study guides20 min read

Simultaneous Equations (Linear & Non-Linear): GCSE & IGCSE Practice Questions with Worked Solutions

Sir Faraz Hassan

8 Jul 2026

Table of Contents▾

Grade 8-9 TargetedFrom linear foundations to the toughest non-linear problems

Simultaneous equations are one of the highest-value topics in GCSE and IGCSE Maths. They appear on every paper, run from a reliable Grade 5 pair all the way up to the hardest Grade 9 problems, and once you can spot the right method they become some of the most dependable marks available. This page gives you 11 exam-style questions, from linear pairs up to non-linear problems involving curves and circles, each with a full worked solution and a mark scheme so you can see exactly where every mark is earned. Try each question yourself first, then reveal the solution to check your method.

What are simultaneous equations?

Simultaneous equations are two or more equations that share the same unknowns and are solved together to find the values that satisfy all of them at once. At GCSE and IGCSE they almost always involve two unknowns, x and y. Linear simultaneous equations, where both equations are straight lines, are solved by elimination or substitution and appear from around Grade 5. Non-linear simultaneous equations, where one equation is a curve containing x² or y², are solved by substitution: this produces a quadratic, which is why they have two solution pairs rather than one. Non-linear pairs are Higher tier, typically Grade 7 to 9.

How to use this page

Try each question on paper first

Give yourself a real attempt before looking at the solution, because that is where the learning happens.

Check the calculator icon

A crossed-out calculator means non-calculator; a plain calculator means a calculator is allowed.

Reveal the worked solution

Open the solution under each question and compare it with your own method, step by step.

Read the mark scheme

See exactly where each method mark (M1) and accuracy mark (A1) is awarded, just like a real examiner.

Come back and redo it

Repeat any question you got wrong a few days later to lock the method in.

Download printable PDF

11 questions with full worked solutions and mark schemes - free PDF

Practice questions

Work through each question, then open the worked solution to check your method. Linear questions come first, then non-linear.

Solve the simultaneous equations.

2x + y = 8

x + 3y = 14

x =

y =

[Total 3 marks]

GRADE5

Show solution & mark schemeHide solution & mark scheme

Question 1 - Exam Solution

Understanding the Question

Given

2x + y = 8

x + 3y = 14

Two linear equations, two unknowns.

Find

The values of

x

and

y

that satisfy both equations.

Plan the Solution

Use elimination.
Multiply (1) by 3 so both equations have $3y$ , then subtract to eliminate $y$ and find $x$ .
Substitute $x$ back to find $y$ .

Worked Solution [3 marks]

Rule - Elimination: make one variable's coefficient equal in both equations, then add or subtract to eliminate that variable.

Step 1: Label the equations.

2x + y = 8 \quad (1)

x + 3y = 14 \quad (2)

(Reason: labelling keeps the working organised.)

Step 2: Match the y-coefficients - multiply (1) by 3.

6x + 3y = 24 \quad (3)

(Reason: Equations (2) and (3) both have

3y

, so subtracting removes

y

Step 3: Subtract (2) from (3) to eliminate y.

(6x + 3y) - (x + 3y) = 24 - 14

5x = 10

x = 2

(Reason:

3y - 3y = 0

, so

y

is eliminated.)

Step 4: Substitute x = 2 into (1) to find y.

2(2) + y = 8

4 + y = 8

y = 4

(Reason: use the simplest equation to find the second variable.)

x = 2, \quad y = 4

Verification

Check 1:

2(2) + 4 = 8

Check 2:

2 + 3(4) = 14

Check 3: both values are simple whole numbers, sensible for a non-calculator pair.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Scale one equation to match a coefficient (×3 to get 3y)	M1	Method - correct elimination setup	✓
Eliminate and solve for the first variable (x = 2)	A1	Accuracy - first value	✓
Substitute back and solve for the second (y = 4)	A1	Accuracy - second value	✓

Full marks: 3/3

Solve the simultaneous equations.

4x + 3y = 17

3x + 5y = 21

x =

y =

[Total 4 marks]

GRADE5

Show solution & mark schemeHide solution & mark scheme

Question 2 - Exam Solution

Understanding the Question

Given

4x + 3y = 17

3x + 5y = 21

Neither coefficient matches yet.

Find

The values of

x

and

y

that satisfy both equations.

Plan the Solution

Use elimination.
Neither coefficient matches yet, so scale both equations.
Multiply (1) by 3 and (2) by 4 so the $x$ -terms both become $12x$ , then subtract to eliminate $x$ and find $y$ .
Substitute back to find $x$ .

Worked Solution [4 marks]

Rule - Elimination: when no coefficient matches, multiply each equation so one variable's coefficients become equal, then add or subtract to eliminate it.

Step 1: Label the equations.

4x + 3y = 17 \quad (1)

3x + 5y = 21 \quad (2)

(Reason: labelling lets us track each multiplication.)

Step 2: Match the x-coefficients - multiply (1) by 3 and (2) by 4.

12x + 9y = 51 \quad (3)

12x + 20y = 84 \quad (4)

(Reason: The LCM of 4 and 3 is 12, so both

x

-terms become

12x

Step 3: Subtract (3) from (4) to eliminate x.

(12x + 20y) - (12x + 9y) = 84 - 51

11y = 33

y = 3

(Reason:

12x - 12x = 0

, so

x

is eliminated.)

Step 4: Substitute y = 3 into (1) to find x.

4x + 3(3) = 17

4x + 9 = 17

4x = 8

x = 2

(Reason: substitute the known value and rearrange to isolate x.)

x = 2, \quad y = 3

Verification

Check 1:

4(2) + 3(3) = 8 + 9 = 17

Check 2:

3(2) + 5(3) = 6 + 15 = 21

Check 3: both values are simple whole numbers, sensible for a non-calculator pair.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Multiply BOTH equations to equate a coefficient (×3 and ×4 → 12x)	M1	Method - set up elimination	✓
Subtract to eliminate x: 11y = 33 → y = 3	A1	Accuracy - first value	✓
Substitute y = 3 back and rearrange (4x + 9 = 17)	M1	Method - find second variable	✓
Second value x = 2	A1	Accuracy - second value	✓

Full marks: 4/4

A cafe charges £7.40 for 3 teas and 2 coffees, and £13.40 for 5 teas and 4 coffees. Work out the cost of 2 teas and 3 coffees.

[Total 4 marks]

GRADE6

Show solution & mark schemeHide solution & mark scheme

Question 3 - Exam Solution

Understanding the Question

Given

3 teas and 2 coffees cost £7.40

5 teas and 4 coffees cost £13.40

Find

The total cost of 2 teas and 3 coffees.

Plan the Solution

Let $t$ = cost of one tea and $c$ = cost of one coffee (in £).
Turn each fact into an equation and solve by elimination.
Then use the prices to cost 2 teas and 3 coffees.

Worked Solution [4 marks]

Rule - Form and solve: turn the words into two equations, solve them simultaneously, then answer the exact question asked.

Step 1: Define variables and write the equations.

3t + 2c = 7.40 \quad (1)

5t + 4c = 13.40 \quad (2)

(Reason:

t

and

c

are the unknown prices; each sentence gives one equation.)

Step 2: Match the c-coefficients - multiply (1) by 2.

6t + 4c = 14.80 \quad (3)

(Reason: Equations (2) and (3) both have

4c

, so subtracting removes

c

Step 3: Subtract (2) from (3) to eliminate c.

(6t + 4c) - (5t + 4c) = 14.80 - 13.40

t = 1.40

(Reason:

4c - 4c = 0

, so

c

is eliminated.)

Step 4: Substitute t = 1.40 into (1) to find c.

3(1.40) + 2c = 7.40

4.20 + 2c = 7.40

2c = 3.20

c = 1.60

So tea = £1.40 and coffee = £1.60.

(Reason: substitute the known value and solve for c.)

Step 5: Answer the question - cost of 2 teas and 3 coffees.

2(1.40) + 3(1.60) = 2.80 + 4.80 = 7.60

(Reason: the question asks for a specific combination, not just the prices - always finish with what was asked.)

£7.60

Verification

Check 1:

3(1.40) + 2(1.60) = 4.20 + 3.20 = 7.40

Check 2:

5(1.40) + 4(1.60) = 7.00 + 6.40 = 13.40

Check 3: £1.40 and £1.60 are realistic cafe prices, so the answer makes sense.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Form two correct equations from the words	M1	Method - translate the problem	✓
Solve simultaneously (correct elimination)	M1	Method - solve the pair	✓
Both prices correct: tea £1.40, coffee £1.60	A1	Accuracy - the two prices	✓
Cost of 2 teas and 3 coffees = £7.60	A1	Accuracy - final answer asked for	✓

Full marks: 4/4

Solve the simultaneous equations.

x^2 + y = 7

y = 2x + 4

x =

y =

x =

y =

[Total 5 marks]

GRADE7

Show solution & mark schemeHide solution & mark scheme

Question 4 - Exam Solution

Understanding the Question

Given

x^2 + y = 7

y = 2x + 4

One curve, one line.

Find

The values of

x

and

y

; a quadratic appears, so expect two solution pairs.

Plan the Solution

Use substitution (the line gives y).
Put it into the quadratic; rearrange to a quadratic equal to 0 and factorise for the two x-values.
Substitute each $x$ back into the line for its $y$ .

Worked Solution [5 marks]

Rule - Substitution (line into curve): replace

y

in the quadratic with the linear expression, solve the resulting quadratic, then find each

y

Step 1: Label the equations.

x^2 + y = 7 \quad (1)

y = 2x + 4 \quad (2)

(Reason: (2) is already

y = \ldots

, which makes substitution natural.)

Step 2: Substitute (2) into (1).

x^2 + (2x + 4) = 7

x^2 + 2x + 4 = 7

(Reason: replacing

y

leaves one equation in

x

only.)

Step 3: Rearrange to a quadratic equal to 0.

x^2 + 2x + 4 - 7 = 0

x^2 + 2x - 3 = 0

(Reason: a quadratic must equal 0 before factorising.)

Step 4: Factorise and solve.

(x + 3)(x - 1) = 0

x = -3

x = 1

(Reason: two numbers with product

-3

and sum

+2

are

+3

and

-1

Step 5: Find each matching y from the line (2).

When

x = 1

y = 2(1) + 4 = 6

→

(1, 6)

When

x = -3

y = 2(-3) + 4 = -2

→

(-3, -2)

(Reason: each

x

pairs with exactly one

y

on the line.)

x = 1, \quad y = 6

x = -3, \quad y = -2

Verification

Check 1 - (1, 6):

1^2 + 6 = 7

Check 2 - (-3, -2):

(-3)^2 + (-2) = 9 - 2 = 7

Check 3: each y came from the line, and a quadratic gives two pairs - both consistent.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute y = 2x + 4 into x² + y = 7	M1	Method - substitution	✓
Form the correct quadratic: x² + 2x - 3 = 0	A1	Accuracy - correct equation	✓
Factorise / solve: (x + 3)(x - 1) = 0	M1	Method - solve the quadratic	✓
Both x-values: x = 1 and x = -3	A1	Accuracy - the two roots	✓
Both complete pairs: (1, 6) and (-3, -2)	A1	Accuracy - matching y-values	✓

Full marks: 5/5

Solve the simultaneous equations.

2x^2 + y^2 = 33

y = x + 3

x =

y =

x =

y =

[Total 5 marks]

GRADE8

Show solution & mark schemeHide solution & mark scheme

Question 5 - Exam Solution

Understanding the Question

Given

2x^2 + y^2 = 33

y = x + 3

A curve and a line.

Find

Values of

x

and

y

; expect two pairs.

Plan the Solution

Substitute $y = x + 3$ into the curve.
Expand the squared bracket, collect, and simplify to a quadratic equal to 0.
Factorise for the two x-values; find each matching y.

Worked Solution [5 marks]

Rule - Substitution (line into curve): replace

y

in the curve, expand and simplify, then solve the quadratic.

Step 1: Label the equations.

2x^2 + y^2 = 33 \quad (1)

y = x + 3 \quad (2)

(Reason: (2) gives

y

directly, so substitute into (1).)

Step 2: Substitute (2) into (1).

2x^2 + (x + 3)^2 = 33

(Reason: replacing

y

leaves an equation in

x

only.)

Step 3: Expand the bracket and collect like terms.

(x + 3)^2 = x^2 + 6x + 9

2x^2 + x^2 + 6x + 9 = 33

3x^2 + 6x + 9 = 33

(Reason: expand

(x + 3)^2

carefully - the middle term

6x

is the one most often missed.)

Step 4: Rearrange to = 0, then simplify.

3x^2 + 6x - 24 = 0

x^2 + 2x - 8 = 0

(divide by 3)

(Reason: subtract 33, then divide by the common factor 3 to make factorising easier.)

Step 5: Factorise and solve.

(x + 4)(x - 2) = 0

x = -4

x = 2

(Reason: two numbers with product

-8

and sum

+2

are

+4

and

-2

Step 6: Find each matching y from the line (2).

When

x = 2

y = 2 + 3 = 5

→

(2, 5)

When

x = -4

y = -4 + 3 = -1

→

(-4, -1)

(Reason: each

x

gives one

y

on the line.)

x = 2, \quad y = 5

x = -4, \quad y = -1

Verification

Check 1 - (2, 5):

2(2)^2 + 5^2 = 8 + 25 = 33

Check 2 - (-4, -1):

2(-4)^2 + (-1)^2 = 32 + 1 = 33

Check 3: both pairs also satisfy the line, and a quadratic gives two pairs - consistent.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute y = x + 3 into 2x² + y² = 33	M1	Method - substitution	✓
Expand (x + 3)² and collect to a quadratic = 0	M1	Method - expand and rearrange	✓
Correct quadratic: x² + 2x - 8 = 0	A1	Accuracy - correct equation	✓
Both x-values: x = 2 and x = -4	A1	Accuracy - the two roots	✓
Both complete pairs: (2, 5) and (-4, -1)	A1	Accuracy - matching y-values	✓

Full marks: 5/5

The line $y = 2x + 1$ and the curve $y = x^2 - 4x + 6$ meet at two points. The distance between the two points is $k\sqrt{5}$ . Find the value of $k$ .

k =

[Total 6 marks]

GRADE9Problem solving

Show solution & mark schemeHide solution & mark scheme

Question 6 - Exam Solution

Understanding the Question

Given

y = 2x + 1

y = x^2 - 4x + 6

A line and a curve; their gap is

k\sqrt{5}

Find

The value of

k

- so find the two crossing points first, then the distance.

Plan the Solution

Set line = curve and solve the quadratic for the two x-values.
Find each y from the line to get A and B.
Use the distance formula for $AB$ , then simplify the surd into $k\sqrt{5}$ form.

Worked Solution [6 marks]

Rule - Intersections, then distance: solve line = curve to find both points, then apply

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Step 1: Set the line equal to the curve (both equal y).

2x + 1 = x^2 - 4x + 6

(Reason: at a crossing point the line and curve share the same x and y.)

Step 2: Rearrange to a quadratic equal to 0.

x^2 - 6x + 5 = 0

(Reason: move every term to one side

(x^2 - 4x + 6 - 2x - 1)

Step 3: Factorise and solve for the x-coordinates.

(x - 1)(x - 5) = 0

x = 1

x = 5

(Reason: two numbers with product

+5

and sum

-6

are

-1

and

-5

Step 4: Find each point's y from the line.

When

x = 1

y = 2(1) + 1 = 3

→

A(1, 3)

When

x = 5

y = 2(5) + 1 = 11

→

B(5, 11)

(Reason: substitute each x into the simpler equation, the line.)

Step 5: Apply the distance formula to A(1, 3) and B(5, 11).

AB = \sqrt{(5 - 1)^2 + (11 - 3)^2}

AB = \sqrt{4^2 + 8^2} = \sqrt{16 + 64}

AB = \sqrt{80}

(Reason: horizontal gap = 4, vertical gap = 8; the distance is the hypotenuse.)

Step 6: Simplify the surd into k√5 form.

\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}

k\sqrt{5} = 4\sqrt{5}

, so

k = 4

(Reason: pull out the largest square factor, 16, to leave √5.)

k = 4

Verification

Check 1 - A(1, 3):

1^2 - 4(1) + 6 = 3

Check 2 - B(5, 11):

5^2 - 4(5) + 6 = 11

Check 3:

4\sqrt{5} \approx 8.94

and

\sqrt{80} \approx 8.94

, and

k = 4

is a whole number, as expected.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Set line = curve and rearrange to x² - 6x + 5 = 0	M1	Method - form the quadratic	✓
Solve for both x: x = 1 and x = 5	A1	Accuracy - the two roots	✓
Coordinates of both points: A(1, 3), B(5, 11)	A1	Accuracy - both points	✓
Apply the distance formula with these coordinates	M1	Method - distance formula	✓
Correct distance: AB = √80	A1	Accuracy - unsimplified surd	✓
Simplify to 4√5, so k = 4	A1	Accuracy - final value of k	✓

Full marks: 6/6

Solve the simultaneous equations.

x^2 + 2y^2 = 18

3x - 2y = x + 6

x =

y =

x =

y =

[Total 5 marks]

GRADE7

Show solution & mark schemeHide solution & mark scheme

Question 7 - Exam Solution

Understanding the Question

Given

x^2 + 2y^2 = 18

3x - 2y = x + 6

A curve and a linear equation that is not yet tidy.

Find

Values of

x

and

y

; expect two pairs.

Plan the Solution

Tidy the linear equation first - collect the $x$ terms and simplify to $x = \ldots$ .
Substitute into the curve, expand and simplify to a quadratic = 0.
Factorise for the two y-values, then find each matching x.

Worked Solution [5 marks]

Rule - Rearrange, then substitute: when the linear equation is untidy, simplify it to

x = \ldots

(or

y = \ldots

) first, then substitute into the curve.

Step 1: Label the equations.

x^2 + 2y^2 = 18 \quad (1)

3x - 2y = x + 6 \quad (2)

(Reason: (2) has

x

on both sides, so it needs tidying before use.)

Step 2: Simplify the linear equation (2).

2x - 2y = 6

(collect the x terms)

x - y = 3

(divide by 2)

x = y + 3 \quad (3)

(Reason:

3x - x = 2x

, then divide by 2 to make it clean.)

Step 3: Substitute (3) into (1).

(y + 3)^2 + 2y^2 = 18

(Reason: replacing

x

leaves an equation in

y

only.)

Step 4: Expand, collect, and rearrange to = 0.

y^2 + 6y + 9 + 2y^2 = 18

3y^2 + 6y - 9 = 0

y^2 + 2y - 3 = 0

(divide by 3)

(Reason: expand

(y + 3)^2 = y^2 + 6y + 9

, then simplify and divide by 3.)

Step 5: Factorise and solve.

(y + 3)(y - 1) = 0

y = -3

y = 1

(Reason: two numbers with product

-3

and sum

+2

are

+3

and

-1

Step 6: Find each matching x from (3).

When

y = 1

x = 1 + 3 = 4

→

(4, 1)

When

y = -3

x = -3 + 3 = 0

→

(0, -3)

(Reason: use the tidy linear

x = y + 3

to pair each

y

with its

x

x = 4, \quad y = 1

x = 0, \quad y = -3

Verification

Check 1 - (4, 1):

4^2 + 2(1)^2 = 16 + 2 = 18

Check 2 - (0, -3):

0^2 + 2(-3)^2 = 0 + 18 = 18

Check 3: both pairs also fit the linear x = y + 3, and a quadratic gives two pairs - consistent.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Rearrange the linear to x = y + 3	M1	Method - tidy the linear first	✓
Substitute into the curve and simplify to a quadratic = 0	M1	Method - substitute and expand	✓
Correct quadratic: y² + 2y - 3 = 0	A1	Accuracy - correct equation	✓
Both y-values: y = 1 and y = -3	A1	Accuracy - the two roots	✓
Both complete pairs: (4, 1) and (0, -3)	A1	Accuracy - matching x-values	✓

Full marks: 5/5

The line $x + y = 1$ crosses the circle $x^2 + y^2 = 25$ at the points A and B. Work out the exact length of AB. Give your answer in its simplest surd form.

AB =

[Total 6 marks]

GRADE9

Show solution & mark schemeHide solution & mark scheme

Question 8 - Exam Solution

Understanding the Question

Given

x + y = 1

x^2 + y^2 = 25

A line cutting a circle at A and B.

Find

The exact length of

AB

in surd form.

Plan the Solution

Rearrange the line to $y = \ldots$ and substitute into the circle.
Solve the quadratic for the two x-values, then find each y to get A and B.
Use the distance formula for $AB$ , then simplify the surd.

Worked Solution [6 marks]

Rule - Intersections, then distance: substitute the line into the circle to find both points, then apply

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Step 1: Label the equations.

x + y = 1 \quad (1)

x^2 + y^2 = 25 \quad (2)

(Reason: the line is easiest to rearrange and substitute.)

Step 2: Rearrange the line and substitute into (2).

y = 1 - x \quad (3)

x^2 + (1 - x)^2 = 25

(Reason: replacing

y

leaves an equation in

x

only.)

Step 3: Expand, collect, and rearrange to = 0.

x^2 + 1 - 2x + x^2 = 25

2x^2 - 2x - 24 = 0

x^2 - x - 12 = 0

(divide by 2)

(Reason: expand

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

, simplify, then divide by 2.)

Step 4: Factorise and solve for the x-coordinates.

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

x = 4

x = -3

(Reason: two numbers with product

-12

and sum

-1

are

-4

and

+3

Step 5: Find each point's y from the line (3).

When

x = 4

y = 1 - 4 = -3

→

A(4, -3)

When

x = -3

y = 1 - (-3) = 4

→

B(-3, 4)

(Reason: substitute each x into the line.)

Step 6: Apply the distance formula to A(4, -3) and B(-3, 4).

AB = \sqrt{(-3 - 4)^2 + (4 - (-3))^2}

AB = \sqrt{(-7)^2 + 7^2} = \sqrt{49 + 49}

AB = \sqrt{98}

(Reason: horizontal gap = 7, vertical gap = 7; the distance is the hypotenuse.)

Step 7: Simplify the surd.

\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}

(Reason: pull out the largest square factor, 49, to leave √2.)

AB = 7\sqrt{2}

Verification

Check 1 - A(4, -3):

4^2 + (-3)^2 = 16 + 9 = 25

Check 2 - B(-3, 4):

(-3)^2 + 4^2 = 9 + 16 = 25

Check 3: both points fit the line

x + y = 1

, and

7\sqrt{2} \approx 9.9 = \sqrt{98}

- consistent.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Rearrange line to y = 1 - x and substitute into the circle	M1	Method - substitution	✓
Expand and simplify to x² - x - 12 = 0	A1	Accuracy - correct quadratic	✓
Coordinates of both points: A(4, -3), B(-3, 4)	A1	Accuracy - both points	✓
Apply the distance formula with these coordinates	M1	Method - distance formula	✓
Correct distance: AB = √98	A1	Accuracy - unsimplified surd	✓
Simplify to AB = 7√2	A1	Accuracy - exact surd form	✓

Full marks: 6/6

The diagram shows an L-shape made from two rectangles. The total area is 34 cm² and the total width along the base is 8 cm. Given that $x$ and $y$ are positive integers, work out the values of $x$ and $y$ .

x =

y =

[Total 5 marks]

GRADE7

Show solution & mark schemeHide solution & mark scheme

Question 9 - Exam Solution

Understanding the Question

Given

Left rectangle:

2x

x

(area

2x^2

Right rectangle:

y

4y

(area

4y^2

Total area 34 cm²; total base 8 cm;

x

and

y

are positive integers.

Find

The values of

x

and

y

Plan the Solution

Write one equation for the total area and one for the total base.
Because $x$ and $y$ must be positive integers, list the few pairs the base allows and test them in the area equation.
This avoids solving a messy quadratic.

Worked Solution [5 marks]

Rule - Form and use the constraint: turn the shape into two equations; when the unknowns must be positive integers, testing the few possible pairs is a complete method.

Step 1: Form the two equations from the shape.

2x^2 + 4y^2 = 34 \quad (1)

(area)

2x + y = 8 \quad (2)

(base)

(Reason: area = sum of the two rectangle areas; base = sum of the two widths.)

Step 2: List the possible integer pairs from (2).

With

2x + y = 8

and positive integers,

x

can only be 1, 2 or 3, giving the pairs (1, 6), (2, 4), (3, 2).

(Reason: if

x \ge 4

then

y \le 0

, which is not allowed.)

Step 3: Test each pair in the area equation (1).

2(1)^2 + 4(6)^2 = 2 + 144 = 146

(too big)

2(2)^2 + 4(4)^2 = 8 + 64 = 72

(too big)

2(3)^2 + 4(2)^2 = 18 + 16 = 34

(correct)

(Reason: we need the area to equal 34 - only the last pair does.)

Step 4: Only (3, 2) satisfies both equations.

The pair (3, 2) fits both the base and the area, so it is the solution.

(Reason: the positive-integer constraint makes this pair unique.)

x = 3, \quad y = 2

Verification

Check 1 - base:

2(3) + 2 = 8

Check 2 - area:

2(3)^2 + 4(2)^2 = 18 + 16 = 34

Check 3: both are positive whole numbers, exactly as the question requires.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Form the area equation: 2x² + 4y² = 34	M1	Method - area from the shape	✓
Form the base equation: 2x + y = 8	M1	Method - base from the shape	✓
Use positive integers to list the possible pairs	M1	Method - apply the constraint	✓
Test in the area equation - (3, 2) works, others rejected	A1	Accuracy - correct testing	✓
State x = 3 and y = 2	A1	Accuracy - final values	✓

Full marks: 5/5

The cuboid shown has a weight of 48 N. When it rests on face A the pressure on the ground is 4 N/m², and when it rests on face B the pressure is 2 N/m². Work out the volume of the cuboid.

Volume = m³

[Total 6 marks]

GRADE8

Show solution & mark schemeHide solution & mark scheme

Question 10 - Exam Solution

Understanding the Question

Given

Width

(x + 2)

, height

y

, depth

(x + 4)

Weight (force) = 48 N.

Face A (front) =

(x + 2)y

, pressure 4 N/m².

Face B (top) =

(x + 2)(x + 4)

, pressure 2 N/m².

Find

The volume of the cuboid.

Plan the Solution

Use the pressure formula (force over area, as a fraction) to turn each pressure into a face area.
Set each face expression equal to its area to get two equations.
Face B involves only $x$ , so solve that quadratic first, then find $y$ , then the volume.

Worked Solution [6 marks]

Rule - Pressure: pressure =

\dfrac{\text{force}}{\text{area}}

, so rearranged, area =

\dfrac{\text{force}}{\text{pressure}}

Step 1: Turn each pressure into a face area.

Face A: area =

\dfrac{48}{4} = 12

Face B: area =

\dfrac{48}{2} = 24

(Reason: dividing the force, 48 N, by each pressure gives that face's area.)

Step 2: Form the two equations from the face expressions.

(x + 2)y = 12 \quad (1)

(face A)

(x + 2)(x + 4) = 24 \quad (2)

(face B)

(Reason: each face area equals its two dimensions multiplied together.)

Step 3: Solve equation (2) - it only involves x.

x^2 + 6x + 8 = 24

x^2 + 6x - 16 = 0

(x + 8)(x - 2) = 0

x = 2

(reject x = -8, a length cannot be negative)

(Reason: expand the brackets, make it equal 0, then factorise.)

Step 4: Substitute x = 2 into (1) to find y.

(2 + 2)y = 12

4y = 12

y = 3

(Reason: now x is known, equation (1) gives y.)

Step 5: Work out the volume = width × height × depth.

V = (x + 2) \times y \times (x + 4)

V = 4 \times 3 \times 6 = 72

m³

(Reason: substitute the found dimensions (4, 3, 6) and multiply.)

Volume = 72 m³

Verification

Check 1 - face A: area =

4 \times 3 = 12

, pressure =

\dfrac{48}{12} = 4

Check 2 - face B: area =

4 \times 6 = 24

, pressure =

\dfrac{48}{24} = 2

Check 3: dimensions 4, 3, 6 are all positive; the smaller face gives the higher pressure - consistent.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Use the pressure relationship (force over area) to get areas 12 and 24	M1	Method - areas from pressure	✓
Form both equations: (x + 2)y = 12 and (x + 2)(x + 4) = 24	M1	Method - two equations	✓
Expand and solve: x² + 6x - 16 = 0	M1	Method - solve the quadratic	✓
x = 2 (reject x = -8)	A1	Accuracy - value of x	✓
y = 3	A1	Accuracy - value of y	✓
Volume = (x + 2)(x + 4)y = 72 m³	A1	Accuracy - final volume	✓

Full marks: 6/6

The line $y = x + 2$ meets the curve $y = -x^2 + 9x - 10$ at the points A and B, as shown. Points C and D lie on the x-axis so that AC and BD are vertical. Work out the area of quadrilateral ABDC.

Area = units²

[Total 6 marks]

GRADE8Problem solving

Show solution & mark schemeHide solution & mark scheme

Question 11 - Exam Solution

Understanding the Question

Given

Line

y = x + 2

Curve

y = -x^2 + 9x - 10

They meet at A and B; C and D lie on the x-axis with AC and BD vertical.

Since AC and BD are parallel, ABDC is a trapezium.

Find

The area of quadrilateral ABDC.

Plan the Solution

Set line = curve and solve to find the two meeting points A and B.
C and D sit directly below A and B on the x-axis, so $AC$ and $BD$ are just the y-values of A and B.
Use the trapezium area formula with those two vertical sides and the gap between them.

Worked Solution [6 marks]

Rule - Trapezium: Area =

\dfrac{1}{2}(a + b)h

, where

a

and

b

are the parallel sides and

h

is the distance between them.

Step 1: Set the line equal to the curve.

x + 2 = -x^2 + 9x - 10

(Reason: at the meeting points the line and curve share the same x and y.)

Step 2: Rearrange to a quadratic equal to 0.

x^2 - 8x + 12 = 0

(Reason: move every term to one side and collect.)

Step 3: Factorise and solve for the x-coordinates.

(x - 2)(x - 6) = 0

x = 2

x = 6

(Reason: two numbers with product

+12

and sum

-8

are

-2

and

-6

Step 4: Find each y from the line.

When

x = 2

y = 2 + 2 = 4

→

A(2, 4)

When

x = 6

y = 6 + 2 = 8

→

B(6, 8)

(Reason: substitute each x into the simpler equation, the line.)

Step 5: Drop C and D to the x-axis and find the sides.

C(2, 0)

and

D(6, 0)

AC = 4

(the

y

of A)

BD = 8

(the

y

of B)

CD = 6 - 2 = 4

(the gap between them)

(Reason: AC and BD are vertical, so their lengths are the y-values of A and B.)

Step 6: Apply the trapezium area formula.

Area =

\dfrac{1}{2}(AC + BD) \times CD

Area =

\dfrac{1}{2}(4 + 8) \times 4

Area =

\dfrac{1}{2} \times 12 \times 4 = 24

units²

(Reason: add the parallel sides, halve, then multiply by the gap.)

Area = 24 units²

Verification

Check 1 - A(2, 4):

-(2)^2 + 9(2) - 10 = -4 + 18 - 10 = 4

Check 2 - B(6, 8):

-(6)^2 + 9(6) - 10 = -36 + 54 - 10 = 8

Check 3: both points are on the curve, and the trapezium has two vertical parallel sides - the method fits the shape.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Set line = curve and rearrange to x² - 8x + 12 = 0	M1	Method - form the quadratic	✓
Solve for both x: x = 2 and x = 6	A1	Accuracy - the two roots	✓
Coordinates: A(2, 4), B(6, 8)	A1	Accuracy - both points	✓
Find the sides: AC = 4, BD = 8, CD = 4	M1	Method - set up the trapezium	✓
Apply Area = ½(AC + BD) × CD	M1	Method - trapezium formula	✓
Area = 24 units²	A1	Accuracy - final area	✓

Full marks: 6/6

Download printable PDF

Print it, work offline, mark yourself against the scheme.

Frequently Asked Questions

Simultaneous equations are two or more equations that share the same unknowns, solved together to find values that satisfy all of them at once. At GCSE and IGCSE they usually involve two unknowns, x and y.

Use elimination (match one variable's coefficient, then add or subtract to remove it) or substitution (make one variable the subject and put it into the other equation). Then find the second variable and check your answer in both original equations.

When one equation is a curve (it has x² or y²), use substitution. Put the linear equation into the quadratic, expand and simplify to a quadratic equal to zero, then factorise. This gives two solution pairs, not one.

Usually two solution pairs, because the line meets the curve at two points. Each x-value pairs with its own y-value, so you always give both pairs.

Linear simultaneous equations appear from around Grade 5. Non-linear (quadratic) simultaneous equations are Higher tier, typically Grade 7 to 9. This resource covers both.

Keep revising

Once you are confident with simultaneous equations, check the full topic lists for GCSE Maths and IGCSE Maths to see what else to revise, and use the GCSE grade boundaries and IGCSE grade boundaries to set your target.

Struggling with simultaneous equations?

Book a free 30-minute intro session and get a clear, step-by-step method that works for your exam board.

Book Free 30-Min Intro

simultaneous-equationsgcseigcsepractice-questionsworked-solutionsalgebranon-linearedexcel

ShareWhatsApp Post

Ready to boost your grades?

Get expert 1-to-1 tutoring in GCSE & IGCSE Maths. Book a free 30-minute intro session to see the difference.

Book Free 30-Min Intro Session

Simultaneous Equations (Linear & Non-Linear): GCSE & IGCSE Practice Questions with Worked Solutions

What are simultaneous equations?

How to use this page

Try each question on paper first

Check the calculator icon

Reveal the worked solution

Read the mark scheme

Come back and redo it

Practice questions

Question 1, Grade 5, Non-calculator

Question 1 - Exam Solution

Question 2, Grade 5, Non-calculator

Question 2 - Exam Solution

Question 3, Grade 6, Calculator allowed

Question 3 - Exam Solution

Question 4, Grade 7, Non-calculator

Question 4 - Exam Solution

Question 5, Grade 8, Non-calculator

Question 5 - Exam Solution

Question 6, Grade 9, Calculator allowed

Question 6 - Exam Solution

Question 7, Grade 7, Non-calculator

Question 7 - Exam Solution

Question 8, Grade 9, Calculator allowed

Question 8 - Exam Solution

Question 9, Grade 7, Non-calculator

Question 9 - Exam Solution

Question 10, Grade 8, Calculator allowed

Question 10 - Exam Solution

Question 11, Grade 8, Calculator allowed

Question 11 - Exam Solution

Frequently Asked Questions

Keep revising

Struggling with simultaneous equations?

Ready to boost your grades?