 # Solve Simultaneous Equations (by multiplying one equation)

In this worksheet, students will learn how to solve simultaneous equations using the elimination method, in which one equation needs to be multiplied. They will also solve contextualised question based on simultaneous equations. 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:   ### QUESTION 1 of 10

A group of people went to the cinema, where they a number of bought adult tickets and children's tickets.

Mr. and Mrs. Baker took their daughters Molly, Polly and Dolly and paid £33, whereas Mr. Butcher took his two sons Ron and Don and paid £19.

One other family went to the cinema - the Fishers.

Can you work out how much Mr. and Mrs. Fisher and their daughter April would pay?

You may recognise this sort of problem as one which can be solved using simultaneous equations.

You may want to review some of the alternative methods for solving simultaneous equations (such as elimination or substitution) before continuing with this activity.

This activity will focus on solving simultaneous equations by multiplying the whole of one equation so that the values of one of the variables cancel each other out

In order to find out how much Mr. and Mrs. Fisher and their daughter will pay, we first need to know the cost of an adult ticket and the cost of a child ticket.

These are known as our variables.

We will use a = cost of an adult and c = cost of a child.

So, the information relating to Mr. and Mrs. Baker and their daughters can be written as: 2a + 3c = 33

Mr. Butcher and his 2 sons become: a + 2c = 19

We now have our two equations, and we need to solve them simultaneously:

2a + 3c = 33   (1)

a + 2c  =  19   (2)

At this point with other pairs of simultaneous equations, we would either add or subtract our equations to eliminate one of the variables.

However, that will not help us here since neither the coefficients of a nor of c are the same.

So what we have to do is match one coefficient by multiplying the whole of one equation.

So, if we take equation (2) and multiply it by 2 we get: 2a + 4c = 38

This means that 2 adults and 4 children would pay £38.

We can call this equation (3) and now place it above equation (1):

2a + 4c = 38   (3)

2a + 3c  =  33   (1)

Now we have the same number of a's in both equations, and the signs in front are the same, so we can subtract (1) from (3):

2a + 4c = 38   (3)

2a + 3c  =  33   (1)

c = 5   (3) - (1)

So, a child's ticket costs £5.

Now, we need to find the cost of an adult ticket, by substituting c = 5 into any of our three equations:

a + 2 x 5 = 19

a + 10 = 19

a = 10.

So an adult ticket costs £10.

Now we can find the cost for the Fisher family (which is two adults and one child):

2 x 9 + 5 = 23, so they would pay £23.

If we compare what we have done here with other simultaneous equation pairs, you will see that we have just added in one extra stage where we multiply one equation by a number, in order to equate the coefficients of one variable.

Other than that, everything else is the same process which you may have practised before.

Now, let's look at another pair of simultaneous equations.

This time we have been provided with just the equations without the real-life problem.

Solve 3x + 2y = 10  (1)

and 7x - y = 29    (2)

Notice that the coefficients of x are different, as are the coefficients of y.

But if we double equation (2), we will have 2y in each equation:

3x + 2y = 10  (1)

7x - y = 29    (2) ×2

14x - 2y = 58  (3)

Since the signs in front of the y's are different, we need to add equations (1) and (3):

3x + 2y = 10  (1)

14x - 2y = 58  (3)

17x = 68

x = 4   [divide by 17]

3 × 4 + 2y = 10  [sub into (1)]

12 + 2y = 10

2y = -2  [subtract 12]

y = -1   [divide by 2]

7 x 4 - -1 = 29  [check in (2)]

28 + 1 = 29 [correct]

So x = 4 and y = -1.

Now you should be ready to try solving some pairs of your own!

This process has many steps and can get complicated, so be sure to have a pen and paper handy to help you.

Remember, that we just need to follow our regular process for solving simultaneous equations with one extra step added - we need to multiply the whole of one of the equations so that we can match one of the coefficients so they cancel each other out.

Consider the pair of equations below:

5a + 3b = 13

7a + 6b = 20

Below you will see the steps to take to solve these simultaneous equations, but not in the correct order.

Match each step to its correct position in the process.

## Column B

1st
10 + 3b = 13
2nd
7 x 2 + 6 x 1 = 20 [check in (2)]
3rd
Multiply (1) by 2
4th
b = 1
5th
a = 2
6th
3a = 6 [(3) - (2)]
7th
3b = 3
8th
5 x 2 + 3b = 13 [sub into (1)]
9th
5a + 3b = 13 (1) and 7a + 6b = 20 (2)
10th
10a + 6b = 26 (3) and 7a + 6b = 20 (2)

Review the pair of equations below:

11x + 3y = 71

5x - y = 37

Below you will see the steps to take to solve these simultaneous equations, but not in the correct order.

Match each step to its correct position in the process.

## Column B

1st
3y = -6
2nd
y = -2
3rd
Multiply (2) by 3
4th
11x + 3y = 71 (1) and 15x - 3y = 111 (3)
5th
5 x 7 - -2 = 37 [check in (2)]
6th
77 + 3y = 71
7th
26x = 182 (1) + (3)
8th
x = 7
9th
11x + 3y = 71 (1) and 5x - y = 37 (2)
10th
11 x 7 + 3y =71 [sub in (1)]

Below you will see all the steps we need to take to solve these equations:

4c + 5d = 17

8c + 3d = 27

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

## Column B

1st
3y = -6
2nd
y = -2
3rd
Multiply (2) by 3
4th
11x + 3y = 71 (1) and 15x - 3y = 111 (3)
5th
5 x 7 - -2 = 37 [check in (2)]
6th
77 + 3y = 71
7th
26x = 182 (1) + (3)
8th
x = 7
9th
11x + 3y = 71 (1) and 5x - y = 37 (2)
10th
11 x 7 + 3y =71 [sub in (1)]

Below you will see all the steps we need to take to solve these equations:

9a + 2b = 41

5a - 4b = 33

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

## Column B

1st
3y = -6
2nd
y = -2
3rd
Multiply (2) by 3
4th
11x + 3y = 71 (1) and 15x - 3y = 111 (3)
5th
5 x 7 - -2 = 37 [check in (2)]
6th
77 + 3y = 71
7th
26x = 182 (1) + (3)
8th
x = 7
9th
11x + 3y = 71 (1) and 5x - y = 37 (2)
10th
11 x 7 + 3y =71 [sub in (1)]

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

## Column B

9x + 2y = 31 and 3x + y = 11
x = 3 and y = 2
5x + 3y = 27 and x + y = 7
x = 3 and y = 4
2x + 7y = 3 and x - y = 6
x = 5 and y= 4
11x + 2y = 63 and 6x + 4y = 46
x = 5 and y = -1

Solve the following pair of equations:

3m - 4n = 13

7m + 2n = 36

Then type your solutions into the spaces below.

## Column B

9x + 2y = 31 and 3x + y = 11
x = 3 and y = 2
5x + 3y = 27 and x + y = 7
x = 3 and y = 4
2x + 7y = 3 and x - y = 6
x = 5 and y= 4
11x + 2y = 63 and 6x + 4y = 46
x = 5 and y = -1

A florist shop sells two types of bouquets: budget and standard.

A budget bouquet costs £5 and a standard bouquet costs £7.

One day, the florist sold 17 bouquets altogether, which cost a total of £109.

Using b = number of budget bouquets and s = number of standard bouquets, choose the two correct equations from the list below which accurately represent the information above.

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

In the previous question, you found the correct equations to solve this problem:

A florist shop sells two types of bouquets: budget and standard.

A budget bouquet costs £5 and a standard bouquet costs £7.

One day, the florist sold 17 bouquets altogether, which cost a total of £109.

Now solve these equations to find the number of budget bouquets and the number of standard bouquets sold.

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

In a cafe, four cups of tea and 3 biscuits costs £7.05, whereas 3 cups of tea and one biscuit costs £4.35.

Using t = cost of a cup of tea in pence and b = cost of a biscuit in pence, form a pair of simultaneous equations.

Then solve these equations by filling in the blanks below.

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

Greg and Hannah both think of whole numbers between 1 and 10.

If Greg's number is multiplied by 5, and double Hannah's number is subtracted, the answer is 14.

If Greg's number is multiplied by 3, and Hannah's number is subtracted, the answer is 9.

Using g = Greg's number and h = Hannah's number, form a pair of simultaneous equations expressing the information above.

Now solve your equations to find the numbers Greg and Hannah originally thought of.

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

• Question 1

Consider the pair of equations below:

5a + 3b = 13

7a + 6b = 20

Below you will see the steps to take to solve these simultaneous equations, but not in the correct order.

Match each step to its correct position in the process.

## Column B

1st
5a + 3b = 13 (1) and 7a + 6b = 2...
2nd
Multiply (1) by 2
3rd
10a + 6b = 26 (3) and 7a + 6b = ...
4th
3a = 6 [(3) - (2)]
5th
a = 2
6th
5 x 2 + 3b = 13 [sub into (1)]
7th
10 + 3b = 13
8th
3b = 3
9th
b = 1
10th
7 x 2 + 6 x 1 = 20 [check in (2)...
EDDIE SAYS
We cannot simply solve this pair of equations, as neither the values of a or b match, which we need to see in order for the variables to cancel each other out. So that we can equate the coefficients of b, we need to double equation (1) and then call our new equation (3). Our symbols relating to b are the same, so we need to subtract (2) from (3) to eliminate b. Then we can find the value of a and substitute this back into equation (1). Now we can solve this equation to find the value of b. Finally, we can substitute both a = 2 and b = 1 into equation (2) to check our answers work and we reach the correct output. Great news - we do, so our method here works! Take a moment to review the Introduction now before we move on to tackle solve some more simultaneous equations.
• Question 2

Review the pair of equations below:

11x + 3y = 71

5x - y = 37

Below you will see the steps to take to solve these simultaneous equations, but not in the correct order.

Match each step to its correct position in the process.

## Column B

1st
11x + 3y = 71 (1) and 5x - y = 3...
2nd
Multiply (2) by 3
3rd
11x + 3y = 71 (1) and 15x - 3y =...
4th
26x = 182 (1) + (3)
5th
x = 7
6th
11 x 7 + 3y =71 [sub in (1)]
7th
77 + 3y = 71
8th
3y = -6
9th
y = -2
10th
5 x 7 - -2 = 37 [check in (2)]
EDDIE SAYS
To equate the coefficients of y, we need to multiply equation (2) by 3 and call this new equation (3). Because the signs related to y are different, we need to add (1) and (3) to eliminate y. Then find the value of x and substitute this back into equation (1). Now we can solve this equation to find the value of y. Finally, we can substitute both x = 7 and y = -2 into equation (2) to check this works and gives the correct output. Did you get those steps in the correct order?
• Question 3

Below you will see all the steps we need to take to solve these equations:

4c + 5d = 17

8c + 3d = 27

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

EDDIE SAYS
Here's the full solution for this one: 4c + 5d = 17 (1) 8c + 3d = 27 (2) Then let's equate the coefficients of c by multiplying (1) by 2: 8c + 10d = 34 (3) 8c + 3d = 27 (2) Our signs here are same, so we need to subtract our equations: (3) - (2) 7d = 7 d = 1 Now let's substitute d = 1 into (1) to find c: 4c + 5 x 1 = 17 4c + 5 = 17 4c = 12 c = 3 Let's check our answers by substituting into (2): 8 x 3 + 3 x 1 = 27 24 + 3 = 27 [correct] Did you manage to fill those gaps with the correct numbers?
• Question 4

Below you will see all the steps we need to take to solve these equations:

9a + 2b = 41

5a - 4b = 33

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

EDDIE SAYS
Here's our full solution this time: 9a + 2b = 41 (1) 5a - 4b = 33 (2) Let's multiply (1) by 2 so that the coefficients of b cancel out: 18a + 4b = 82 (3) As our signs related to b are different, we need to add our equations here: (2) + (3) 23a = 115 a = 5 Now let's substitute this value for a into (1) to find b: 9 × 5 + 3b = 41 45 + 3b = 41 3b = -4 b = -2 We can now check both these values in (2) to ensure they give the correct output: 5 × 5 - 4 × -2 = 33 25 + 8 = 33 [correct] How's it going so far? Are you getting the hang of this process?
• Question 5

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

## Column B

9x + 2y = 31 and 3x + y = 11
x = 3 and y = 2
5x + 3y = 27 and x + y = 7
x = 3 and y = 4
2x + 7y = 3 and x - y = 6
x = 5 and y = -1
11x + 2y = 63 and 6x + 4y = 46
x = 5 and y= 4
EDDIE SAYS
For this question, you could try solving each pair of equations one at a time, but a quicker method would be to substitute each pair of solutions into each pair of equations, until you get a match. So with the first pair of equations, the solutions are x = 3 and y = 2 since: 9 × 3 + 2 × 2 = 31 and 3 × 3 + 2 = 11 For the second pair, x = 3 and y = 4 work since: 5 × 3 + 3 × 4 = 27 and 3 + 4 = 7 For the third pair, x = 5 and y = -1 work since: 2 × 5 + 7 × -1 = 3 and 5 - -1 = 6 Finally, for the last pair, x = 5 and y = 4 work since: 11 × 5 + 2 × 4 = 63 and 6 × 5 + 4 × 4 = 46 That was a bit of a shortcut, wasn't it?
• Question 6

Solve the following pair of equations:

3m - 4n = 13

7m + 2n = 36

Then type your solutions into the spaces below.

EDDIE SAYS
Let's layout our equations to start: 3m - 4n = 13 (1) 7m + 2n = 36 (2) Then we can multiply (2) by 2 so that the values of n match and cancel each other out: 14m + 4n = 72 (3) As the signs attached to n differ here, we need to add our equations: (1) + (3) 17m = 85 m = 5 Next let's substitute m = 5 into (2) to find the value of n: 7 × 5 + 2n = 36 35 + 2n = 36 2n = 1 n = 0.5 Then we can check these values work in (1) and give the correct output: 3 × 5 - 4 × 0.5 = 13 15 - 2 = 13 (correct) How similar was your working to ours above? Don't forget to use a pen and paper to keep track of your working. Did you get all those steps correct?
• Question 7

A florist shop sells two types of bouquets: budget and standard.

A budget bouquet costs £5 and a standard bouquet costs £7.

One day, the florist sold 17 bouquets altogether, which cost a total of £109.

Using b = number of budget bouquets and s = number of standard bouquets, choose the two correct equations from the list below which accurately represent the information above.

5b + 7s = 109
b + s = 17
EDDIE SAYS
We know that the total number of bouquets bought was 17, so that means that b + s = 17 is one equation. The cost of the budget bouquets is 5b and the cost of the standard bouquets is 7s. When these are added, the total is 109 so this means: 5b + 7s = 109 Did you choose the correct two equations? Let's move on to solve these now in the next question...
• Question 8

In the previous question, you found the correct equations to solve this problem:

A florist shop sells two types of bouquets: budget and standard.

A budget bouquet costs £5 and a standard bouquet costs £7.

One day, the florist sold 17 bouquets altogether, which cost a total of £109.

Now solve these equations to find the number of budget bouquets and the number of standard bouquets sold.

EDDIE SAYS
So the correct equations for us to use are: b + s = 17 (1) 5b + 7s = 109 (2) To equate coefficients, we could either multiply equation (1) by 5 or 7. Here we will use 5. b + s = 17 (1 × 5) 5b + 5s = 85 (3) As our signs relating to s are the same, we need to subtract our equations here: (2) - (3) 2s = 24 s = 12 Let's substitute this value for s into (1) to find b: b + 12 = 17 b = 5 So 5 budget and 12 standard bouquets were sold.
• Question 9

In a cafe, four cups of tea and 3 biscuits costs £7.05, whereas 3 cups of tea and one biscuit costs £4.35.

Using t = cost of a cup of tea in pence and b = cost of a biscuit in pence, form a pair of simultaneous equations.

Then solve these equations by filling in the blanks below.

EDDIE SAYS
The important thing to remember here is to switch our prices from £s to p. So the equations we need to use are: 4t + 3b = 705 (1) 3t + b = 435 (2) Then we can multiply (2) by 3 so that the coefficients of b balance: 3t + b = 435 (2) ×3 9t + 3b = 1305 (3) Our symbols attached to b are the same here, so we need to subtract our equations: (3) - (1) 5t = 600 t = 120 Now let's substitute this value for t into (2) to find b: 3 x 120 + b = 435 360 + b = 435 b = 75 We can check both these values in (1) to ensure we are correct: 4 x 120 + 3 x 75 = 705 480 + 225 = 705 [correct] So a cup of tea costs £1.20 and a biscuit costs 75p. Has this question made you ready for something to eat and drink?
• Question 10

Greg and Hannah both think of whole numbers between 1 and 10.

If Greg's number is multiplied by 5, and double Hannah's number is subtracted, the answer is 14.

If Greg's number is multiplied by 3, and Hannah's number is subtracted, the answer is 9.

Using g = Greg's number and h = Hannah's number, form a pair of simultaneous equations expressing the information above.

Now solve your equations to find the numbers Greg and Hannah originally thought of.

EDDIE SAYS
So the equations we need to work with are: 5g - 2h = 14 (1) 3g - h = 9 (2) Can you see how these relate to the data provided? Let's multiply (2) by 2 so that the coefficients of h match: 3g - h = 9 (2) × 3 6g - 2h = 18 (3) Since the signs in front of h are the same, we need to subtract our equations: (3) - (1) g = 4 Next we can substitute this value for g into (2) to find the value of h: 3 × 4 - h = 9 12 - h = 9 h = 3 So Greg's number was 4 and Hannah's was 3. That's it - you've finished! How was that? Why not move on to learn about how to use graphs to solve simultaneous equations now?
---- 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 