AQA GCSE Mathematics 8300: Solving equations and inequalities – Study Guide

This study guide teaches subsection 3.2.3, Solving equations and inequalities (specification references A17 to A22), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Linear equations, factorising quadratics, linear/linear simultaneous equations, forming equations and linear inequalities are Foundation content, so they appear at both tiers. Completing the square, the quadratic formula, linear/quadratic simultaneous equations, iteration (A20) and quadratic inequalities are Higher tier only and are labelled below.

The Algebra study guide gives the overview of all of Topic 2; this page goes deeper on 3.2.3 only. Use it with the solving equations revision notes and the solving equations practice questions. The course hub is AQA GCSE Mathematics, the printable checklist lists every statement, and the free diagnostics show where to start.

What this subsection covers

Spec	What you must be able to do	Tier
A17	Solve linear equations in one unknown, including brackets and the unknown on both sides; find approximate solutions from a graph	Both
A18	Solve quadratic equations by factorising; find approximate solutions from a graph	Both
A18	Quadratics that need rearranging; completing the square; the quadratic formula	Higher tier only
A19	Solve two linear simultaneous equations algebraically; find approximate solutions from a graph	Both
A19	Linear/quadratic simultaneous equations	Higher tier only
A20	Find approximate solutions numerically by iteration, using suffix notation	Higher tier only
A21	Translate situations into expressions or formulae; derive equations, solve them and interpret the answer	Both
A22	Solve linear inequalities in one variable and show them on a number line	Both
A22	Linear inequalities in two variables, quadratic inequalities, set notation, regions on a graph	Higher tier only

Any of this can appear on the non-calculator Paper 1 or the calculator Papers 2 and 3.

A17: Linear equations

Do the same operation to both sides until the unknown is alone. The specification note says “including use of brackets”, so expect to expand first.

Method

1. Expand any brackets.
2. Collect the unknown on the side with the larger coefficient.
3. Collect the numbers on the other side.
4. Divide by the coefficient.

Worked example 1. Solve 4(2x − 3) = 3(x + 6).

8x - 12 = 3x + 18      expand both brackets
5x - 12 = 18           subtract 3x from both sides
5x = 30                add 12
x = 6                  divide by 5
Check: 4(12 - 3) = 36 and 3(6 + 6) = 36

From a graph. To solve 2x + 1 = 6, draw y = 2x + 1 and y = 6 and read the x-coordinate where they cross. Graph readings are approximate.

A18: Quadratic equations

Get zero on one side of ax² + bx + c = 0 before you do anything else.

By factorising (both tiers)

If two factors multiply to give 0, one of them must be 0.

Worked example 2. Solve x² + 3x − 28 = 0.

Need two numbers that multiply to -28 and add to +3: +7 and -4
(x + 7)(x - 4) = 0
x + 7 = 0  or  x - 4 = 0
x = -7  or  x = 4

Two special cases come up often:

• Common factor: x² − 9x = 0 gives x(x − 9) = 0, so x = 0 or x = 9. Never divide both sides by x; you lose the solution x = 0.
• Difference of two squares: x² − 49 = 0 gives (x − 7)(x + 7) = 0, so x = 7 or x = −7.

Equations that need rearranging (Higher tier only)

Worked example 3. Solve 3x² = 5x + 2.

3x^2 - 5x - 2 = 0                rearrange so one side is 0
(3x + 1)(x - 2) = 0              factorise ax^2 + bx + c
x = -1/3  or  x = 2

Completing the square (Higher tier only)

Write x² + bx + c as (x + b/2)² − (b/2)² + c, then solve.

Worked example 4. Solve x² − 6x + 4 = 0, giving exact answers.

x^2 - 6x + 4 = (x - 3)^2 - 9 + 4 = (x - 3)^2 - 5
(x - 3)^2 - 5 = 0
(x - 3)^2 = 5
x - 3 = ±√5
x = 3 + √5  or  x = 3 - √5

The same form gives the turning point of y = x² − 6x + 4 at (3, −5) (reference A11).

The quadratic formula (Higher tier only)

For ax² + bx + c = 0, where a ≠ 0:

x = (−b ± √(b² − 4ac)) / (2a)

The specification’s appendix lists this among the formulae you must know: it is not given in the exam.

Worked example 5. Solve 3x² + 7x − 5 = 0, giving answers to 2 decimal places.

a = 3, b = 7, c = -5
b^2 - 4ac = 49 - 4(3)(-5) = 49 + 60 = 109
x = (-7 ± √109) / 6
x = 0.57  or  x = -2.91   (2 d.p.)

Write the substitution down first: it can earn method marks even if the final answer is wrong.

Approximate solutions from a graph (both tiers)

The solutions of ax² + bx + c = 0 are the x-coordinates where y = ax² + bx + c crosses the x-axis (the roots). To solve x² − 2x − 4 = 1, draw y = x² − 2x − 4 and the line y = 1 and read off where they cross: about x = −1.4 and x = 3.4. (The exact values are 1 ± √6, which round to −1.45 and 3.45.) The readings should sit equally either side of the line of symmetry, x = 1.

A19: Simultaneous equations

Linear/linear by elimination (both tiers)

Worked example 6. Solve 3x + 2y = 17 and 5x − 2y = 7.

The y terms are +2y and -2y, so ADD the equations
8x = 24
x = 3
Substitute into the first: 9 + 2y = 17, so 2y = 8, y = 4
Check in the second: 15 - 8 = 7

When no coefficients match, multiply one or both equations first.

Worked example 7. Solve 2x + 3y = 12 and 5x + 4y = 23.

Multiply the first by 4:   8x + 12y = 48
Multiply the second by 3:  15x + 12y = 69
Subtract:                  7x = 21, so x = 3
Substitute: 6 + 3y = 12, so y = 2

Rule of thumb: same signs, subtract; different signs, add.

Linear/quadratic by substitution (Higher tier only)

Rearrange the linear equation to make one letter the subject, then substitute into the quadratic.

Worked example 8. Solve x² + y² = 25 and y = x + 1.

x^2 + (x + 1)^2 = 25
x^2 + x^2 + 2x + 1 = 25
2x^2 + 2x - 24 = 0
x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x = -4  or  x = 3
y = x + 1 gives y = -3  or  y = 4
Solutions: x = -4, y = -3  and  x = 3, y = 4

Pair each x with its own y.

From a graph (both tiers)

The solution is the point where the two graphs cross. A line can meet a curve twice, once (touching) or not at all.

A20: Iteration (Higher tier only)

Iteration finds an approximate solution by feeding each answer back into a formula. The specification note asks for suffix notation in recursive formulae: xₙ₊₁ means the next value and xₙ the current one.

Step 1: locate the root by a change of sign. For x³ − 5x − 3 = 0, let f(x) = x³ − 5x − 3.

f(2) = 8 - 10 - 3 = -5
f(3) = 27 - 15 - 3 = 9
The sign changes, so there is a solution between x = 2 and x = 3

Step 2: rearrange into an iterative formula.

x^3 = 5x + 3
x = ∛(5x + 3)
So xₙ₊₁ = ∛(5xₙ + 3)

Step 3: iterate. Start with x₀ = 2.5. On a calculator, type 2.5, press =, then type ∛(5 × ANS + 3) and press = repeatedly.

x₁ = 2.4933...
x₂ = 2.4915...
x₃ = 2.4910...

The values are settling at 2.49 to 2 decimal places. To confirm, check the sign change across the rounding interval: f(2.485) = −0.0796 and f(2.495) = 0.0564. The sign changes, so the solution is 2.49 to 2 d.p.

Write every iterate to at least 4 decimal places.

A21: Forming and solving equations

Turn words or a diagram into algebra, solve, then answer in context. The specification note includes geometrical problems and problems set in context.

Worked example 9 (geometry). A rectangle is 4 cm longer than it is wide. Its area is 45 cm². Find its width.

Let the width be x cm, so the length is (x + 4) cm
x(x + 4) = 45
x^2 + 4x - 45 = 0
(x + 9)(x - 5) = 0
x = -9  or  x = 5
A length cannot be negative, so the width is 5 cm

The rejection step is part of “interpret the solution”. Say why you reject −9.

Worked example 10 (two equations). 3 adult tickets and 2 child tickets cost £39. 2 adult tickets and 5 child tickets cost £48. Find the price of each.

Let an adult ticket be £a and a child ticket £c
3a + 2c = 39
2a + 5c = 48
Multiply the first by 2:  6a + 4c = 78
Multiply the second by 3: 6a + 15c = 144
Subtract: 11c = 66, so c = 6
3a + 12 = 39, so a = 9
Adult £9, child £6

Define your letters and finish with a sentence that answers the question.

A22: Inequalities

Linear inequalities in one variable (both tiers)

Solve them like equations, with one extra rule: if you multiply or divide by a negative number, reverse the inequality sign.

Worked example 11. Solve 5 − 2x < 11.

-2x < 6
x > -3      dividing by -2 reverses < to >

Worked example 12. Find the integers n that satisfy −3 ≤ 2n + 1 < 7.

-4 ≤ 2n < 6       subtract 1 from all three parts
-2 ≤ n < 3        divide all three parts by 2
Integers: -2, -1, 0, 1, 2

Number lines. The specification sets the convention: an open circle for a strict inequality (< or >) and a closed circle for an included boundary (≤ or ≥). For −2 ≤ n < 3, draw a closed circle at −2, an open circle at 3, and a line joining them.

Set notation (Higher tier only)

x > −3 can be written {x : x > −3}, read as “the set of x such that x is greater than −3”. Two separate parts join with the union symbol ∪.

Quadratic inequalities (Higher tier only)

Find the roots, sketch the parabola, then read off which part you need.

Worked example 13. Solve x² − x − 12 > 0.

x^2 - x - 12 = (x - 4)(x + 3), roots x = -3 and x = 4
y = x^2 - x - 12 is a U-shape, so it is above the x-axis
outside the roots
x < -3  or  x > 4
Set notation: {x : x < -3} ∪ {x : x > 4}

For x² ≤ 16, the answer is −4 ≤ x ≤ 4. Writing only x ≤ 4 misses the negative root.

Regions on a graph (Higher tier only)

An inequality in two variables describes a region. Draw each boundary line: dashed for a strict inequality and solid for an included one, as the specification requires. Then shade or label the region.

For y ≥ 1, y < 2x and x + y ≤ 6: y = 1 and x + y = 6 are solid lines, and y = 2x is dashed. The point (3, 3) is in the region, because it lies on the solid line x + y = 6. The point (2, 4) is not, because it lies on the dashed line y = 2x and 4 < 4 is false.

Calculator or not?

On Paper 1 you must do everything by hand, including exact surd answers. On Papers 2 and 3 use the calculator for the quadratic formula and iteration, but still write the substitution and each iterate.

Common errors

• Expanding 3(x + 6) as 3x + 6.
• Dividing by x in x² = 9x and losing x = 0.
• Writing −b as b in the quadratic formula when b is already negative.
• Forgetting the ± when square-rooting in completing the square.
• Not reversing the sign when dividing an inequality by a negative.
• Using a closed circle for < or a solid line for a strict inequality.
• Giving x < 5 for a quadratic inequality that needs two parts.

Next steps

Condense this page with the revision notes, then test yourself with the practice questions. For the rest of Topic 2, use the Algebra revision notes and Algebra practice questions.

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.3 Solving equations and inequalities (A17 to A22), with the Appendix on mathematical formulae.