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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.

'Solve Simultaneous Equations (by multiplying one equation)' 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 cinema, where they bought adult tickets and child 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. Can you work out how much Mr and Mrs Fisher and their daughter April would pay?'

If you have completed the activity 'Simple Simultaneous Equations' then you will recognise this sort of problem as one which can be solved using simultaneous equations. If you have not completed that activity yet then you should go back and complete that one before continuing with this.

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 variables and we use a = cost of an adult and c = cost of a child. So, Mr and Mrs Baker and their daughters becomes 2a + 3c = 33. Mr Butcher and his 2 sons becomes a + 2c = 19. We now have our two equations, and we need to solve them simultaneously.

2a + 3c = 33   (1)

a + 2c  =  19   (2)

In the previous activity, at this point, 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 make them the same by multiplying the whole of one equation. So, if we take equation (2) and multiply it by 2 we get 2a + 4c = 38 (it means that 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 ticket costs £5.

Now, to find the cost of an adult ticket, substitute c = 5 into any of the three equations, usually the 'easiest' one.

a + 2 x 5 = 19  [sub into (2)]

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 you now compare what we have done here with the previous activity, 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.

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

Solve  3x + 2y = 10  (1)

and      7x - y = 29    (2)

Notice the coefficients of x are different, as are the coefficients of y. But if we double equation (2) we will have 2y in each

3x + 2y = 10  (1)

 7x - y = 29    (2) x 2

14x - 2y = 58  (3)

Since the signs in front of the y's are different we add equations (1) and (3).

3x + 2y = 10  (1)

14x - 2y = 58  (3)

 17x       = 68

     x       = 4   [divide by 17]

3 x 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 some. Remember, its the same as before with just one extra step. 

 

 

 

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

5a + 3b = 13

7a + 6b = 20

but not in the correct order.

Match each step to its correct position.

 

Column A

Column B

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

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

11x + 3y = 71

5x - y = 37

but not in the correct order.

Match each step to its correct position.

Column A

Column B

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

Below you will see all the steps to solving the equations 

4c + 5d = 17   
8c + 3d = 27  

Fill in the blanks to make the solution complete.

Column A

Column B

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

Below you will see all the steps to solving the equations 

9a + 2b = 41

5a - 4b = 33

Fill in the blanks to make the solution complete.

Column A

Column B

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

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

Column A

Column B

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

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

3m - 4n = 13

7m + 2n = 36

Column A

Column B

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

A florist shop sells two types of bouquet '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 those below for this problem? You do not have to solve the equations.

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

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

A florist shop sells two types of bouquet '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. Do your working on paper and fill your answers into the blanks.

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. 

Solve the 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 and fill in the blanks below. 

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

7b + 5s = 109

b - s = 17

5b + 7s = 109

b + s = 17

5b - 7s = 109

  • Question 1

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

5a + 3b = 13

7a + 6b = 20

but not in the correct order.

Match each step to its correct position.

 

CORRECT ANSWER

Column A

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
To equate the coefficients of b we double equation (1) and call it (3) Then we subtract (2) from (3) to eliminate b. Then find the value of a and substitute this back into equation (1). Solve this equation to find the value of b. Finally, sub both a = 2 and b = 1 into equation 2 to check it gives the correct answer.
  • Question 2

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

11x + 3y = 71

5x - y = 37

but not in the correct order.

Match each step to its correct position.

CORRECT ANSWER

Column A

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 multiply equation (2) by 3 and call it (3) Because the signs are different we add (1) and (3) 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 = 7 and y = -2 into equation 2 to check it gives the correct answer.
  • Question 3

Below you will see all the steps to solving the equations 

4c + 5d = 17   
8c + 3d = 27  

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
Here's the full solution. 4c + 5d = 17 (1) 8c + 3d = 27 (2) Equate coefficients of c by multiplying (1) by 2 8c + 10d = 34 (3) 8c + 3d = 27 (2) Signs are same so subtract 7d = 7 [(3) - (2)] d = 1 Substitute d = 1 into (1) 4c + 5 x 1 = 17 4c + 5 = 17 4c = 12 c = 3 Check by substituting into (2) 8 x 3 + 3 x 1 = 27 24 + 3 = 27 [correct]
  • Question 4

Below you will see all the steps to solving the equations 

9a + 2b = 41

5a - 4b = 33

Fill in the blanks to make the solution complete.

CORRECT ANSWER
EDDIE SAYS
Here's the full solution 9a + 2b = 41 (1) 5a - 4b = 33 (2) 18a + 4b = 82 (3) [(1) x 2] 23a = 115 (2) + (3) a = 5 9 x 5 + 3b = 41 [sub into (1)] 45 + 3b = 41 3b = -4 b = -2 5 x 5 - 4 x -2 = 33 [check in (2)] 25 + 8 = 33 [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

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, 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 = 2 since 9 x 3 + 2 x 2 = 31 and 3 x 3 + 2 = 11 For the second pair x = 3 and y = 4 work since 5 x 3 + 3 x 4 = 27 and 3 + 4 = 7 For the 3rd pair x = 5 and y = -1 work since 2 x 5 + 7 x -1 = 3 and 5 - -1 = 6 Finally, for the last pair x = 5 and y = 4 work since 11 x 5 + 2 x 4 = 63 and 6 x 5 + 4 x 4 = 46
  • Question 6

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

3m - 4n = 13

7m + 2n = 36

CORRECT ANSWER
EDDIE SAYS
3m - 4n = 13 (1) 7m + 2n = 36 (2) Multiply (2) by 2 14m + 4n = 72 (3) 17m = 85 (1) + (3) m = 5 Sub m = 5 into (2) 7 x 5 + 2n = 36 35 + 2n = 36 2n = 1 n = 0.5 Check in (1) 3 x 5 - 4 x 0.5 = 13 15 - 2 = 13 (correct) How did you do with this one? Did you get all the steps correct?
  • Question 7

A florist shop sells two types of bouquet '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 those below for this problem? You do not have to solve the equations.

CORRECT ANSWER
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 and when these are added the total is 109 so this means 5b + 7s = 109.
  • Question 8

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

A florist shop sells two types of bouquet '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. Do your working on paper and fill your answers into the blanks.

CORRECT ANSWER
EDDIE SAYS
So the correct equations 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. 5b + 5s = 85 (3) Put (3) below (2) and subtract 5b + 7s = 109 (2) 5b + 5s = 85 (3) 2s = 24 (2) - (3) s = 12 Sub into (1) 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. 

Solve the equations by filling in the blanks below.

CORRECT ANSWER
EDDIE SAYS
So the equations are 4t + 3b = 705 (1) 3t + b = 435 (2) Multiply (2) by 3 then subtract (1) 9t + 3b = 1305 (3) 4t + 3b = 705 (1) 5t = 600 t = 120 Sub into (2) 3 x 120 + b = 435 360 + b = 435 b = 75 Check in (1) 4 x 120 + 3 x 75 = 705 480 + 225 = 705 [correct] So a cup of tea costs £1.20 and a biscuit costs 75p Making 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 and fill in the blanks below. 

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

CORRECT ANSWER
EDDIE SAYS
So the equations are 5g - 2h = 14 (1) 3g - h = 9 (2) Multiply (2) by 2 6g - 2h = 18 (3) Since the signs in front of h are the same we subtract. 6g - 2h = 18 (3) 5g - 2h = 14 (1) g = 4 (3) - (1) 3 x 4 - h = 9 [sub into (2)] 12 - h = 9 h = 3 So Greg's number was 4 and Hannah's was 3. That's it. You've finished. How was it?
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