The smart way to improve grades

Comprehensive & curriculum aligned

Affordable pricing from £10/month

Solve Simultaneous Equations (by elimination)

In this worksheet, students will learn how to solve simple simultaneous equations using the elimination method. They will also solve contextualised question based on simultaneous equations.

'Solve Simultaneous Equations (by elimination)' worksheet

Key stage:  KS 4

GCSE Subjects:   Maths

GCSE Boards:   Pearson Edexcel, OCR, Eduqas, AQA

Curriculum topic:   Algebra

Curriculum subtopic:   Solving Equations and Inequalities, Algebraic Equations

Difficulty level:  

Worksheet Overview

QUESTION 1 of 10

Some people went to London Zoo, where they bought adult tickets and child tickets. Mr and Mrs Brown took their daughter Libby and paid £68, whereas Mr and Mrs White took their three sons, Tom, Dick and Harry and paid £100. Can you work out how much Ms Black and her children Jack and Maisy would have to pay?

Problems like this are common in maths books and exams. There are a number of ways you can solve them, such as trial and improvement, but here we want to look at a method where we turn the information into two equations and solve them at the same time. This method is known as 'solving simultaneous equations'

In order to find out how much Ms Black and her two children will pay we first need to know the cost of an adult ticket and the cost of a child ticket. These are known as variables and we use a letter to stand for each one. In many simultaneous equations, x and y are used but you can use any letter and here we will use a = cost of an adult and c = cost of a child. So, Mr and Mrs Brown and their daughter becomes 2 adults and one child paying £68 or 2a + c = 68. Mr and Mrs White and their 3 sons becomes 2 adults and 3 children paying £100 or 2a + 3c = 100. We now have our two equations, and we need to find values for a and c which work in both equations at the same time (simultaneously).

If we line up the equations, one above the other with the one with the biggest values on top it will help.

2a + 3c = 100

2a + c  =  68

We can see that both equations have 2a but the top one has 2c more than the bottom one and the cost is £32 more, so 2c = 32. If 2 children cost £32 then one must cost £16, so c =16. Knowing this we can know work out a. If 2c + c = 68 and c = 16 then 2c + 16 = 68. We can solve this equation easily by subtracting 16 from both sides giving 2a = 52, and dividing by 2 gives a = 26. So an adult ticket costs £26. Now we have our two ticket costs it is a good idea to check they work for the other family (equation). We had 2a + 3c = 100 and substituting a = 26 and c= 16 gives 2 x 26 + 3 x 16 = 52 + 48 = 100, which is correct. 

Now we can find the cost for the Black family which is one adult and 2 children. 26 + 2 x 16 = 58, so they would pay £58.

Now, let's look at another pair of simultaneous equations. This time just the equations without the real-life problem and we will look at a shorter way of showing the working.

Solve  5x + y = 17

and     3x + y = 11

Let's number our equations (1) and (2) with the one with bigger values on top then subtract equation (2) from (1)

5x + y = 17   (1)

3x + y = 11   (2)

2x       =  6    (1) - (2)

  x       =  3    (divide by 2)

Substitute x = 3 into (2) and solve the equation [you could use equation (1) if you prefer]

3 x 3 + y = 11

9 + y = 11

y = 2

So x = 3 and y = 2

Check these work with equation (1) [use the equation you did not use to find the value of y]

5 x 3 + 2 = 17 (which is correct)

Let's look at one more problem.

5x - y = 17  (1)

2x + y = 11  (2)

Notice that the y's have the same coefficients (number in front) and so this is the variable which will be eliminated when we combine the equations. But because the signs in front of them are different (one positive and one negative) this time we add the equations.

7x    =  28   (1) + (2)

  x =    4   [divide by 7]

2 x 4 + y = 11   [substitute into (2)]

8 + y = 11

y = 3

Check in (1)

5 x 4 - 3 = 17 [correct]

So x = 4 and y = 3.

Remember the variable with equal coefficients will be eliminated when the equations are combined. If the signs in front are the same you subtract if they are different you add. (Remember Same Sign Subtract - SSS)

 

 

 

Look at the following pairs of simultaneous equations and decide if you would add or subtract them to eliminate one of the variables. You do not need to solve the equations.

 

 AddSubtract
4x + y = 13 and 2x + y = 7
3x + 2y = 19 and 5x - 2y = 23
3x - 4y = 5 and 3x + 2y = 26
x - 2y = 45 and 6x - 2y = 56

Below, you will see the steps to solving the simultaneous equations

8x + 3y = 19 

3x - 3y = 3 

but not in the correct order.

Match each step to its correct position.

Column A

Column B

1st
3y = 3
2nd
3x2 - 3x1 = 3 [check in (2)]
3rd
16 + 3y = 19
4th
8x2 + 3y = 19 [sub into (1)]
5th
11x = 22 (1) + (2)
6th
y = 1
7th
x = 2
8th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)

Below you will see all the steps to solving the equations 

2x + 6y = 14

2x - 2y = -2

Fill in the blanks to make the solution complete.

Column A

Column B

1st
3y = 3
2nd
3x2 - 3x1 = 3 [check in (2)]
3rd
16 + 3y = 19
4th
8x2 + 3y = 19 [sub into (1)]
5th
11x = 22 (1) + (2)
6th
y = 1
7th
x = 2
8th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)

Below you will see all the steps to solving the equations 

3a - 4b = 20

6a - 4b = 32

Fill in the blanks to make the solution complete.

Column A

Column B

1st
3y = 3
2nd
3x2 - 3x1 = 3 [check in (2)]
3rd
16 + 3y = 19
4th
8x2 + 3y = 19 [sub into (1)]
5th
11x = 22 (1) + (2)
6th
y = 1
7th
x = 2
8th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)

Match the following pairs of simultaneous equations to their correct solutions.

Column A

Column B

4x + y = 13 and 2x + y = 7
x = 5 and y= 2
2x + 4y = 18 and 2x + 2y = 8
x = -1 and y = 5
7x - 4y = 22 and 4x + 4y = 0
x = 2 and y = -2
7x - 2y = 31 and 5x - 2y = 21
x = 3 and y = 1

On a piece of paper solve the following pair of equations and type your solutions into the blanks.

7p + 3q = 26

3p + 3q = 18

Column A

Column B

4x + y = 13 and 2x + y = 7
x = 5 and y= 2
2x + 4y = 18 and 2x + 2y = 8
x = -1 and y = 5
7x - 4y = 22 and 4x + 4y = 0
x = 2 and y = -2
7x - 2y = 31 and 5x - 2y = 21
x = 3 and y = 1

Ted buys 3 ties and two shirts for £100

Kieran buys 3 ties and 5 shirts for £169

Which of the following pairs of equations are the correct ones for this problem? You do not have to solve the equations.

3s + 2t = 100 and 3s + 5t = 169

3t + 2s = 100 and 3t + 5s = 169

3t - 2s = 100 and 3t + 5s = 169

3t + 2s = 100 and 3t - 5s = 169

In question 7 you found the correct equations to solve the problem

Ted buys 3 ties and two shirts for £100

Kieran buys 3 ties and 5 shirts for £169

Now solve these equations to find the cost of a tie and the cost of a shirt. Do your working on paper and fill your answers into the blanks.

3s + 2t = 100 and 3s + 5t = 169

3t + 2s = 100 and 3t + 5s = 169

3t - 2s = 100 and 3t + 5s = 169

3t + 2s = 100 and 3t - 5s = 169

Jack and Jill are each asked to think of a number between -5 and +5. When Jack's number is multiplied by 5 and Jill's number is subtracted the answer is -16. But when Jack's number is multiplied by 3 and Jill's number subtracted the answer is -10. Using x for Jack's number and y for Jill's number write this information as a pair of simultaneous equations by filling in the blanks below. 

Then solve your equations on paper and fill in the solutions in the blanks.

3s + 2t = 100 and 3s + 5t = 169

3t + 2s = 100 and 3t + 5s = 169

3t - 2s = 100 and 3t + 5s = 169

3t + 2s = 100 and 3t - 5s = 169

Claire is buying office stationery. She can buy 400 pencils and 250 erasers for £71 or she can buy 550 pencils and 250 erasers for £84.50.

Use this information to set up a pair of simultaneous equations by filling in the blanks below. Use p = cost of a pencil in pence and e = cost of an eraser in pence

Then solve your equations on paper and fill in the solutions in the blanks to find the cost of a pencil and the cost of an eraser.

3s + 2t = 100 and 3s + 5t = 169

3t + 2s = 100 and 3t + 5s = 169

3t - 2s = 100 and 3t + 5s = 169

3t + 2s = 100 and 3t - 5s = 169

  • Question 1

Look at the following pairs of simultaneous equations and decide if you would add or subtract them to eliminate one of the variables. You do not need to solve the equations.

 

CORRECT ANSWER
 AddSubtract
4x + y = 13 and 2x + y = 7
3x + 2y = 19 and 5x - 2y = 23
3x - 4y = 5 and 3x + 2y = 26
x - 2y = 45 and 6x - 2y = 56
EDDIE SAYS
In the first one, there is a +y in each equation. The signs are the same so we subtract. In the second there is a +2y and a -2y so the signs are different and we add. In the third, each starts with a 3x which are both positive so we subtract. In the last, both have a -2y. Same sign subtract. Hopefully, you got these OK. Now to solve them......
  • Question 2

Below, you will see the steps to solving the simultaneous equations

8x + 3y = 19 

3x - 3y = 3 

but not in the correct order.

Match each step to its correct position.

CORRECT ANSWER

Column A

Column B

1st
8x + 3y = 19 (1) and 3x - 3y = 3 ...
2nd
11x = 22 (1) + (2)
3rd
x = 2
4th
8x2 + 3y = 19 [sub into (1)]
5th
16 + 3y = 19
6th
3y = 3
7th
y = 1
8th
3x2 - 3x1 = 3 [check in (2)]
EDDIE SAYS
8x + 3y = 19 (1) 3x - 3y = 3 (2) The signs in front of y are different so we add the equations. 11x = 22 (1) + (2) x = 2 [divide by 11] Then we sub x = 2 into equation (2) 8x2 + 3y = 19 And then solve 16 + 3y = 19 3y = 3 y = 1 Finally, we check our solutions by substituting them into (2) 3x2 - 3x1 = 3 [correct] Did you get them all in the right order?
  • Question 3

Below you will see all the steps to solving the equations 

2x + 6y = 14

2x - 2y = -2

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
We start by subtracting the equations since the signs in front of the 2x are the same. Be careful here as when subtracting a negative you are adding. 8y = 16 (1) - (2) y = 2 [divide by 8] Now sub y = 2 into (1) 2x + 6 x 2 = 14 [sub in (1)] 2x + 12 = 14 [solve] 2x = 2 x = 1 2 x 1 - 2 x 2 = -2 [check in (2)] 2 - 4 = -2 [correct]
  • Question 4

Below you will see all the steps to solving the equations 

3a - 4b = 20

6a - 4b = 32

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
We start by subtracting the equations since both equations have -4b. SInce we are subtracting it is helpful if equation (2) is on top as it has the larger values. 6a - 4b = 32 (2) 3a - 4b = 20 (1) 3a = 12 (2) - (1) a = 4 [divide by 3] 3 x 4 - 4b = 20 [sub in (1)] 12 - 4b = 20 [-12 from both sides, don't forget the '-' in front of 4b] -4b = 8 b = -2 [divide by -4] 6 x 4 - 4 x -2 = 32 [check in (2)] 24 + 8 = 32 [remember -4 x -2 = +8] How's it going so far? Getting the hang of them?
  • Question 5

Match the following pairs of simultaneous equations to their correct solutions.

CORRECT ANSWER

Column A

Column B

4x + y = 13 and 2x + y = 7
x = 3 and y = 1
2x + 4y = 18 and 2x + 2y = 8
x = -1 and y = 5
7x - 4y = 22 and 4x + 4y = 0
x = 2 and y = -2
7x - 2y = 31 and 5x - 2y = 21
x = 5 and y= 2
EDDIE SAYS
For this question, you could try solving each pair of equations, but a quicker method would be to try substituting each pair of solutions into each pair of equations until you get the correct answer. So with the first pair of equations, the solutions are x = 3 and y = 1 since 4 x 3 + 1 = 13 and 2 x3 + 1 = 7. For the second pair x = -1 and y = 5 work since 2 x -1 + 4 x 5 = 18 and 2 x -1 + 2 x 5 = 8 For the 3rd pair x = 2 and y = -2 work since 7 x 2 - 4 x -2 = 22 and 4 x 2 + 4 x -2 = 0 Finally, for the last pair x = 5 and y = 2 work since 7 x 5 - 2 s = 31 and 5 x 5 - 2 x 2 = 21
  • Question 6

On a piece of paper solve the following pair of equations and type your solutions into the blanks.

7p + 3q = 26

3p + 3q = 18

CORRECT ANSWER
EDDIE SAYS
7p + 3q = 26 (1) 3p + 3q = 18 (2) 3q is the same in both so we subtract 4p = 8 p = 2 Sub into (2) 3x2 +3q = 18 6 + 3q = 18 3q = 12 q = 4 7 x 2 + 3 x 4 = 26 [check in (1)] 14 + 12 = 26 [correct] How did you do with this one? Did you get all the steps correct?
  • Question 7

Ted buys 3 ties and two shirts for £100

Kieran buys 3 ties and 5 shirts for £169

Which of the following pairs of equations are the correct ones for this problem? You do not have to solve the equations.

CORRECT ANSWER
3t + 2s = 100 and 3t + 5s = 169
EDDIE SAYS
Ted buys 3 ties and two shirts for £100. Using t = cost of a tie and s = cost of a shirt we have 3t + 2s = 100. Kieran buys 3 ties and 5 shirts for £169. This becomes 3t + 5s = 169. These are the correct equations.
  • Question 8

In question 7 you found the correct equations to solve the problem

Ted buys 3 ties and two shirts for £100

Kieran buys 3 ties and 5 shirts for £169

Now solve these equations to find the cost of a tie and the cost of a shirt. Do your working on paper and fill your answers into the blanks.

CORRECT ANSWER
EDDIE SAYS
So the correct equations are 3t + 2s = 100 (1) 3t + 5s = 169 (2) Both have 3t so we subtract after swapping them over. 3t + 5s = 169 (2) 3t + 2s = 100 (1) 3s = 69 s = 23 3t + 2 x 23 = 100 [sub in (1)] 3t + 46 = 100 3t = 54 t = 18 3 x 18 + 5 x 23 = 169 [check in 2] 54 + 115 = 169 [correct] So a tie costs £18 and a shirt costs £23.
  • Question 9

Jack and Jill are each asked to think of a number between -5 and +5. When Jack's number is multiplied by 5 and Jill's number is subtracted the answer is -16. But when Jack's number is multiplied by 3 and Jill's number subtracted the answer is -10. Using x for Jack's number and y for Jill's number write this information as a pair of simultaneous equations by filling in the blanks below. 

Then solve your equations on paper and fill in the solutions in the blanks.

CORRECT ANSWER
EDDIE SAYS
So the equations are 5x - y = -16 (1) 3x - y = -10 (2) 2x = -6 (1) - (2) x = -3 3 x -3 - y = -10 [sub in (2)] -9 - y = -10 -y = -1 [add 9] y = 1 5 x -3 - 1 = 16 [check in 1] -15 - 1 = -16 [correct]
  • Question 10

Claire is buying office stationery. She can buy 400 pencils and 250 erasers for £71 or she can buy 550 pencils and 250 erasers for £84.50.

Use this information to set up a pair of simultaneous equations by filling in the blanks below. Use p = cost of a pencil in pence and e = cost of an eraser in pence

Then solve your equations on paper and fill in the solutions in the blanks to find the cost of a pencil and the cost of an eraser.

CORRECT ANSWER
EDDIE SAYS
So the equations are 400p + 250e = 7100 (1) 550p + 250e = 8450 (2) 400p + 250e = 7100 (1) [put (1) underneath for subtracting] 150p = 1350(2) - (1) p = 9 400 x 9 + 250e = 7100 [sub in (1)] 3600 + 250e = 7100 250e = 3500 e = 14 550 x 9 + 250 x 14 = 8450 [check in 2] 4950 + 3500 = 8450 [correct] That's it. You've finished. How was it?
---- OR ----

Sign up for a £1 trial so you can track and measure your child's progress on this activity.

What is EdPlace?

We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.

Get started
laptop

Start your £1 trial today.
Subscribe from £10/month.