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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

A group of people went to London Zoo, where they bought a number of adult tickets and childrens' 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.

The final family group attending were the Blacks.

Can you work out how much Ms. Black, with 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 will 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's 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 we can use any letter we choose.

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 in them, 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 these two facts, 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 that they work for the other family (the equation we have not yet used). 

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 have been given without the real-life problem.

 

Solve  5x + y = 17 and 3x + y = 11

 

Let's number our equations (1) and (2) with the 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

 

Let's check these values work with equation (1) [i.e. the equation we did not use to find our values]

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

 

 

 

Let's look at one more problem to check you have it.

 

Solve  5x - y = 17 and 2x + y = 11

 

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.

 

Because the signs in front of the equations are different (one positive and one negative) this time we need to add them:

7x = 28   (1) + (2)

x = 4   [divide by 7]

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

8 + y = 11

y = 3

 

Let's check this in (1):

5 x 4 - 3 = 17 [correct]

So x = 4 and y = 3.

 

 

 

TOP TIPS:


Remember the variable with equal coefficients will be eliminated when the equations are combined.

 

If the signs in front are the same, you subtract, but if they are different then you add.

Remember the catchphrase: Same Sign Subtract (SSS) to help you commit this rule to memory. 

 

 

 

In this activity, we will find the values of variable by solving pairs of simultaneous equations.

 

You may want to have a pen and paper handy to record your working and so you can compare it to our maths teacher's method. 

Look at the pairs of simultaneous equations below.

 

Then decide if you would add or subtract them to eliminate one of the variables.

 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

Consider these equations: 

 

8x + 3y = 19 

3x - 3y = 3 

 

Below, you will see the steps required to solve these simultaneous equations.

 

Place them in the correct order by matching each stage to its correct position.

Column A

Column B

1st
x = 2
2nd
11x = 22 (1) + (2)
3rd
(8 × 2) + 3y = 19
4th
y = 1
5th
16 + 3y = 19
6th
(3 × 2) - (3 × 1) = 3
7th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)
8th
3y = 3

Consider these equations: 

 

2x + 6y = 14

2x - 2y = -2

 

Fill in the blanks below to complete the solution and solve these equations. 

Column A

Column B

1st
x = 2
2nd
11x = 22 (1) + (2)
3rd
(8 × 2) + 3y = 19
4th
y = 1
5th
16 + 3y = 19
6th
(3 × 2) - (3 × 1) = 3
7th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)
8th
3y = 3

Consider these equations: 

 

3a - 4b = 20

6a - 4b = 32

 

Fill in the blanks below to complete the solution and solve these equations. 

Column A

Column B

1st
x = 2
2nd
11x = 22 (1) + (2)
3rd
(8 × 2) + 3y = 19
4th
y = 1
5th
16 + 3y = 19
6th
(3 × 2) - (3 × 1) = 3
7th
8x + 3y = 19 (1) and 3x - 3y = 3 (2)
8th
3y = 3

Match each pair of simultaneous equations below to their correct solutions.

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 = 5 and y= 2
7x - 2y = 31 and 5x - 2y = 21
x = 2 and y = -2

Solve the following pair of equations and type your solutions into the blanks below:

 

7p + 3q = 26

3p + 3q = 18

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 = 5 and y= 2
7x - 2y = 31 and 5x - 2y = 21
x = 2 and y = -2

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 correctly represent this problem?

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 the last question, we found the correct equations to solve this 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.

 

Fill your answers into the blanks below. 

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 to represent 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 type your solutions in the blanks below.

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 to represent the cost of a pencil in pence, and e to represent the cost of an eraser in pence.

 

Then solve your equations on paper and type your 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 pairs of simultaneous equations below.

 

Then decide if you would add or subtract them to eliminate one of the variables.

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
We need to remember our catchphrase here: Same Sign Subtract (SSS) So we need to decide overall if the equations have the same sign (and so would be subtracted), or different signs (and so would be added). In the first equation, there is a '+ y' in each option. These signs are the same so we need to subtract. In the second, there is '+ 2y' and a '-2y' so the signs are different, which means we would add them. In the third, each starts with a '3x' which are both positive, so we would subtract. In the last, both equations have a '- 2y'. They have the Same Sign, so we need to Subtract them. Hopefully, you applied our rule successfully. Now let's move on to solve them...
  • Question 2

Consider these equations: 

 

8x + 3y = 19 

3x - 3y = 3 

 

Below, you will see the steps required to solve these simultaneous equations.

 

Place them in the correct order by matching each stage 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
(8 × 2) + 3y = 19
5th
16 + 3y = 19
6th
3y = 3
7th
y = 1
8th
(3 × 2) - (3 × 1) = 3
EDDIE SAYS
1) Let's start by listing our new equations: 8x + 3y = 19 (1) 3x - 3y = 3 (2) 2) The signs in front of y are different, so we need to add these equations: 11x = 22 (1) + (2) 3) Then we need to divide by 11 to find the value of x: x = 2 4) Then we substitute x = 2 into equation (2): (8 × 2) + 3y = 19 5) Let's simplify this: 16 + 3y = 19 6) And solve: 3y = 3 7) y = 1 8) Finally, we need to check our solutions by substituting them into (2): (3 × 2) - (3 × 1) = 3 [correct] Did you get all those steps in the right order?
  • Question 3

Consider these equations: 

 

2x + 6y = 14

2x - 2y = -2

 

Fill in the blanks below to complete the solution and solve these equations. 

CORRECT ANSWER
EDDIE SAYS
Since the signs in front of the '2x' are the same, we will be subtracting the equations. Be careful here, as when we subtract a negative, we actually need to add it. (1) - (2): 8y = 16 y = 2 [divide by 8] Now substitute y = 2 into (1): 2x + 6 × 2 = 14 2x + 12 = 14 Solve: 2x = 2 x = 1 Check in (2): 2 x 1 - 2 x 2 = -2 2 - 4 = -2 [correct] How did you get on with that one?
  • Question 4

Consider these equations: 

 

3a - 4b = 20

6a - 4b = 32

 

Fill in the blanks below to complete the solution and solve these equations. 

CORRECT ANSWER
EDDIE SAYS
Sine both the equations have '- 4b', their signs are the same, therefore we need to subtract them. Since we are subtracting, it is helpful to put equation (2) on top, as it has the larger values. 6a - 4b = 32 (2) 3a - 4b = 20 (1) (2) - (1): 3a = 12 a = 4 [divide by 3] Substitute in (1): 3 x 4 - 4b = 20 12 - 4b = 20 [-12 from both sides] -4b = 8 b = -2 [divide by -4] Check in (2): 6 x 4 - 4 x -2 = 32 24 + 8 = 32 [Remember -4 x -2 = +8] which is correct How's it going so far? Are you getting the hang of these?
  • Question 5

Match each pair of simultaneous equations below 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, we could try solving each pair of equations one at a time. However, a quicker method would be to try substituting each pair of solutions into each pair of equations until we find the correct matches. So with the first pair of equations, the solutions are x = 3 and y = 1 since: 4 × 3 + 1 = 13 and 2 × 3 + 1 = 7 For the second pair, x = -1 and y = 5 work since: 2 × -1 + 4 × 5 = 18 and 2 × -1 + 2 × 5 = 8 For the 3rd pair, x = 2 and y = -2 work since: 7 × 2 - 4 × -2 = 22 and 4 × 2 + 4 × -2 = 0 Finally, for the last pair, x = 5 and y = 2 work since: 7 × 5 - 2 × 2 = 31 and 5 × 5 - 2 × 2 = 21 That was much simpler, wasn't it?
  • Question 6

Solve the following pair of equations and type your solutions into the blanks below:

 

7p + 3q = 26

3p + 3q = 18

CORRECT ANSWER
EDDIE SAYS
Let's line up our equations: 7p + 3q = 26 (1) 3p + 3q = 18 (2) We have two positive values (3q) here which are the same, so we will need to subtract our equations: (1) - (2) = 4p = 8 p = 2 Let's substitute this value of p into (2): 3 × 2 +3q = 18 6 + 3q = 18 3q = 12 q = 4 Now we can check these two values work in (1): 7 x 2 + 3 x 4 = 26 14 + 12 = 26 [correct] Did you use a pen and paper to write our your full working?
  • 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 correctly represent this problem?

CORRECT ANSWER
3t + 2s = 100 and 3t + 5s = 169
EDDIE SAYS
Ted buys 3 ties and two shirts for £100. We are using t to represent the cost of a tie and s to represent the cost of a shirt. So we know that: 3t + 2s = 100 Kieran buys 3 ties and 5 shirts for £169. Using algebra, this fact becomes: 3t + 5s = 169 These are the correct two equations which we will need to work with to solve this problem. Now let's move on to find out how much a tie or a shirt costs...
  • Question 8

In the last question, we found the correct equations to solve this 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.

 

Fill your answers into the blanks below. 

CORRECT ANSWER
EDDIE SAYS
So the correct equations are: 3t + 2s = 100 (1) 3t + 5s = 169 (2) Both equations have the same positive values (3t) so we need to subtract here. Equation (2) has larger values, so let's place this on top: (2) - (1) = 3t + 5s = 169 (2) 3t + 2s = 100 (1) Then let's solve to find s: 3s = 69 s = 23 Let's substitute this value into (1): 3t + (2 × 23) = 100 3t + 46 = 100 3t = 54 t = 18 Now we can check these values work in (2): 3 × 18 + 5 × 23 = 169 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 to represent 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 type your solutions in the blanks below.

CORRECT ANSWER
EDDIE SAYS
So our starting equations are: 5x - y = -16 (1) 3x - y = -10 (2) Both values of x are positive here, so we will need to subtract our equations: (1) - (2) = 2x = -6 x = -3 Now let's substitute this value into (2): 3 × -3 - y = -10 -9 - y = -10 -y = -1 [add 9] y = 1 Finally, let's check these values work in (1): 5 x -3 - 1 = 16 -15 - 1 = -16 [correct] Did you get all of those numbers into the correct blanks?
  • 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 to represent the cost of a pencil in pence, and e to represent the cost of an eraser in pence.

 

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

CORRECT ANSWER
EDDIE SAYS
So our starting equations need to be: 400p + 250e = 7100 (1) 550p + 250e = 8450 (2) Both values of p are positive, so we will be subtracting our equations. Position (1) underneath for subtracting as its smaller: (2) - (1) = 150p = 1350 p = 9 Let's substitute this value into (1): 400 x 9 + 250e = 7100 3600 + 250e = 7100 250e = 3500 e = 14 We can check both values are correct using (2): 550 x 9 + 250 x 14 = 8450 4950 + 3500 = 8450 [correct] That's it - you've finished this activity! How was it? Don't forget that you can earn marks for your working in problems with complex processes, so always write down what you are doing. Are you feeling more confident now and ready for a challenge? Why not try our Level 2 activity on simultaneous equations?
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