A19, A21AlgebraFoundation & HigherGrades 4–8

Simultaneous Equations

Simultaneous equations are two equations that share the same unknowns and are true at the same time. One equation with two unknowns has endless solutions; two together usually pin down exactly one pair of values. The skill is to combine them so one unknown disappears.

50 min Video by Maths Genie AQA GCSE Maths

The big picture

Every linear equation in $x$ and $y$ is a straight line, and every point on the line is a solution. Two lines usually cross at exactly one point — and that point is the solution to both equations at once.

There are two algebraic methods. $Elimination$ adds or subtracts the equations to remove a letter. $Substitution$ replaces one letter using the other equation. Elimination is usually quickest for two linear equations; substitution is essential when one equation is quadratic.

Solving Simultaneous EquationsCurated by Kepler, not produced by us. Watch it, then read the notes below.

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What you'll learn

Solve two linear simultaneous equations by elimination
Solve simultaneous equations by substitution
Form and solve simultaneous equations from a context
Solve one linear and one quadratic equation simultaneously (Higher)

Elimination

Make the coefficients of one letter the same size, multiplying one or both equations if needed. If the signs are the same, $subtract$ the equations; if they are different, $add$ them. ("Same Sign Subtract.")

Solve for the remaining letter, substitute it back into either original equation to find the other, and check in the equation you did not use.

1The

y

coefficients are

+ 2

and

- 2

: different signs, so add.

8 x = 24

, so

x = 3

2Substitute into the first:

9 + 2 y = 16

, so

y = 3.5

3Check in the second:

15 - 7 = 8

✓

Answer

x = 3

y = 3.5

Your turnSame method, new numbers — try it before moving on.

Solve $4 x + y = 23$ and $2 x - y = 7$ .

Your answer

Type it like x = 2, y = -1.

Your turnSame method, new numbers — try it before moving on.

Solve $3 x + 4 y = 18$ and $2 x + 3 y = 13$ .

Your answer

Type it like x = 2, y = -1.

Tip — Label the equations ① and ② and write what you did, e.g. "① × 2 − ②". It makes your method easy to follow and to check.

Substitution

If one equation already gives a letter on its own, such as $y = 2 x - 1$ , replace that letter in the other equation. This leaves one equation in one unknown.

1Substitute:

3 x + 2 (2 x - 1) = 12

2Expand:

7 x - 2 = 12

, so

x = 2

3Then

y = 2 (2) - 1 = 3

Answer

x = 2

y = 3

Your turnSame method, new numbers — try it before moving on.

Solve $x = 3 y + 2$ and $2 x - 5 y = 7$ .

Your answer

Type it like x = 2, y = -1.

Forming simultaneous equations

Choose a letter for each unknown quantity. Each separate piece of information gives one equation. Solve, then answer in context — with units.

1Let

a

be the adult price and

c

the child price:

2 a + 3 c = 41

and

a + 4 c = 38

2Double the second:

2 a + 8 c = 76

. Subtract the first:

5 c = 35

, so

c = 7

3Substitute:

a + 28 = 38

, so

a = 10

AnswerAdult £10, child £7

Your turnSame method, new numbers — try it before moving on.

3 pens and 2 rulers cost £4.30. 1 pen and 2 rulers cost £2.10. How much does one pen cost, in pounds?

Your answer

The graphical view

The solution of two simultaneous linear equations is the point where their graphs intersect. Reading the intersection from an accurate graph gives an approximate solution.

Parallel lines never meet, so the equations have no solution; identical lines have infinitely many.

y = 2 x - 1

and

y = 5 - x

cross at

(2, 3)

, so

x = 2

y = 3

solves both.

Linear and quadratic (Higher)

When one equation is quadratic, rearrange the linear equation to make one letter the subject and substitute it into the quadratic. This gives a quadratic equation — usually with two solutions.

Find the matching value of the other letter for $each$ solution, using the linear equation. The answers come in pairs: they are the two points where a line crosses a curve.

1Substitute:

x^{2} + (x + 1)^{2} = 13

2Expand:

2 x^{2} + 2 x + 1 = 13

, so

2 x^{2} + 2 x - 12 = 0

, i.e.

x^{2} + x - 6 = 0

3Factorise:

(x + 3) (x - 2) = 0

, so

x = - 3

x = 2

4Pair up using

y = x + 1

y = - 2

y = 3

Answer

x = 2, y = 3

and

x = - 3, y = - 2

Your turnSame method, new numbers — try it before moving on.

Solve $y = 2 x - 3$ and $y = x^{2} - 6$ . Give the two $x$ -values.

Your answer

Separate with a comma.

Think like an examiner

Check your pair of values in the equation you did not substitute into. If it fails, there is an arithmetic slip — find it before moving on.
For linear–quadratic questions, full marks need both pairs of $(x, y)$ , correctly matched.
In context questions, the final mark is for answering the question: "the adult ticket costs £10".

Methods

Same Sign Subtract

Otherwise add, to eliminate a letter.

y = m x + c \to substitute

Use substitution when a letter is already the subject.

solution = intersection

Where the graphs cross.

Watch out for these

Adding when the signs are the same:

2 x + 3 y = 13

and

2 x + y = 7

added gives

4 x + 4 y = 20

Same sign subtract:

2 y = 6

Multiplying only one side of an equation:

2 x + y = 5

becomes

4 x + 2 y = 5

Multiply every term:

4 x + 2 y = 10

Subtracting a negative incorrectly:

3 y - (- 2 y) = y

3 y - (- 2 y) = 5 y

Pairing

x = - 3

with

y = 3

in a linear–quadratic answer.

Substitute each

x

into the linear equation to get its own

y

Stretch yourself

The line $y = k x - 2$ passes through the points $(3, 7)$ and $(a, 13)$ . The line $y = 2 x + b$ also passes through $(a, 13)$ . Find $k$ , $a$ and $b$ .

Hint — Use $(3, 7)$ to find $k$ first. Then each remaining fact gives one equation.

Questions students ask

Which letter should I eliminate?

Whichever needs the least work. If one letter already has matching (or opposite) coefficients, eliminate that one straight away.

What if I get a fraction or a decimal?

That is fine — simultaneous equations do not always have whole-number solutions. Keep exact fractions until the final answer.

Why do linear–quadratic questions have two answers?

A straight line can cross a curve such as a circle or parabola at two points. Each crossing point is a solution pair. (If the line is a tangent there is one; if it misses, there are none.)

Key takeaways

Elimination: match a coefficient, then add (different signs) or subtract (same signs).
Substitution: replace a letter using the other equation.
Each fact in a context gives one equation; answer in context.
The solution is where the graphs intersect.
Linear and quadratic: substitute the linear into the quadratic and give both pairs.

Exam-style questions

Practise it the way it is examined

3 original questions · 12 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

Solve the simultaneous equations $4 x - 3 y = 11$ and $2 x + y = 13$ .[3]

Question 2Grade 54 marks

A café sells cups of tea and slices of cake. 4 teas and 3 cakes cost £14.70. 2 teas and 5 cakes cost £16.80.

Work out the cost of one tea and the cost of one slice of cake.[4]

Question 3Grade 85 marks

Solve algebraically the simultaneous equations $x^{2} + y^{2} = 25$ and $y = 2 x - 5$ .[5]

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Simultaneous Equations

Elimination

Substitution

Forming simultaneous equations

The graphical view

Linear and quadratic (Higher)

Practise it the way it is examined

Skill drill

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Related topics

Simultaneous Equations

Elimination

Substitution

Forming simultaneous equations

The graphical view

Linear and quadratic (Higher)

Practise it the way it is examined

Skill drill

Practise this topic

Related topics