Secondary Worksheets & Practice Questions

15 Simultaneous Equations Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included

June 26, 2023 | 3 min read

Beki Christian

Simultaneous equations questions involve systems of equations in which there are two or more unknowns. When solving simultaneous equations, we are finding solutions which work for all of the equations in the system. Simultaneous equations can be solved graphically or algebraically.

Here, you will find a selection of simultaneous equations questions of varying difficulty, from questions suitable for KS3 students through to some tricky exam questions! The focus will be on word problems and problem solving questions.

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How to solve simultaneous equations

In KS3 and KS4, we work with systems with two equations and two unknowns. There are three methods for solving simultaneous equations: graphically, by elimination or by substitution.

Solving simultaneous equations graphically

simultaneous equations question solving graphically — Example of a Third Space Learning GCSE revision slide on solving simultaneous equations graphically

If we have two simultaneous equations, we can solve them by plotting their graphs. Once plotted, the solution or solutions will be the point or points of intersection of the two lines.

15 Simultaneous Equations Questions And Practice Problems (KS3 & KS4) Worksheet

Download this free worksheet on simultaneous equations. This set of 15 simultaneous equations questions and answer key will help you prepare for GCSE Maths

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Solving simultaneous equations by elimination

When solving simultaneous equations algebraically, elimination is a great method to use if we have two linear equations. Elimination works by making the coefficient of one of the variables the same in each equation and then subtracting one equation from the other to eliminate that variable.

For example, let’s solve the simultaneous equations

\begin{aligned} 3a + 2b = 19\\ 4a + 3b = 27\end{aligned}

Step 1: Use multiplication to make the coefficient of one of the variables the same in both equations.

Multiplying the first equation by $3,$ we get $9a+6b=57.$

Multiplying the second equation by $2,$ we get $8a+6b=54.$

The coefficient of $b$ for both equations is now $6.$

Step 2: Subtract one equation from the other.

\begin{aligned} 9a + 6b = 57\\ -8a + 6b = 54\\ a = 3\end{aligned}

Step 3: Substitute into one of the original equations to work out the other value.

\begin{aligned} 3a + 2b = 19\\ 3 \times 3 + 2b = 19\\ 9 + 2b = 19\\ 2b = 10\\ b = 5\end{aligned}

Step 4: Check your answer by substituting both values into the other original equation.

\begin{aligned} 4a + 3b = 27\\ 4\times 3 + 3 \times 5 = 12 + 15 = 27\end{aligned}

This works, so our solution is correct.

$a=3$ and $b=5$

Solving simultaneous equations by substitution

simultaneous equations question substitution — Example of a Third Space Learning GCSE revision slide on solving simultaneous equations using substitution

Substitution is a useful method to use when one or more of the equations is not linear. Substitution works by making one of the variables the subject of one of the equations and then substituting this into the other equation.

For example, let’s solve the simultaneous equations

\begin{aligned} 3x - y - 10 = 0\\ x^{2} + y^{2} = 100\end{aligned}

Step 1: Make one of the variables the subject of one of the equations.

3x-y-10=0

Rearranging this we get

y=3x-10

Step 2: Substitute this into the other equation.

\begin{aligned} x^{2} + y^{2} = 100\\ x^{2} + (3x - 10)^{2} = 100\\ x^{2} + 9x^{2} - 60x + 100 = 100\\ 10x^{2} - 60x + 100 = 100\end{aligned}

Step 3: Solve the equation.

\begin{aligned} 10x^{2} - 60x + 100 = 100\\ 10x^{2} - 60x = 0\\ x^{2} - 6x = 0\\ x(x - 6) = 0\end{aligned}

$x=0$ or $x=6$

Step 4: Substitute into one of the original equations to find the other value.

y=3x-10

When $x=0$

y=3\times0-10=-10

When $x=6$

y=3\times6-10=8

The solutions are $x=0$ and $y=-10$ or $x=6$ and $y=8$ .

Step 5: Check your answer by substituting into the other original equation.

x^{2}+y^{2}=100

0^{2}+(-10)^{2}=0+100=100

6^{2}+8^{2}=36+64=100

These both equal $100$ , so our solutions are correct.

KS3 simultaneous equations questions

Simultaneous equations will first be covered towards the end of KS3. To begin with, students are taught how to solve simultaneous equations graphically and are introduced to the method of elimination.

Solving simultaneous equations questions graphically

x=1, \ y=3

x = 4, \ y = 4

x = 3, \ y = 1

x = 4, \ y = 2.5

The equations have been plotted on the graph.

The solution to the simultaneous equations is the point at which the lines meet: $x = 3, \ y = 1.$

x=2, \ y=4

x = 3, \ y = -6

x = 4, \ y = 5

x = 0, \ y = 0

We can plot the line $y=4$

Simultaneous Equations Questions Blog Image 3

Then the solution to the simultaneous equations is the point of intersection of the two lines: $x=2, \ y=4.$

Linear simultaneous equations

m = 10, \ n = 4

m = 14, \ n = 8

m = 10, \ n = 12

m = 28, \ n = 22

For the sum, the equation is $m+n=22.$

We can solve these equations by elimination.

The coefficient of $m$ is $1$ in both equations.

Subtracting them we get

$\begin{aligned} m+n&=22\\ -~m-n&=6\\ \hline 2n&=16 \end{aligned}$

Therefore $n=16 \div 2=8$

Substituting back into one of the original equations

$m + n = 22$

\begin{aligned} m+8&=22\\ m&=14 \end{aligned}

We can check our answer by substituting into the other equation

$\begin{aligned} &m-n=6 \\ &14-8=6 \end{aligned}$

This works, so our answer is correct.

The two numbers are $14$ and $8.$

35p

45p

40p

30p

We can write two equations here

$\begin{aligned} &2a + 3b = 90\\ &3a + b = 65 \end{aligned}$

We need to make the coefficients of either a or b the same.

Multiplying the second equation by $3$ we get $9a + 3b = 195.$

Subtracting one equation from the other

\begin{aligned} &9a+3b&=195\\ -~&2a+3b&=90\\ \hline &7a&=105 \end{aligned}

Therefore $a=105\div7=15$

Substituting into one of the original equations

\begin{aligned} 3a+b&=65\\ 3 \times 15+b&=65\\ 45+b&=65\\ b&=20 \end{aligned}

We can check by substituting into the other original equation

2a+3b=90

2 \times 15+3 \times 20=30+60=90

This works, so our solution is correct.

The cost of an apple is $15p$ and the cost of a banana is $20p.$ Therefore the cost of an apple and a banana is $35p.$

KS4 simultaneous equations questions

Simultaneous equations are covered more extensively in GCSE maths.

During KS4, the method of elimination is revisited and solving equations graphically is extended to include solving pairs of simultaneous equations where one equation may be a quadratic equation or another non-linear equation such as the equation of a circle. The method of substitution is also introduced as a way of solving quadratic simultaneous equations algebraically.

Simultaneous equations is a part of the national curriculum and is examined by all exam boards including Edexcel, AQA and OCR. The questions included below are great practice for students working towards GCSEs and many are of a similar style to those found on past papers.

Simultaneous equations also feature at A Level, so it is important for those wanting to study maths at KS5 to be confident in each of the methods discussed here.

Solving simultaneous equations questions graphically

5. Lucy has $£15$ pocket money saved up. Lucy gets another $£2$ each week.

Plot a line on the axis below to show how Lucy’s money will increase over time.

Simultaneous Equations Questions Blog Image 4

Charlie has no money saved. Charlie gets $£5$ pocket money each week.

Plot a line showing Charlie’s money on the same set of axes.

Use your lines to determine how many weeks it will be until Lucy and Charlie have the same amount of money and the amount they will each have at that time.

$8$ weeks $£40$

$3$ weeks $£21$

$5$ weeks $£20$

$5$ weeks $£25$

Simultaneous Equations Questions Blog Image 5

The lines intersect at $5$ weeks and $£25.$

Therefore, Lucy and Charlie will have the same amount of money in $5$ weeks and they will both have $£25.$

$x = 2$ and $y = 3$ or $x = -2$ and $y = 3$

$x = 0$ and $y = -1$ or $x = 3$ and $y = 8$

$x = 3$ and $y = 8$

$3 = 0$ and $y = 0$ or $y = 3$ and $y = 9$

Simultaneous Equations Questions Blog Image 7

The points of intersection are $(0, \ -1)$ and $(3, 8)$ , so the solutions to the simultaneous equations are $x = 0$ and $y =-1$ or $x = 3$ and $y = 8.$

Linear simultaneous equations

12g

150g

84g

126g

If we call the weight of the box B and the weight of each chocolate C, then we can write two equations:

\begin{aligned} &B + 12C = 294\\ &B + 5C = 210 \end{aligned}

We can solve these using the elimination method.

The coefficient of B is the same in both equations, so we can go ahead and subtract one from the other:

\begin{aligned} B+12&C&=294\\ -~B+5&C&=210\\ \hline 7&C&=84 \end{aligned}

Therefore $C=84\div 7=12$

Substituting this into one of the original equations

$\begin{aligned} B+5C&=210\\ B+5\times 12&=210\\ B+60&=210\\ B&=150 \end{aligned}$

We can check this by substituting these values into the other original equation

$B + 12C = 294$

$150+12 \times 12=150+144=294$

This works, so our answer is correct.

The box weighs $150g$ and the chocolates weigh $12g$ each.

$300$ adult tickets, $200$ child tickets

$450$ adult tickets, $150$ child tickets

$200$ adult tickets, $500$ child tickets

$350$ adult tickets, $250$ child tickets

We can write two equations.

If we call the number of adult tickets A and the number of child tickets C then

\begin{aligned} &A + C = 600\\ &5A + 3C = 2500 \end{aligned}

We can solve these using the elimination method.

Multiplying the first equation by $3$ we get $3A + 3C = 1800.$

Subtracting one equation from the other

\begin{aligned} &5A+3C&=2500\\ -~&3A+3C&=1800\\ \hline &2A&=700 \end{aligned}

Therefore $A=700 \div 2=350$

Substituting this into one of the original equations

$\begin{aligned} A+C&=600\\ 350+C&=600\\ C&=250 \end{aligned}$

We can check our answer by substituting into the other original equation

5A + 3C = 2500

5 \times 350 + 3 \times 250 = 1750+750=2500

This works, so our answer is correct.

$350$ adult tickets and $250$ child tickets were sold.

24

28

30

26

There is one row and one column containing only lightning and sun symbols.

From these we can write the following equations

\begin{aligned} &3L + S = 22\\ &2L + 2S = 24 \end{aligned}

We can solve these using the elimination method.

Multiplying the first equation by $2$ gives us $6L + 2S = 44.$

Subtracting one of these from the other

\begin{aligned} &6L+2S&=44\\ -~&2L+2S&=24\\ \hline &4L&=20 \end{aligned}

Therefore $L=20\div 4=5$

Substituting into one of the original equations

$\begin{aligned} 3L+S&=22\\ 3 \times 5 + S&=22\\ 15+S&=22\\ S&=7 \end{aligned}$

Using a different column: $2L+2M=28.$

\begin{aligned} 2\times 5+2M&=28\\ 10+2M&=28\\ 2M&=18\\ M&=9 \end{aligned}

The lightning symbol is worth $5$ , the sun symbol is worth $7$ and the moon symbol is worth $9.$

We can check this by trying different rows or columns.

The missing total is $5 + 7 + 9 + 9 = 30.$

46

16

5

20

Since it is a rectangle, we can write two equations

\begin{aligned} &2y + 1 = 2x – y\\ &2x + 2y = 5x – 3y \end{aligned}

We can solve these using the elimination method.

Firstly, rearranging each equation we get

\begin{aligned} &2x – 3y = 1\\ &3x – 5y = 0 \end{aligned}

Next, we multiply the first equation by $5$ and the second equation by $3$ to make the coefficients of $y$ the same

\begin{aligned} &10x – 15y = 5\\ &9x – 15y = 0 \end{aligned}

Subtracting one from the other

$\begin{aligned} &10x-15y&=5\\ -~&9x-15y&=0\\ \hline &x&=5 \end{aligned}$

Substituting this into one of the original equations

$\begin{aligned} 2x-3y&=1\\ 2\times 5-3y&=1\\ 10-3y&=1\\ 3y&=9\\ y&=3 \end{aligned}$

We can check our answer by substituting these values into the other original equation

3x-5y = 0

3\times 5-5\times 3=15-15=0

This works, so our solution is correct.

We can now calculate the length and width of the rectangle.

\begin{aligned} &\text{Length: } 2\times3+1=7\\ &\text{Width: }2 \times 5+2\times 3=16 \end{aligned}

Non-linear simultaneous equations questions

Length $12cm$ , width $4cm$

Length $24cm$ , width $2cm$

Length $6cm$ , width $10cm$

Length $8cm$ , width $6cm$

If we call the length of the rectangle $a$ and the width $b$ then

\begin{aligned} &ab = 48\\ &2a + 2b = 32 \end{aligned}

We can solve these using the substitution method.

Firstly, we rearrange one equation to make one of the variables the subject

\begin{aligned} 2a+2b&=32\\ a+b&=16\\ a&=16-b \end{aligned}

Then we substitute this into the other equation

\begin{aligned} ab&=48\\ (16-b)b&=48\\ 16b-b^{2}&=48\\ b^{2}-16b+48&=0\\ (b-4)(b-12)&=0 \end{aligned}

$b = 4$ or $b = 12$

Substituting these back into one of the original equations

ab = 48

If $b = 4, \ 4a = 48$ so $a = 12$

If $b = 12, \ 12a = 48$ so $a = 4$

The side lengths are $4$ and $12.$

Checking this for the perimeter

2 \times 4 + 2 \times 12 = 8+24=32

This works, so our answer is correct

2

7

9

\sqrt{32}

From this information we can write the equations

\begin{aligned} &a + b = 16\\ &a^{2}-b^{2} = 32 \end{aligned}

We can solve these using substitution.

First, make one of the variables the subject of one of the equations

\begin{aligned} a+b&=16\\ a&=16-b \end{aligned}

Next, substitute this into the other equation

\begin{aligned} a^{2}-b^{2}&=32\\ (16-b)^{2}-b^{2}&=32\\ 256-32b+b^{2}-b^{2}&=32\\ 256-32b&=32\\ 32b&=224\\ b&=7 \end{aligned}

Substituting this into one of the original equations

\begin{aligned} a+b&=16\\ a+7&=16\\ a&=9 \end{aligned}

We can check this by substituting into the other original equation

\begin{aligned} &a^{2}-b^{2} = 32\\\\ &9^{2}-7^{2} = 81-49 = 32 \end{aligned}

This works, so our solution is correct.

The difference between $a$ and $b$ is $9-7 = 2.$

$(0, -5)$ and $(3, 4)$

$(0, 5)$ and $(3, 4)$

$(3, 4)$ and $(10, 25)$

$(-3, -4)$ and $(0, -5)$

The coordinates of the points of intersection are the solutions to the simultaneous equations.

We can solve these using substitution.

We know that $y = 3x-5,$ so substituting this into the other equation

\begin{aligned} x^{2}+y^{2}&=25\\ x^{2}+(3x-5)^{2}&=25\\ x^{2}+9x^{2}-30x+25&=25\\ 10x^{2}-30x&=0\\ 10x(x-3)&=0 \end{aligned}

$x=0$ or $x=3$

Substituting these values into one of the original equations

y = 3x-5

When $x = 0$

y=3\times 0 -5=-5

When $x = 3$

y=3\times 3-5=4

The points of intersection are $(0, -5)$ and $(3, 4).$

We can check these values by substituting them into the equation for the circle

\begin{aligned} &x^{2}+y^{2}=25\\ &0^{2}+(-5)^{2}=0+25=25\\ &3^{2}+4^{2}=16+9=25 \end{aligned}

Both solutions work, so our answer is correct.

Hard GCSE simultaneous equations questions

61

11

5 \sqrt{2}

\sqrt{10}

To find the points of intersection of the lines, we solve the equations simultaneously.

We can do this using substitution.

First, rearrange one equation so that one of the variables is the subject

\begin{aligned} x-y+3&=0\\ y&=x+3 \end{aligned}

Now, substitute into the other equation

\begin{aligned} x+3&=2x^{2}-13x+15\\ 2x^{2}-14x+12&=0\\ x^{2}-7x+6&=0\\ (x-1)(x-6)&=0 \end{aligned}

$x=1$ or $x=6$

Substituting these into one of the original equations

y=x+3

When $x = 1, \ y = 4$
When $x = 6, \ y = 9$

The points of intersection are $(1, 4)$ and $(6, 9).$

To find the length of the line PQ, we use Pythagoras theorem

\begin{aligned} PQ^{2}&=(6-1)^{2}+(9-4)^{2}\\ PQ^{2}&=25+25\\ PQ^{2}&=50\\ PQ&=\sqrt{50}\\ PQ&=5 \sqrt{2} \end{aligned}

The length of the line is $5 \sqrt{2}$

$x = 2$ and $y = 7$ or $x = 5$ and $y = 1$

$\frac{3}{2}$ and $y = 8$ or $x = 2$ and $x = 7$

$x = 1$ and $y = 9$ or $x = 5$ and $y = 1$

$\frac{16}{3}$ and $\frac{1}{3}$ or $x = 2$ and $y = 7$

This can be solved using substitution.

First, rearrange one equation so that one of the variables is the subject

y = 11-2x

Next, substitute this into the other equation

\begin{aligned} 3x^{2}+2xy+y^{2}&=89\\ 3x^{2}+2x(11-2x)+(11-2x)^{2}&=89\\ 3x^{2}+22x-4x^{2}+121-44x+4x^{2}&=89\\ 3x^{2}-22x+32&=0\\ (3x-16)(x-2)&=0 \end{aligned}

$x=\frac{16}{3}$ or $x=2$

Substituting these into one of the original equations

2x+y=11

When $x=\frac{16}{3}:$

\begin{aligned} \\ 2 \times \frac{16}{3}+y&=11\\ y&=11-\frac{32}{3}\\ y&=\frac{1}{3} \end{aligned}

When $x=2$

\begin{aligned} \\ 2 \times 2+y&=11\\ y&=11-4\\ y&=7 \end{aligned}

The solutions are $x=\frac{16}{3}$ and $y=\frac{1}{3}$ or $x=2$ and $y=7$ .

We can check these solutions by substituting them back into the other original equation

3x^{2}+2xy+y{2}=89

\begin{aligned} 3 \times 2^{2}+2\times 2 \times 7 + 7^{2} &=89\\ 12+28+49&=89\\ 89&=89 \end{aligned}

\begin{aligned} 3 \times (\frac{16}{3})^{2}+2\times \frac{16}{13} \times \frac{1}{3} + (\frac{1}{3})^{2} &=89\\\\ \frac{768}{9}+\frac{32}{9}+\frac{1}{19}&=89\\\\ \frac{801}{9}&=89\\\\ 89&=89 \end{aligned}

Both solutions work, so our answers are correct.

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