1MA1 · A19, A21AlgebraFoundation & HigherGrades 4–8

Simultaneous Equations

Three coffees and two cakes cost £13.20; two coffees and five cakes cost £17.60. One equation alone cannot tell you the price of a coffee, because there are two unknowns. With two equations that must both be true at once — $\textbf{simultaneous equations}$ — there is enough information to find both.

55 min Equations & Inequalities

The big picture

The strategy is always to eliminate one unknown, leaving a single equation you already know how to solve. Elimination does this by adding or subtracting the equations; substitution does it by replacing one unknown with an expression for it.

Graphically, each equation is a line (or curve), and the solution is where they cross. That explains why a linear–quadratic pair can have two solutions — a line can cut a curve twice — and why Edexcel Higher questions on circles and lines are really simultaneous equations.

What you'll learn

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

Elimination

Multiply one or both equations so that one unknown has the same coefficient (ignoring sign) in both.

If the signs are the same, subtract the equations; if different, add them. One unknown disappears.

Solve for the remaining unknown, substitute back into either original equation, and check in the other.

Example 1Solve

3x + 2y = 16

and

5x - 3y = 14

1Make the

y

coefficients match: multiply the first by 3 and the second by 2:

9x + 6y = 48

and

10x - 6y = 28

2Signs differ, so add:

19x = 76

, giving

x = 4

3Substitute into

3x + 2y = 16

12 + 2y = 16

, so

y = 2

4Check in the second:

20 - 6 = 14

✓.

Answer

x = 4

y = 2

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

Solve $2x + 5y = 24$ and $3x - 2y = 17$ .

Your answer

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

Show hint

Multiply the first by 2 and the second by 5, then add.

Show worked solution

Solution

1 $4x + 10y = 48$ and $15x - 10y = 85$ .
2Add: $19x = 133$ , so $x = 7$ ; then $14 + 5y = 24$ , so $y = 2$ .

Answer: $x = 7$ , $y = 2$

Tip — "Same Signs Subtract" — if the matching terms have the same sign, subtract; otherwise add.

Substitution

If one equation already gives one unknown in terms of the other — like $y = 2x - 3$ — substitute it into the other equation.

Substitution is also the method for linear–quadratic pairs, so it is worth being fluent with it.

Example 1Solve

y = 2x - 3

and

4x - 3y = 1

1Substitute:

4x - 3(2x - 3) = 1

4x - 6x + 9 = 1

, so

-2x = -8

and

x = 4

3Then

y = 2(4) - 3 = 5

Answer

x = 4

y = 5

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

Solve $y = 3x + 1$ and $2x + y = 11$ .

Your answer

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

Show worked solution

Solution

1 $2x + 3x + 1 = 11$ , so $5x = 10$ .
2 $x = 2$ and $y = 7$ .

Answer: $x = 2$ , $y = 7$

Forming simultaneous equations

Define two letters for the two unknowns, then write one equation for each piece of information.

Answer the question in context, with units.

Example 1Three coffees and two cakes cost £13.20; two coffees and five cakes cost £17.60. Find the cost of a coffee and of a cake.

1Let a coffee cost

c

pounds and a cake

k

pounds:

3c + 2k = 13.20

and

2c + 5k = 17.60

2Multiply the first by 2 and the second by 3:

6c + 4k = 26.40

and

6c + 15k = 52.80

3Same signs, so subtract:

11k = 26.40

, giving

k = 2.40

4Substitute:

3c + 4.80 = 13.20

, so

3c = 8.40

and

c = 2.80

AnswerA coffee costs £2.80 and a cake costs £2.40.

Linear and quadratic pairs

Rearrange the linear equation to make one unknown the subject, substitute into the quadratic, and solve the resulting quadratic.

Each solution for one unknown gives a partner value from the linear equation — so solutions come in pairs.

Two solutions means the line crosses the curve twice; one repeated solution means it is a tangent; no real solutions means they do not meet.

Example 1Solve

x^{2} + y^{2} = 25

and

y = x + 1

1Substitute:

x^{2} + (x + 1)^{2} = 25

, so

2x^{2} + 2x + 1 = 25

2x^{2} + 2x - 24 = 0

, so

x^{2} + x - 12 = 0

and

(x + 4)(x - 3) = 0

x = 3

gives

y = 4

;

x = -4

gives

y = -3

Answer

(3, 4)

and

(-4, -3)

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

The line $y = x - 1$ meets the circle $x^2 + y^2 = 13$ at two points. Find their $x$ -coordinates.

Your answer

Separate with a comma.

Show worked solution

Solution

1 $x^2 + (x - 1)^2 = 13$ gives $2x^2 - 2x - 12 = 0$ .
2 $x^2 - x - 6 = 0$ , so $(x - 3)(x + 2) = 0$ .

Answer: $x = 3$ or $x = -2$

Why this works

The two answers are the two points where the line $y = x + 1$ cuts the circle of radius 5. Pairing each $x$ with its own $y$ matters: $(3, -3)$ is not a solution.

Think like an examiner

Show the multiplied equations before adding or subtracting.
Give paired solutions clearly, e.g. $x = 3, y = 4$ or $(3, 4)$ .
Check solutions in the equation you did not substitute into.

Simultaneous equations

\text{Match coefficients, then add or subtract}

Elimination.

y = \dots \text{ into the other equation}

Substitution.

\text{solutions} = \text{intersection points}

Graphical meaning.

Watch out for these

Subtracting when the matching terms have opposite signs.

Opposite signs: add.

Multiplying only one term of an equation.

Multiply every term, including the right-hand side.

Mixing up the pairs in a linear–quadratic answer.

Substitute each

x

into the linear equation to find its own

y

Stopping after finding one unknown.

Substitute back to find the other.

Stretch yourself

The line $y = 2x + k$ is a tangent to the curve $y = x^{2} + 3$ . Find the value of $k$ and the coordinates of the point of contact.

Hint — Solve simultaneously to get a quadratic in $x$ . A tangent meets the curve exactly once, so the quadratic has one repeated root.

Reveal solution

SolutionSetting

x^{2} + 3 = 2x + k

gives

x^{2} - 2x + (3 - k) = 0

. For a single repeated root this must be a perfect square,

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

, so

3 - k = 1

and

k = 2

. The repeated root is

x = 1

, giving

y = 1 + 3 = 4

. So

k = 2

and the point of contact is

(1, 4)

. Check: the line

y = 2x + 2

passes through

(1, 4)

✓.

\blacksquare

Questions students ask

Elimination or substitution?

Elimination is usually quicker for two linear equations; substitution is better when one equation already gives

y = \dots

, and is required for linear–quadratic pairs.

How can I check my answer?

Substitute both values into the equation you did not use for the final substitution. Both sides should match.

Can simultaneous equations have no solution?

Yes. Two parallel lines never meet, and a line may miss a curve altogether.

Key takeaways

Elimination: match coefficients, then add (different signs) or subtract (same signs).
Substitution: replace one unknown with its expression in terms of the other.
Form equations from context with clearly defined unknowns.
For linear–quadratic pairs, substitute the linear into the quadratic; solutions come in pairs.
Solutions are the intersection points of the graphs.

Exam-style questions

Practise it the way it is examined

3 original questions · 11 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 $5x + 2y = 11$ and $3x - 4y = 4$ .[3]
Show mark scheme
- M1Correct method to eliminate one variable
- A1 $x = 2$
- A1 $y = 0.5$
Answer: $x = 2$ , $y = \frac{1}{2}$

Question 2Grade 54 marks

3 adult tickets and 2 child tickets cost £36.50. 2 adult tickets and 5 child tickets cost £50.

Work out the cost of an adult ticket and the cost of a child ticket.[4]
Show mark scheme
- P1Forms two equations
- P1Correct process to eliminate a variable
- A1Adult £7.50
- A1Child £7
Answer: Adult £7.50, child £7

Question 3Grade 84 marks

Solve algebraically $y = 2x + 1$ and $y = x^2 - 2x + 4$ .[4]
Show mark scheme
- M1 $x^2 - 2x + 4 = 2x + 1$
- M1 $x^2 - 4x + 3 = 0$
- A1 $x = 1$ , $y = 3$
- A1 $x = 3$ , $y = 7$
Answer: $(1, 3)$ and $(3, 7)$

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.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 4 lessons