Key Stage 4 maths · Algebra

GCSE quadratic and simultaneous equations, and inequalities

Solving equations is the core of GCSE algebra. In Years 10 and 11 pupils solve quadratic equations by factorising, find approximate solutions from graphs, solve two simultaneous equations in two unknowns, and solve linear inequalities, showing the answer on a number line. These methods are used again and again, in word problems, geometry and later study, so accuracy and checking matter.

Quadratics, simultaneous equations, inequalities, explained

A quadratic equation has an x² term and can have two solutions. To solve one by factorising, pupils first rearrange so that one side is zero, then factorise, then use a simple fact: if two numbers multiply to give zero, at least one of them must be zero. So (x + 5)(x − 3) = 0 means x + 5 = 0 or x − 3 = 0, giving x = −5 or x = 3. Equations such as x² − 25 = 0 have two solutions as well, x = 5 and x = −5. Pupils always substitute their answers back to check.

Graphs give approximate solutions. The solutions of x² − 3x − 10 = 0 are the x-values where the graph of y = x² − 3x − 10 crosses the x-axis, at x = −2 and x = 5. To solve a related equation such as x² − 3x − 10 = 2, pupils draw the horizontal line y = 2 on the same axes and read the x-values where the line meets the curve. Because a graph is read by eye, these answers are approximate, often to one decimal place.

Simultaneous equations are two equations that must both be true at once, with two unknowns. For 2x + y = 7 and 3x − y = 8, adding the equations makes the y terms cancel, giving 5x = 15, so x = 3, and then 2 × 3 + y = 7 gives y = 1. This is the elimination method. Pupils may need to multiply one or both equations first so that a pair of terms matches. Substitution, where one equation is rearranged to give y and put into the other, is an alternative. On a graph, each equation is a straight line and the solution is the point where the lines cross, which pupils can use to find approximate solutions.

Linear inequalities are solved just like equations, with one important difference: multiplying or dividing both sides by a negative number reverses the inequality sign. So −3x < 12 gives x > −4. For 2x + 3 ≤ 11, subtracting 3 and dividing by 2 gives x ≤ 4. The solution set is shown on a number line, with an open circle for < or > (the end value is not included) and a filled circle for ≤ or ≥ (the end value is included). A pupil might also be asked for the whole numbers that satisfy something like −2 < x ≤ 3, which are −1, 0, 1, 2 and 3.

These skills combine in problems set in context. A rectangle’s length is 3 cm more than its width x and its area is 40 cm², so x(x + 3) = 40. That becomes x² + 3x − 40 = 0, which factorises to (x + 8)(x − 5) = 0. Both x = −8 and x = 5 solve the equation, but a width cannot be negative, so the rectangle is 5 cm by 8 cm. Deciding which answers make sense is part of the work.

Words your child will hear

Quadratic equation: An equation where the highest power of the unknown is 2, such as x² + 2x − 15 = 0.
Solution (or root): A value of the unknown that makes the equation true.
Simultaneous equations: Two or more equations with the same unknowns that are solved together.
Elimination: Adding or subtracting two equations so that one unknown cancels out.
Substitution: Replacing a letter with its value or with an expression in terms of another letter.
Inequality: A statement that uses <, >, ≤ or ≥ rather than =, such as x ≤ 4.
Solution set: All the values that make an inequality true, often shown on a number line.

What the National Curriculum says

In Key Stage 4, pupils should be taught to:

Solve quadratic equations algebraically by factorising; find approximate solutions using a graph
Solve 2 simultaneous equations in 2 variables (linear/linear) algebraically; find approximate solutions using a graph
Solve linear inequalities in 1 variable; represent the solution set on a number line

Objectives from the National Curriculum in England, © Crown copyright, used under the Open Government Licence v3.0.

A worked example

Solve x² + 2x = 15, and check your solutions.

A quadratic must have zero on one side before it can be factorised. Subtract 15 from both sides.
```
x² + 2x − 15 = 0
```
Factorise. We need two numbers that multiply to −15 and add to +2. These are +5 and −3.
```
(x + 5)(x − 3) = 0
```
If two things multiply to give zero, one of them must be zero. So either the first bracket is zero or the second is.
```
x + 5 = 0   or   x − 3 = 0
```
Solve each of these small equations.
```
x = −5   or   x = 3
```
Check both in the original equation. (−5)² + 2 × (−5) = 25 − 10 = 15. And 3² + 2 × 3 = 9 + 6 = 15. Both work.

x = −5 or x = 3

Common mistakes and how to help

Factorising before getting zero on one side, such as x(x + 2) = 15 so x = 15 or x = 13

The rule that one factor must be zero works only when the product is zero. With 15 on the right, the brackets could be many different numbers.

What helps: Always move everything to one side first, so the equation reads “… = 0”. Ask “is the other side zero?” before writing any brackets.

Finding only one solution, or losing one by dividing by x

To solve x² = 3x, many pupils divide both sides by x to get x = 3 and lose the solution x = 0. Others forget that the factorised form has two brackets and so two answers.

What helps: Subtract to get x² − 3x = 0, factorise to x(x − 3) = 0, and read off both answers: x = 0 or x = 3. Avoid dividing both sides by x, because x could be 0.

Mistakes with signs when eliminating a variable

To remove 2y from 3x + 2y = 12 and x + 2y = 8, pupils add the equations instead of subtracting. A pair of terms with the same sign must be subtracted, and a pair with opposite signs must be added.

What helps: Say it out loud: “same signs subtract, different signs add”. Always substitute both values back into both original equations as a final check.

Not reversing the inequality sign when dividing by a negative number

Inequality solving looks exactly like equation solving, so −3x < 12 becomes x < −4 by dividing both sides by −3 without a thought.

What helps: Test a value. If x = −5 then −3x = 15, which is not less than 12, so x < −4 cannot be right. The correct answer, x > −4, is true for x = 0.

How to help at home

Check by substituting. Make it routine for your child to put every answer back into the original equation. It takes seconds and catches nearly every slip. For a quadratic, check both solutions.
Number riddles. “Two numbers add to 10 and differ by 4.” Your child can solve it by trial first (7 and 3) and then write x + y = 10 and x − y = 4 and solve it with algebra. Both routes should agree.
Shopping puzzles. Two coffees and one cake cost £8, and one coffee and two cakes cost £8.50. Ask for the price of each (coffee £2.50, cake £3), then work out the algebra together.
Inequalities in daily life. “I need at least £60 and have £25, and I save £7 a week.” Ask your child to write 25 + 7w ≥ 60, solve it (w ≥ 5) and say what it means: at least 5 weeks. Height limits and age limits also make easy number-line examples.

Builds on

Key Stage 3: Algebraic notation and formulae
Key Stage 3: Solving linear equations
Key Stage 3: Graphs of linear and quadratic functions
Key Stage 3: Sequences and the nth term