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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.

'Solve Simultaneous Equations (by multiplication)' 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 the theatre, where they bought adult tickets and child 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. Can you work out how much Mr Sheriff and his two daughters would pay?'

If you have completed previous 'Simultaneous Equations' activities then you will recognise this sort of problem. This activity builds on ideas established in two previous activities so if you have not completed those activities yet then you should go back and complete them before continuing with this.

In order to find out how much Mr and Mrs Sheriff and their daughters 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 = 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, but before we do this we need to have the same coefficient of a or c. 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 that will 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 ticket, substitute a = 37 into any of the four equations, usually the 'easiest' one.

2 x 37 + 3c = 143  [sub into (2)]

74 + 3c = 143

3c = 69

c = 23 

So a child 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 the previous activity, you will see that we multiply 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 would have been just as good a method.

Now, let's look at another pair of simultaneous equations. This time just the equations without the real-life problem.

Solve  2x + 5y = -2  (1)

and      3x - 4y = 20  (2)

 

To equate coefficients of 'y' multiply (1) by 4 and (2) by 5

 

2x + 5y = -2  (1) x 4

3x - 4y = 20  (2) x 5

8x + 20y = -8  (3)

15x - 20y = 100  (4)

Since the signs in front of the y's are different we 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  [sub 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.

Now you should be ready to try some. Remember, its the same as before but just look for what are the best numbers to multiply each equation by.

 

 

 

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

6x + 5y = 13

7x + 2y = 19

but not in the correct order.

Match each step to its correct position.

 

Column A

Column B

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

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

9m - 2n = -23

5m - 9n = 3

but not in the correct order.

Match each step to its correct position.

Column A

Column B

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

Below you will see all the steps to solving the equations 

5g + 3h = 27

4g + 5h = 32

Fill in the blanks to make the solution complete.

Column A

Column B

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

Below you will see all the steps to solving the equations 

7p - 3q = 15

5p + 2q = 19

Fill in the blanks to make the solution complete.

Column A

Column B

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

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

Column A

Column B

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

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

8r + 5s = 41

9r + 11s = 30

Column A

Column B

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

A group of friends go into a coffee shop and buys 3 cappuccinos and 4 americanos for £16.80. A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.  

Using c = cost of a cappuccino in pence and a = cost of an americano in pence, choose the two correct equations from those below for this problem? You do not have to solve the equations.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

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

A group of friends go into a coffee shop and buys 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. Do your working on paper and fill your answers into the blanks.

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 = grams of fat in a cake and p = grams of fat in a bag of peanuts form two equations by filling in the blanks below.

Solve the equations on paper and fill in the solutions below.

How many grams of fat would be in 5 cakes and 2 bags of peanuts? Fill in the blank below.

 

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 = Kate's number and j = Joe's number, form a pair of simultaneous equations and fill in the blanks below. 

Now, on paper, solve your equations to find the numbers Kate and Joe thought of and fill in the blanks below.

3c + 4a = 1680

3c + 4a = 16.80

3c + 4a = 1715

4c + 3a = 17.15

4c + 3a = 1715

4c + 3a = 1680

  • Question 1

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

6x + 5y = 13

7x + 2y = 19

but not in the correct order.

Match each step to its correct position.

 

CORRECT ANSWER

Column A

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 [sub into (1)]
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 we can multiply (1) by 2 and (2) by 5 and call them (3) and (4). Then we subtract (3) from (4) to eliminate y. Then find the value of x and substitute this back into equation (1). Solve this equation to find the value of y. Finally, sub both x = 3 and y = -1 into equation 2 to check it gives the correct answer.
  • Question 2

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

9m - 2n = -23

5m - 9n = 3

but not in the correct order.

Match each step to its correct position.

CORRECT ANSWER

Column A

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 - ...
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) and (2) by 2 and call it (4) Because the signs are the same (both negative) we subtract (4) from (3) to eliminate n. Then find the value of m and substitute this back into equation (2). Solve this equation to find the value of n. Finally, sub both m = -3 and n -2 into equation (1) to check it gives the correct answer.
  • Question 3

Below you will see all the steps to solving the equations 

5g + 3h = 27

4g + 5h = 32

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
Here's the full solution. 4g + 5h = 32 (1) 5g + 3h = 27 (2) Equate 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] Signs are the same so subtract. 13h = 52 h = 4 Sub h = 4 into (1) 5g + 3 x 4 = 27 5g + 12 = 27 5g = 15 g = 3 Check in (2) 4 x 3 + 5 x 4 = 32 [correct] So g = 3 and h = 4 How are you doing?
  • Question 4

Below you will see all the steps to solving the equations 

7p - 3q = 15

5p + 2q = 19

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
Here's the full solution. 7p - 3q = 15 (1) x2 5p + 2q = 19 (2) x3 14p - 6q = 30 (3) 15p + 6q = 57 (4) Since the signs in front of 6y are different we add the equations. 29p = 87 p = 3 [divide by 29] Sub in (2) 5 x 3 + 2q = 19 15 + 2q = 19 2q = 4 q = 2 Check in (1) 7 x 3 - 3 x 2 = 15 [correct] 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

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, 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 = 5 and y = 2 since 8 x 5 + 3 x 2 = 46 and 5 x 5 + 2 x 2 = 29 For the second pair x = 2 and y = -3 work since 13 x 2 - 7 x -3 = 47 and 7 x 2 - 9 x -3 = 41 For the 3rd pair x = -2 and y = 4 work since 6 x -2 - 5 x 4 = -32 and 5 x -2 + 4 x 4 = 6 Finally, for the last pair x = -2 and y = 5 work since 6 x -2 - 3 x 5 = - 27 and 5 x -2 + 4 x 5 = 10. Did you get all these correct?
  • Question 6

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

8r + 5s = 41

9r + 11s = 30

CORRECT ANSWER
EDDIE SAYS
8r + 5s = 41 (1) x11 9r + 11s = 30 (2) x5 88r + 55s = 451 (3) 45r + 55s = 150 (4) 43r = 301 r = 7 8 x 7 + 5s = 41 [sub into (1)] 56 + 5s = 41 5s = -15 s = -3 9 x 7 + 11 x -3 = 30 [check in (2)] 63 - 33 = 30 [correct] How did you do with this one? Did you get all the steps correct?
  • Question 7

A group of friends go into a coffee shop and buys 3 cappuccinos and 4 americanos for £16.80. A second group of friends buy 4 cappuccinos and 3 americanos for £17.15.  

Using c = cost of a cappuccino in pence and a = cost of an americano in pence, choose the two correct equations from those below for this problem? You do not have to solve the equations.

CORRECT ANSWER
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.
  • Question 8

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

A group of friends go into a coffee shop and buys 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. Do your working on paper and fill your answers into the blanks.

CORRECT ANSWER
EDDIE SAYS
So the correct equations are 3c + 4a = 1680 (1) 4c + 3a = 1715 (2) We can either equate coefficients of a or c. Here we will equate a. 3c + 4a = 1680 (1) x3 4c + 3a = 1715 (2) x4 9c + 12a = 5040 (3) 16c + 12a = 6860 (4) 7c = 1820 (4) - (3) c = 260 Sub into (1) 3 x 260 + 4a = 1680 780 + 4a = 1680 4a = 900 a = 225 Check in (2) 4 x 260 + 3 x 225 = 1715 1040 + 675 = 1715 [correct] So a cappuccino is £2.60 and an americano is £2.25
  • 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 = grams of fat in a cake and p = grams of fat in a bag of peanuts form two equations by filling in the blanks below.

Solve the equations on paper and fill in the solutions below.

How many grams of fat would be in 5 cakes and 2 bags of peanuts? Fill in the blank below.

 

CORRECT ANSWER
EDDIE SAYS
So the equations are 2c + 7p = 115 (1) 5c + 3p = 99 (2) 10c + 35p = 575 (3) [(1) x 5] 10c + 6p = 198 (4) [(2) x 2] 29p = 377 (3) - (4) p = 13 Sub into (1) 2c + 7 x 13 = 115 2c + 91 = 115 2c = 24 c = 12 So 4 cakes and 2 bags of peanuts contain 4 x 12 + 2 x 13 = 48 + 26 = 74g of fat.
  • 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 = Kate's number and j = Joe's number, form a pair of simultaneous equations and fill in the blanks below. 

Now, on paper, solve your equations to find the numbers Kate and Joe thought of and fill in the blanks below.

CORRECT ANSWER
EDDIE SAYS
The equations are 3k + 5j = 27.9 (1) 2k - 2j = 9 (2) 6k + 10j = 55.8 (3) [(1) x 2] 10k - 10j = 45 (4) [(2) x 5] 16k = 100.8 k = 6.3 2 x 6.3 - 2j = 9 [sub into (2)] 12.6 - 2j = 9 -2j = -3.6 j = 1.8 3 x 6.3 + 5 x 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?
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