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. $\textbf{Elimination}$ adds or subtracts the equations to remove a letter. $\textbf{Substitution}$ 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, $\textbf{subtract}$ the equations; if they are different, $\textbf{add}$ 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.

Example 1Solve

3x + 2y = 16

and

5x - 2y = 8

1The

y

coefficients are

+2

and

-2

: different signs, so add.

8x = 24

, so

x = 3

2Substitute into the first:

9 + 2y = 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 $4x + y = 23$ and $2x - y = 7$ .

Your answer

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

Show worked solution

Solution

1Add the equations: $6x = 30$ , so $x = 5$ .
2Substitute: $20 + y = 23$ , so $y = 3$ .

Answer: $x = 5$ , $y = 3$

Example 2Solve

2x + 3y = 13

and

5x + 2y = 16

1Make the

y

coefficients equal: multiply the first by 2 and the second by 3.

4x + 6y = 26

and

15x + 6y = 48

2Same sign, so subtract:

11x = 22

, giving

x = 2

3Substitute:

4 + 3y = 13

, so

y = 3

Answer

x = 2

y = 3

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

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

Your answer

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

Show hint

Multiply the first equation by 2 and the second by 3 to match the $x$ terms.

Show worked solution

Solution

1 $\times 2$ : $6x + 8y = 36$ . $\times 3$ : $6x + 9y = 39$ .
2Subtract: $y = 3$ .
3Substitute: $3x + 12 = 18$ , so $x = 2$ .

Answer: $x = 2$ , $y = 3$

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 $y = 2x - 1$ , replace that letter in the other equation. This leaves one equation in one unknown.

Example 1Solve

y = 2x - 1

and

3x + 2y = 12

1Substitute:

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

2Expand:

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

Your answer

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

Show worked solution

Solution

1 $2(3y + 2) - 5y = 7$ .
2 $6y + 4 - 5y = 7$ , so $y = 3$ .
3 $x = 3(3) + 2 = 11$ .

Answer: $x = 11$ , $y = 3$

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.

Example 1Two adult tickets and three child tickets cost £41. One adult ticket and four child tickets cost £38. Find the cost of each ticket.

1Let

a

be the adult price and

c

the child price:

2a + 3c = 41

and

a + 4c = 38

2Double the second:

2a + 8c = 76

. Subtract the first:

5c = 35

, so

c = 7

3Substitute:

a + 28 = 38

, so

a = 10

Answer

Adult £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

Show worked solution

Solution

1 $3p + 2r = 4.30$ and $p + 2r = 2.10$ .
2Subtract: $2p = 2.20$ , so $p = 1.10$ .

Answer: £1.10

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 = 2x - 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 $\textbf{each}$ solution, using the linear equation. The answers come in pairs: they are the two points where a line crosses a curve.

Example 1Solve

y = x + 1

and

x^2 + y^2 = 13

1Substitute:

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

2Expand:

2x^2 + 2x + 1 = 13

, so

2x^2 + 2x - 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 $y = 2x - 3$ and $y = x^2 - 6$ . Give the two $x$ -values.

Your answer

Separate with a comma.

Show hint

Set $x^2 - 6 = 2x - 3$ and rearrange to $= 0$ .

Show worked solution

Solution

1 $x^2 - 6 = 2x - 3$ , so $x^2 - 2x - 3 = 0$ .
2 $(x - 3)(x + 1) = 0$ : $x = 3$ or $x = -1$ .
3The points are $(3, 3)$ and $(-1, -5)$ .

Answer: $x = -1$ or $x = 3$

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

\text{Same Sign Subtract}

Otherwise add, to eliminate a letter.

y = mx + c \to \text{substitute}

Use substitution when a letter is already the subject.

\text{solution} = \text{intersection}

Where the graphs cross.

Watch out for these

Adding when the signs are the same:

2x + 3y = 13

and

2x + y = 7

added gives

4x + 4y = 20

Same sign subtract:

2y = 6

Multiplying only one side of an equation:

2x + y = 5

becomes

4x + 2y = 5

Multiply every term:

4x + 2y = 10

Subtracting a negative incorrectly:

3y - (-2y) = y

3y - (-2y) = 5y

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

Reveal solution

SolutionFrom

(3, 7)

7 = 3k - 2

, so

k = 3

. Then

13 = 3a - 2

, so

a = 5

. Finally

13 = 2(5) + b

, so

b = 3

\blacksquare

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 $\quad 4x - 3y = 11 \quad$ and $\quad 2x + y = 13$ .[3]
Show mark scheme
- M1Correct method to eliminate a variable, e.g. $2x + y = 13$ multiplied by 3 and added: $10x = 50$
- A1 $x = 5$
- A1 $y = 3$
Answer: $x = 5$ , $y = 3$

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]
Show mark scheme
- M1Forms both equations: $4t + 3c = 14.70$ and $2t + 5c = 16.80$
- M1Correct elimination, e.g. $2 \times$ second $-$ first: $7c = 18.90$
- A1Cake £2.70 (accept 270p)
- A1Tea £1.65
Answer: Tea £1.65, cake £2.70

Question 3Grade 85 marks

Solve algebraically the simultaneous equations $\quad x^2 + y^2 = 25 \quad$ and $\quad y = 2x - 5$ .[5]
Show mark scheme
- M1Substitutes the linear into the quadratic: $x^2 + (2x - 5)^2 = 25$
- M1Expands and simplifies to a quadratic $= 0$ : $5x^2 - 20x = 0$
- M1Solves their quadratic: $5x(x - 4) = 0$
- A1 $x = 0, y = -5$
- A1 $x = 4, y = 3$
Answer: $(0, -5)$ and $(4, 3)$

Examiner note

Dividing $5x^2 - 20x = 0$ by $x$ loses the solution $x = 0$ . Factorise instead.

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Solve $x + y = 10$ and $x - y = 4$ .
- $x = 7, y = 3$
- $x = 3, y = 7$
- $x = 6, y = 4$
- $x = 14, y = -4$
Show answer and explanation
Answer: $x = 7, y = 3$
Add: $2x = 14$ .
To eliminate $y$ from $2x + 3y = 12$ and $x + 3y = 9$ you should…
- subtract the equations
- add the equations
- multiply both by 3
- divide by $y$
Show answer and explanation
Answer: subtract the equations
The $y$ terms have the same sign.
Solve $y = 3x$ and $x + y = 12$ .
- $x = 3, y = 9$
- $x = 4, y = 12$
- $x = 9, y = 3$
- $x = 2, y = 6$
Show answer and explanation
Answer: $x = 3, y = 9$
$x + 3x = 12$ .
Solve $3x + 2y = 19$ and $x + 2y = 9$ .
- $x = 5, y = 2$
- $x = 2, y = 5$
- $x = 4, y = 2.5$
- $x = 7, y = 1$
Show answer and explanation
Answer: $x = 5, y = 2$
Subtract: $2x = 10$ .
Solve $2x + 5y = 1$ and $3x - 2y = 11$ .
- $x = 3, y = -1$
- $x = -2, y = 1$
- $x = 3, y = 1$
- $x = 1, y = -4$
Show answer and explanation
Answer: $x = 3, y = -1$
Eliminate $y$ : $4x + 10y = 2$ and $15x - 10y = 55$ ; add to get $19x = 57$ .
Where do $y = x + 2$ and $y = 8 - x$ intersect?
- $(3, 5)$
- $(5, 3)$
- $(2, 6)$
- $(4, 6)$
Show answer and explanation
Answer: $(3, 5)$
$x + 2 = 8 - x$ gives $x = 3$ .
How many solutions do $y = 2x + 1$ and $y = 2x - 4$ have?
- none
- one
- two
- infinitely many
Show answer and explanation
Answer: none
Same gradient, different intercepts: the lines are parallel.
Solve $y = x$ and $y = x^2 - 6$ .
- $(3, 3)$ and $(-2, -2)$
- $(3, 3)$ only
- $(-3, -3)$ and $(2, 2)$
- $(6, 6)$ and $(0, 0)$
Show answer and explanation
Answer: $(3, 3)$ and $(-2, -2)$
$x^2 - x - 6 = 0$ gives $x = 3$ or $x = -2$ .

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