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

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.

We can plot the line y=4

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

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
 

 
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.

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

Therefore a=105\div7=15

Substituting into one of the original equations

We can check by substituting into the other original equation

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.

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.

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.

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

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:

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.

We can write two equations.

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

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

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

This works, so our answer is correct.

350 adult tickets and 250 child tickets were sold.

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

From these we can write the following equations

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

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.

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.

Since it is a rectangle, we can write two equations

We can solve these using the elimination method.

Firstly, rearranging each equation we get

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

 
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

This works, so our solution is correct.

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

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

We can solve these using the substitution method.

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

Then we substitute this into the other equation

b = 4 or b = 12

Substituting these back into one of the original equations

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

This works, so our answer is correct

From this information we can write the equations

We can solve these using substitution.

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

Next, substitute this into the other equation

Substituting this into one of the original equations

We can check this by substituting into the other original equation

This works, so our solution is correct.

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

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

x=0 or x=3

Substituting these values into one of the original equations

When x = 0

When x = 3

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

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

Both solutions work, so our answer is correct.

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

Now, substitute into the other equation

x=1 or x=6

Substituting these into one of the original equations

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

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

This can be solved using substitution.

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

Next, substitute this into the other equation

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

Substituting these into one of the original equations

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

When x=2

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

Both solutions work, so our answers are correct.