 # Solve Simultaneous Equations (by multiplication)

In this worksheet, students will learn how to solve simultaneous equations using the elimination method, in which both equations 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

Some people went to the theatre, where they bought a number of adult and children's tickets.

Mrs. Mayor took her two parents and her four children and paid £203, whereas Mr. and Mrs. Clarke took their three children and paid £143.

One other family went to the theatre - the Sheriffs.

Can you work out how much Mr. Sheriff and his two daughters would pay?

You may already recognise questions like this.

This activity builds on ideas established other levels of simultaneous equations activities, so if you have not completed those activities yet or are not feeling totally confident with this concept, then you may want to complete them before continuing with this.

In order to find out how much Mr. Sheriff and his daughters will pay, we first need to know the cost of an adult ticket and the cost of a children's ticket.

These are known as variables.

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

So, the Mayor family becomes 3a + 4c = 203, whereas the Clarke family becomes 2a + 3c = 143.

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

3a + 4c = 203   (1)

2a + 3c = 143   (2)

At this point, we want to eliminate one of the variables by adding or subtracting the equations.

We do this by ensuring that either the coefficients (numbers) attached to a or c cancel each other out.

There is no whole number we can multiply equation (2) by that will do this, so this time we need to multiply equation (1) by one number and equation (2) by a different number to achieve this.

If we want to equate the coefficient of c, we can multiply (1) by 3 and (2) by 4 which will give us 12c in both:

9a + 12c = 609   (3)

8a + 12c = 572   (4)

Since both equations have been changed we label the new equations (3) and (4).

Now we have the same number of a's in both equations and the signs in front are the same so we can subtract (4) from (3):

9a + 12c = 609   (3)

8a + 12c = 572   (4)

a = 37   (3) - (4)

So, an adult ticket costs £37.

Now, to find the cost of a child's ticket, substitute a = 37 into any of the four equations:

2 x 37 + 3c = 143  [substitute into (2)]

74 + 3c = 143

3c = 69

c = 23

So, a child's ticket costs £23.

Now we can find the cost for the Sheriff family which is one adult and two children:

37 + 2 x 23 = 83 so they would pay £83.

If you now compare what we have done here with solving simultaneous equations by elimination, you will see that in this example we multiplied both equations in order to equate the coefficients of one variable.

Other than that, everything else is the same.

We could have equally multiplied equation (1) by 2 and equation (2) by 3 which would have given us 6a in both and then we could have eliminated a.

This method would have been equally valid and enabled us to reach the correct answer too.

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

This time we will use just the equations without the real-life problem.

Solve 2x + 5y = -2  (1)

and 3x - 4y = 20  (2)

To equate coefficients of y, we need to multiply (1) by 4 and (2) by 5:

2x + 5y = -2  (1) × 4

3x - 4y = 20  (2) × 5

8x + 20y = -8  (3)

15x - 20y = 100  (4)

Since the signs in front of the y's are different, we need to add equations (3) and (4):

8x + 20y = -8  (3)

15x - 20y = 100  (4)

23x =  92   (3) + (4)

x = 4  [divide by 23]

2 x 4 + 5y = -2  [substitute into (1)]

8 + 5y = -2

5y = -10  [subtract 8]

y = -2   [divide by 5]

3 x 4 - 4 x -2 = 20  [check in (2)]

12 - -8  = 20 [correct]

So x = 4 and y = -2.

Hopefully now you are feeling ready to solve some simultaneous equations of your own.

As these methods are long and complex, you will want to have a pen and paper handy to record your working.

Remember we are looking for a way to eliminate one of the coefficients, so just look for the best numbers to multiply each equation by so they match.

Consider this pair of equations:

6x + 5y = 13

7x + 2y = 19

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

Correctly order this process by matching each step to its correct position.

## Column B

1st
6 x 3 + 5y = 13 [substitute into (1)]
2nd
12x + 10y = 26 (3) and 35x + 10y = 95
3rd
y = -1
4th
18 + 5y = 13
5th
x = 3
6th
23x = 69 [(4) - (3)]
7th
7 x 3 + 2 x -1 = 19 [check in (2)]
8th
Multiply (1) by 2 and (2) by 5
9th
5y = -5
10th
6x + 5y = 13 (1) and 7x + 2y = 19 (2)

Consider this pair of equations:

9m - 2n = -23

5m - 9n = 3

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

Correctly order this process by matching each step to its correct position.

## Column B

1st
-15 - 9n = 3
2nd
m = -3
3rd
5 x -3 - 9n = 3 [sub in (2)]
4th
-9n = 18
5th
71m = -213
6th
n = -2
7th
Multiply (1) by 9 and (2) by 2
8th
9m - 2n = -23 (1) and 5m - 9n = 3 (2)
9th
81m - 18n = -207 (3) and 10m - 18n = 6 (4)
10th
9 x -3 - 2 x -2 = -23 [check in (1)]

Look at this new pair of equations:

5g + 3h = 27

4g + 5h = 32

Fill in the blanks in the solution below to complete the steps for solving this pair of simultaneous equations.

## Column B

1st
-15 - 9n = 3
2nd
m = -3
3rd
5 x -3 - 9n = 3 [sub in (2)]
4th
-9n = 18
5th
71m = -213
6th
n = -2
7th
Multiply (1) by 9 and (2) by 2
8th
9m - 2n = -23 (1) and 5m - 9n = 3 (2)
9th
81m - 18n = -207 (3) and 10m - 18n = 6 (4)
10th
9 x -3 - 2 x -2 = -23 [check in (1)]

Review this new pair of equations:

7p - 3q = 15

5p + 2q = 19

Fill in the blanks in the solution below to complete the steps for solving this pair of simultaneous equations.

## Column B

1st
-15 - 9n = 3
2nd
m = -3
3rd
5 x -3 - 9n = 3 [sub in (2)]
4th
-9n = 18
5th
71m = -213
6th
n = -2
7th
Multiply (1) by 9 and (2) by 2
8th
9m - 2n = -23 (1) and 5m - 9n = 3 (2)
9th
81m - 18n = -207 (3) and 10m - 18n = 6 (4)
10th
9 x -3 - 2 x -2 = -23 [check in (1)]

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

## Column B

8x + 3y = 46 and 5x + 2y = 29
x = -2 and y = 4
13x - 7y = 47 and 7x - 9y = 41
x = 5 and y = 2
6x - 5y = -32 and 5x + 4y = 6
x = -2 and y = 5
6x - 3y = -27 and 5x + 4y = 10
x = 2 and y = -3

Solve the following pair of equations:

8r + 5s = 41

9r + 11s = 30

Then type your solutions into the blanks below.

## Column B

8x + 3y = 46 and 5x + 2y = 29
x = -2 and y = 4
13x - 7y = 47 and 7x - 9y = 41
x = 5 and y = 2
6x - 5y = -32 and 5x + 4y = 6
x = -2 and y = 5
6x - 3y = -27 and 5x + 4y = 10
x = 2 and y = -3

A group of friends go into a coffee shop and buy 3 cappuccinos and 4 americanos for £16.80.

A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.

Using c to represent the cost of a cappuccino in pence and a to represent the cost of an americano in pence, choose the two equations from the options below which correctly express this problem algebraically.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

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

A group of friends go into a coffee shop and buy 3 cappuccinos and 4 americanos for £16.80.

A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.

Now solve these equations to find the cost of a cappuccino and the cost of an americano and fill your answers in the blanks below.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

Two cakes and 7 bags of peanuts contain 115g of fat, whereas five cakes and three bags of peanuts contain 99g of fat.

Using c to represent grams of fat in a cake and p to represent grams of fat in a bag of peanuts, form two equations by filling in the blanks below.

Then solve these equations and fill in the values for c and p below.

Finally, work our how many grams of fat would be in 5 cakes and 2 bags of peanuts.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

Kate and Joe both think of numbers between 1 and 10.

Kate's number is greater than Joe's.

If Kate's number is tripled, and five times Joe's number is added, the answer is 27.9.

If both numbers are doubled, the difference is 9.

Using k to represent Kate's number and j to represent Joe's number, form two equations by filling in the blanks below.

Then solve these equations and fill in the values for and j below to show the numbers that Kate and Joe thought of.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

• Question 1

Consider this pair of equations:

6x + 5y = 13

7x + 2y = 19

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

Correctly order this process by matching each step to its correct position.

## Column B

1st
6x + 5y = 13 (1) and 7x + 2y = 1...
2nd
Multiply (1) by 2 and (2) by 5
3rd
12x + 10y = 26 (3) and 35x + 10y...
4th
23x = 69 [(4) - (3)]
5th
x = 3
6th
6 x 3 + 5y = 13 [substitute into...
7th
18 + 5y = 13
8th
5y = -5
9th
y = -1
10th
7 x 3 + 2 x -1 = 19 [check in (2...
EDDIE SAYS
To equate the coefficients of y (i.e. ensure they cancel each other other), we can multiply (1) by 2 and (2) by 5. We will call these new equations (3) and (4). As the signs attached to the y's are the same, we need to subtract (3) from (4) to eliminate y. We should 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, substitute both x = 3 and y = -1 into equation (2) to check this gives us the correct answer. Did you follow those steps through? Review the Introduction now if you need to before you move on to tackle the rest of this activity.
• Question 2

Consider this pair of equations:

9m - 2n = -23

5m - 9n = 3

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

Correctly order this process by matching each step to its correct position.

## Column B

1st
9m - 2n = -23 (1) and 5m - 9n =...
2nd
Multiply (1) by 9 and (2) by 2
3rd
81m - 18n = -207 (3) and 10m - 18...
4th
71m = -213
5th
m = -3
6th
5 x -3 - 9n = 3 [sub in (2)]
7th
-15 - 9n = 3
8th
-9n = 18
9th
n = -2
10th
9 x -3 - 2 x -2 = -23 [check in (...
EDDIE SAYS
To equate the coefficients of n, we multiply equation (1) by 9 and call it (3), then multiply (2) by 2 and call it (4). Because the signs are the same (both negative), we need to subtract (4) from (3) to eliminate n. Then we need to find the value of m and substitute this back into equation (2). We can then solve this equation to find the value of n. Finally, we must substitute both m = -3 and n -2 into equation (1) to check it gives us the correct answer. Are you getting the hang of this process now?
• Question 3

Look at this new pair of equations:

5g + 3h = 27

4g + 5h = 32

Fill in the blanks in the solution below to complete the steps for solving this pair of simultaneous equations.

EDDIE SAYS
Here's the steps we need to take to solve this pair: 4g + 5h = 32 (1) 5g + 3h = 27 (2) Equate the coefficients of g by multiplying (1) by 5 and (2) by 4: 20g + 25h = 160 (3) [(2) x 5] 20g + 12h = 108 (4) [(1) x 4] The signs related to g are the same (+) so we need to subtract: (3) - (4) 13h = 52 h = 4 Substitute h = 4 into (1) to find g: 5g + 3 × 4 = 27 5g + 12 = 27 5g = 15 g = 3 Check these two values work in (2): 4 x 3 + 5 x 4 = 32 [correct] So g = 3 and h = 4. Does that make sense?
• Question 4

Review this new pair of equations:

7p - 3q = 15

5p + 2q = 19

Fill in the blanks in the solution below to complete the steps for solving this pair of simultaneous equations.

EDDIE SAYS
Here's the full solution: 7p - 3q = 15 (1) ×2 5p + 2q = 19 (2) ×3 14p - 6q = 30 (3) 15p + 6q = 57 (4) Since the signs in front of 6y are different, we need to add our equations: 29p = 87 p = 3 [divide by 29] Let's substitute this value for p into (2): 5 x 3 + 2q = 19 15 + 2q = 19 2q = 4 q = 2 Check both these values work in (1): 7 x 3 - 3 x 2 = 15 [correct] So p = 3 and q = 2.
• Question 5

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

## Column B

8x + 3y = 46 and 5x + 2y = 29
x = 5 and y = 2
13x - 7y = 47 and 7x - 9y = 41
x = 2 and y = -3
6x - 5y = -32 and 5x + 4y = 6
x = -2 and y = 4
6x - 3y = -27 and 5x + 4y = 10
x = -2 and y = 5
EDDIE SAYS
For this question, you could try solving each pair of equations one at a time, but a quicker method is to substitute each pair of solutions into each pair of equations, until you find a match. So with the first pair of equations, the solutions are x = 5 and y = 2 since: 8 × 5 + 3 × 2 = 46 and 5 × 5 + 2 × 2 = 29 For the second pair, x = 2 and y = -3 work since: 13 × 2 - 7 × -3 = 47 and 7 × 2 - 9 × -3 = 41 For the third pair, x = -2 and y = 4 work since: 6 × -2 - 5 × 4 = -32 and 5 × -2 + 4 × 4 = 6 Finally, for the last pair, x = -2 and y = 5 work since: 6 × -2 - 3 × 5 = - 27 and 5 × -2 + 4 × 5 = 10 That was a much quicker method, wasn't it? Did you match all the equations and solutions accurately?
• Question 6

Solve the following pair of equations:

8r + 5s = 41

9r + 11s = 30

Then type your solutions into the blanks below.

EDDIE SAYS
Your working should look something like this: 8r + 5s = 41 (1) ×11 9r + 11s = 30 (2) ×5 88r + 55s = 451 (3) 45r + 55s = 150 (4) Same signs, so subtract: 43r = 301 r = 7 Substitute into (1) to find s: 8 × 7 + 5s = 41 56 + 5s = 41 5s = -15 s = -3 Check both values work in (2): 9 × 7 + 11 × -3 = 30 63 - 33 = 30 [correct] How close was your working to what is written here? Make sure you are using a pen and paper to record your steps and so you can compare easily.
• Question 7

A group of friends go into a coffee shop and buy 3 cappuccinos and 4 americanos for £16.80.

A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.

Using c to represent the cost of a cappuccino in pence and a to represent the cost of an americano in pence, choose the two equations from the options below which correctly express this problem algebraically.

3c + 4a = 1680
4c + 3a = 1715
EDDIE SAYS
The first group buy 3 cappuccinos (3c) and 4 americanos (4a) and pay £16.80 (1680p) so: 3c + 4a = 1680 The second group buy 4 cappuccinos (4c) and 3 americanos (3a) and pay £17.15 (1715p) so: 4c + 3a = 1715 If we were to solve this pair of simultaneous equations, we could accurately work out the cost of a cappuccino and americano (the values of c and a).
• Question 8

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

A group of friends go into a coffee shop and buy 3 cappuccinos and 4 americanos for £16.80.

A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.

Now solve these equations to find the cost of a cappuccino and the cost of an americano and fill your answers in the blanks below.

EDDIE SAYS
So the correct equations we need to work with are: 3c + 4a = 1680 (1) 4c + 3a = 1715 (2) We can choose to either equate the coefficients of a or c. Let's choose a: 3c + 4a = 1680 (1) ×3 4c + 3a = 1715 (2) ×4 9c + 12a = 5040 (3) 16c + 12a = 6860 (4) 7c = 1820 [(4) - (3)] c = 260 Sub into (1): 3 × 260 + 4a = 1680 780 + 4a = 1680 4a = 900 a = 225 Check in (2): 4 × 260 + 3 × 225 = 1715 1040 + 675 = 1715 [correct] So a cappuccino costs £2.60 and an americano costs £2.25. Did you remember to express these solutions as decimals (£) rather than whole numbers (p)?
• Question 9

Two cakes and 7 bags of peanuts contain 115g of fat, whereas five cakes and three bags of peanuts contain 99g of fat.

Using c to represent grams of fat in a cake and p to represent grams of fat in a bag of peanuts, form two equations by filling in the blanks below.

Then solve these equations and fill in the values for c and p below.

Finally, work our how many grams of fat would be in 5 cakes and 2 bags of peanuts.

EDDIE SAYS
So the equations we need to use are: 2c + 7p = 115 (1) 5c + 3p = 99 (2) To find the values of p and c, we need to follow these steps: 10c + 35p = 575 (3) [(1) × 5] 10c + 6p = 198 (4) [(2) × 2] 29p = 377 (3) - (4) p = 13 Sub into (1): 2c + 7 × 13 = 115 2c + 91 = 115 2c = 24 c = 12 Then we need to add these values into the sum for 4 cakes and 2 bags of peanuts: 4 × 12 + 2 × 13 = 48 + 26 = 74g of fat How did you get on with that complex, multi-step problem? Don't be disheartened if you made an error or two. These questions are very challenging, which is why examiners will award marks for your working as well as the correct answer.
• Question 10

Kate and Joe both think of numbers between 1 and 10.

Kate's number is greater than Joe's.

If Kate's number is tripled, and five times Joe's number is added, the answer is 27.9.

If both numbers are doubled, the difference is 9.

Using k to represent Kate's number and j to represent Joe's number, form two equations by filling in the blanks below.

Then solve these equations and fill in the values for and j below to show the numbers that Kate and Joe thought of.

EDDIE SAYS
One of the most challenging things about this question, was to work out the starting equations. The equations we need to use are: 3k + 5j = 27.9 (1) 2k - 2j = 9 (2) To solve, let's equate j: 6k + 10j = 55.8 (3) [(1) × 2] 10k - 10j = 45 (4) [(2) × 5] 16k = 100.8 k = 6.3 2 × 6.3 - 2j = 9 [sub into (2)] 12.6 - 2j = 9 -2j = -3.6 j = 1.8 3 × 6.3 + 5 × 1.8 = 27.9 [check in (1)] 18.9 + 9 = 27.9 [correct] So Kate's number was 6.3 and Joe's was 1.8. That's it - you've finished! How was it? If this was tricky, why not try our Level 1 activity on simultaneous equations? If you are ready for a challenge, give our Level 3 activity a go!
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