 2x − 15 = 02Factorise(x − 5)(x + 3) = 03One bracket must be 0x − 5 = 0 or x + 3 = 04Solve each onex = 5 or x = −3
A product is zero only if one of its factors is zero. That is why the right side must be 0.

Quadratic equation — an equation where the highest power of x is x², written ax² + bx + c = 0. It usually has two solutions (roots).

  • EXT Always make one side zero first. Then factorise (see Algebraic manipulation), set each bracket = 0 and solve.
Worked example

EXT Solve x² − 2x − 15 = 0.

  1. factorise: (x − 5)(x + 3) = 0
  2. x − 5 = 0 or x + 3 = 0
  3. x = 5 or x = −3
Worked example

EXT Solve 2x² + 5x − 3 = 0.

  1. a × c = −6, numbers that add to 5: +6 and −1
  2. split: 2x² + 6x − x − 3 = 0
  3. 2x(x + 3) − 1(x + 3) = 0
  4. (2x − 1)(x + 3) = 0
  5. x = 12 or x = −3
Worked example

EXT Solve x² = 7x.

  1. make one side zero: x² − 7x = 0
  2. factorise: x(x − 7) = 0
  3. x = 0 or x = 7
Common mistakeEXT Dividing both sides of x² = 7x by x. That loses the solution x = 0. Always move everything to one side and factorise instead.

Roots on a graph

xy-2-11234-5-4-3-2-11234560y = x² − 2x − 3y = (x + 1)(x − 3)roots x = −1 and x = 3(where the graph crosses y = 0)turning point (1, −4)on the line of symmetry x = 1y-intercept (0, −3)(put x = 0)
The solutions of a quadratic are where its graph crosses the x-axis. The turning point lies halfway between them.
  • EXT The roots are the x-intercepts (where y = 0). The graph is symmetrical about the vertical line through the turning point.
  • EXT The turning point is at the midpoint of the roots: (−1 + 3) ÷ 2 = 1.

The quadratic formula

For ax² + bx + c = 0x =−b ± √(b² − 4ac)2aSolve 2x² + 3x − 4 = 0a = 2, b = 3, c = −4x = (−3 ± √(9 + 32)) ÷ 4x = (−3 ± √41) ÷ 4x = 0.851 or x = −2.3511. write a, b and c with their signs (here c = −4)2. put b² − 4ac in brackets: 3² − 4 × 2 × (−4) = 413. the ± gives two answers: use + then − (3 s.f.)
Write down a, b and c with their signs, substitute carefully, and use ± to get both roots.
x = −b ± √(b² − 4ac)2a
for ax² + bx + c = 0. Use brackets for any negative number
  • EXT Use the formula when the quadratic does not factorise or the question asks for answers to decimal places or significant figures.
  • EXT Put b² − 4ac in brackets on the calculator and keep the whole fraction line covered: (−b ± √(…)) ÷ (2a).
Worked example

EXT Solve 2x² + 3x − 4 = 0. Give answers to 3 decimal places.

  1. a = 2, b = 3, c = −4
  2. b² − 4ac = 9 − 4 × 2 × (−4) = 9 + 32 = 41
  3. x = −3 ± √414
  4. x = 0.851 or x = −2.351
Worked example

EXT Solve x² − 4x − 3 = 0 to 3 significant figures.

  1. a = 1, b = −4, c = −3
  2. b² − 4ac = 16 + 12 = 28
  3. x = 4 ± √282
  4. x = 4.65 or x = −0.646
Common mistakeEXT Writing −b as −4 when b = −4. If b is negative, −b is positive: −(−4) = +4.

How many solutions?

b² − 4ac > 0two different rootscrosses the x-axis twiceb² − 4ac = 0one repeated roottouches the x-axisb² − 4ac < 0no real rootsmisses the x-axis
The sign of b² − 4ac, the part under the square root, tells you how many times the graph meets the x-axis.
  • EXT b² − 4ac > 0: two different roots. b² − 4ac = 0: one repeated root. b² − 4ac < 0: the square root of a negative number does not exist, so no real roots.

Completing the square

x²3x3x9x3x3missing corner:3 × 3 = 9(x + 3)² = x² + 6x + 9so x² + 6x = (x + 3)² − 9rule: halve the number in front of x,then subtract its squarex² + 6x + 2 = (x + 3)² − 9 + 2= (x + 3)² − 7minimum point (−3, −7)
Half the coefficient of x goes inside the bracket, and its square is subtracted so the value does not change.
  • EXT Form: x² + bx + c = (x + b2)² − (b2)² + c. The turning point is (−b2, q) when the answer is (x + p)² + q.
Worked example

EXT Write x² + 6x + 2 in the form (x + p)² + q.

  1. halve 6: p = 3
  2. (x + 3)² = x² + 6x + 9, so subtract 9
  3. x² + 6x + 2 = (x + 3)² − 9 + 2
  4. = (x + 3)² − 7
Worked example

EXT Write x² − 8x + 3 in the form (x + p)² + q.

  1. halve −8: p = −4
  2. (x − 4)² = x² − 8x + 16, so subtract 16
  3. = (x − 4)² − 16 + 3
  4. = (x − 4)² − 13
Worked example

EXT Solve x² + 6x − 2 = 0 by completing the square. Give answers to 3 s.f.

  1. (x + 3)² − 9 − 2 = 0
  2. (x + 3)² = 11
  3. x + 3 = ±√11
  4. x = −3 ± 3.3166…
  5. x = 0.317 or x = −6.32
Worked example

EXT Write 2x² + 8x + 5 in the form a(x + p)² + q.

  1. take out 2 from the x terms: 2(x² + 4x) + 5
  2. complete the square: 2[(x + 2)² − 4] + 5
  3. = 2(x + 2)² − 8 + 5
  4. = 2(x + 2)² − 3 (turning point (−2, −3))

Problems that lead to quadratics

Worked example

EXT A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm². Find x.

  1. area: x(x + 3) = 40
  2. x² + 3x − 40 = 0
  3. (x + 8)(x − 5) = 0
  4. x = −8 or x = 5
  5. a length cannot be negative, so x = 5: the rectangle is 5 cm by 8 cm

4Inequalities

Number lines

x > 2−3−2−1012345x ≤ −1−3−2−1012345−2 ≤ x < 3−3−2−1012345open circle = not included (< or >) filled circle = included (≤ or ≥)
An open circle means the number itself is not included. A filled circle means it is included.
  • Symbols: > greater than, < less than, ≥ greater than or equal to, ≤ less than or equal to.
  • Solve like an equation, then show the answer on a number line: open circle for < and >, filled circle for ≤ and ≥.
  • A double inequality such as −2 ≤ x < 3 gives a segment on the number line.
Worked example

Solve 3x − 4 ≤ 11 and show the answer on a number line.

  1. add 4: 3x ≤ 15
  2. divide by 3: x ≤ 5
  3. filled circle at 5 and an arrow pointing left
Worked example

Solve −2 < 2x + 1 ≤ 9. List the integer values of x.

  1. subtract 1 from all three parts: −3 < 2x ≤ 8
  2. divide all three by 2: −1.5 < x ≤ 4
  3. integers: −1, 0, 1, 2, 3, 4

Multiplying or dividing by a negative

Why the sign flips−5−4−3−2−101234524−2−42 < 4 is truemultiply by −1: the order reverses−2 > −4Solve −2x < 6divide both sides by −2and FLIP the signx > −3test x = 0: 0 < 6 ✓test x = −4: 8 < 6 ✗(−4 is not > −3)
A negative number turns the number line around, so the inequality sign must turn around too.

Flip rule — when you multiply or divide an inequality by a negative number, reverse the sign: < becomes > and ≤ becomes ≥.

Worked example

Solve −2x < 6.

  1. divide both sides by −2 and flip the sign
  2. x > −3
  3. test x = 0: −2(0) = 0, and 0 < 6 ✓
Worked example

Solve 5 − 3x ≥ 14.

  1. subtract 5: −3x ≥ 9
  2. divide by −3 and flip: x ≤ −3
Memory trickTo avoid flipping, move the x term to the positive side instead: 5 − 3x ≥ 14 becomes 5 − 14 ≥ 3x, so −9 ≥ 3x, so −3 ≥ x, which is x ≤ −3.

Graphical inequalitiesEXTENDED

xy-2-1123456-11234560Ry ≤ x + 2 (solid)y > 1 (dashed)x + y < 6 (dashed)solid line: ≤ or ≥dashed line: < or >test a point, e.g. (2, 2):2 ≤ 4 ✓ 2 > 1 ✓ 4 < 6 ✓so R is the triangle
Draw each boundary line, solid for ≤ or ≥ and dashed for < or >, then shade or label the region that satisfies all the inequalities.
  • EXT Boundary lines: use solid lines for ≤ and ≥ (points on the line count) and dashed lines for < and >.
  • EXT Which side? Test a point that is not on the line, such as (0, 0) or a point inside the region.
Worked example

EXT The region R is bounded by y = x + 2 (solid), y = 1 (dashed) and x + y = 6 (dashed). Write down the three inequalities.

  1. test the point (2, 2) inside R
  2. y = x + 2 is solid and 2 ≤ 4, so y ≤ x + 2
  3. y = 1 is dashed and 2 > 1, so y > 1
  4. x + y = 6 is dashed and 4 < 6, so x + y < 6

Quadratic inequalitiesEXTENDED

xy-3-2-11234-7-6-5-4-3-2-1123450x² − x − 6 < 0below the x-axis−2 < x < 3x² − x − 6 > 0above the x-axisx < −2 or x > 3roots: x = −2, 3(open circles: < and >do not include them)
Find the roots, sketch the curve, and read off where it is below the axis (less than 0) or above it (greater than 0).
  • EXT Method: (1) solve the equation to find the roots; (2) sketch the parabola (U shape when the x² term is positive); (3) read the x-values where the graph is below or above the axis.
Worked example

EXT Solve x² − x − 6 < 0.

  1. factorise: (x − 3)(x + 2) = 0, so roots x = 3 and x = −2
  2. U-shaped curve, and we want y below the axis
  3. −2 < x < 3
Worked example

EXT Solve x² − 9 ≥ 0.

  1. roots: x = −3 and x = 3
  2. U-shaped curve, and we want y on or above the axis
  3. x ≤ −3 or x ≥ 3
Common mistakeEXT Writing "x < −3 or x > 3" as −3 > x > 3. That is impossible. Two separate regions need the word "or".

5Test yourself

Test yourself — cover the answers first

Quick check

  1. Solve 5x − 7 = 18.
  2. Solve 3(x + 4) = 27.
  3. Solve 9 − 2x = 3x − 1.
  4. Solve x4 + 3 = 8.
  5. Solve x + y = 9 and x − y = 3.
  6. Solve y = 2x and x + y = 12.
  7. Solve 4x − 1 > 11.
  8. Solve −3x ≤ 12.
  9. List the integers that satisfy −2 ≤ x < 2.
  10. EXT Solve x² − 5x + 6 = 0.
  11. EXT Solve x² = 9x.
  12. EXT Solve 2x² − 7x + 3 = 0.
  13. EXT Solve x² + 2x − 4 = 0, giving answers to 2 decimal places.
  14. EXT Write x² + 10x + 7 in the form (x + p)² + q.
  15. EXT Solve x² − 16 < 0.
  16. EXT Solve x² − 4 ≥ 0.

Exam-style

  1. Solve 4(x − 3) = 2x + 7. [3]
  2. Solve the simultaneous equations 3x + 2y = 7 and 5x − 2y = 9. [3]
  3. 2 adult tickets and 3 child tickets cost £35. 1 adult ticket and 4 child tickets cost £30. Find the cost of one adult ticket and one child ticket. [4]
  4. Solve 3 − 2x < 11 and show the solution on a number line. [3]
  5. EXT Solve x² − 3x − 10 = 0. [3]
  6. EXT Solve 3x² + 4x − 5 = 0, giving your answers correct to 2 decimal places. [4]
  7. EXT Write x² − 6x + 1 in the form (x − p)² − q, and hence solve x² − 6x + 1 = 0 to 2 decimal places. [4]
  8. EXT A rectangle has width x cm and length (x + 4) cm. The area is 45 cm². Find the dimensions. [4]
  9. EXT Solve x² − 2x − 8 > 0. [3]
Show answers
  1. x = 5
  2. x = 5
  3. x = 2
  4. x = 20
  5. x = 6, y = 3
  6. x = 4, y = 8
  7. x > 3
  8. x ≥ −4
  9. −2, −1, 0, 1
  10. x = 2 or x = 3
  11. x = 0 or x = 9
  12. x = 3 or x = 0.5
  13. x = 1.24 or x = −3.24
  14. (x + 5)² − 18
  15. −4 < x < 4
  16. x ≤ −2 or x ≥ 2
  17. 4x − 12 = 2x + 7
    2x = 19
    x = 9.5
  18. add the equations: 8x = 16, so x = 2
    6 + 2y = 7, so y = 0.5
  19. let adult = a and child = c: 2a + 3c = 35 and a + 4c = 30
    a = 30 − 4c, so 2(30 − 4c) + 3c = 35
    60 − 5c = 35, so c = 5
    a = 30 − 20 = 10: adult £10, child £5
  20. −2x < 8
    divide by −2 and flip the sign: x > −4
    open circle at −4 with an arrow pointing right
  21. (x − 5)(x + 2) = 0
    x = 5 or x = −2
  22. a = 3, b = 4, c = −5: b² − 4ac = 16 + 60 = 76
    x = (−4 ± √76) ÷ 6
    x = 0.79 or x = −2.12
  23. (x − 3)² − 9 + 1 = (x − 3)² − 8
    (x − 3)² = 8
    x = 3 ± √8
    x = 5.83 or x = 0.17
  24. x(x + 4) = 45, so x² + 4x − 45 = 0
    (x + 9)(x − 5) = 0
    x = 5 (reject −9)
    5 cm by 9 cm
  25. roots from (x − 4)(x + 2) = 0: x = 4 and x = −2
    U-shaped graph, above the axis outside the roots
    x < −2 or x > 4
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Linear equationsSimultaneous equationsQuadratic equationsInequalities
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