 # Simultaneous Equations Using Substitution

In this worksheet, students will learn how to solve one linear and one quadratic simultaneous equation using the substitution method. 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

In the activity 'Simultaneous Equations - substitution' we learnt how to use the substitution method to solve two linear simultaneous equations. If you have not completed that activity yet you should do so before continuing with this one, as in this activity we extend this idea to deal with situations where one equation is linear but the other is not (usually quadratic). The method of substitution will be the same but the problems will become more complicated and lengthy. Here is an example to demonstrate how to go about this.

Solve the simultaneous equations

x - y = 1   (1)

x² + y² = 13  (2)

In this question equation (1) is linear and (2) is quadratic.  We always substitute the linear into the quadratic equation, but first, we must rearrange (1) so that either x or y is the subject? In this case, because equation (2) contains and y²  it does not matter which so let's make x the subject.

Adding 'y' to both sides gives us

x = 1 + y  (1)

Now we can substitute this into (2)

(1 + y)² + y² = 13  (2)

Now, we need to expand the brackets and simplify

1 + 2y + y² + y² = 13

2y² + 2y - 12 = 0

This is now a quadratic equation in y and needs to be solved using any of the methods for solving quadratics. Firstly we can divide the whole equation by 2 to make life easier.

y² + y - 6 = 0

This equation will factorise into two brackets (always try to factorise before you turn to any other method for solving quadratics, and usually they will factorise).

(y + 3)(y - 2) = 0

Either y + 3 = 0 or y - 2 = 0

So y = -3 or y = 2

So we have two possible values for y and each will have a corresponding value for x. To find the x-values always substitute the y-values into the linear equation.

When y = -3  x = 1 + -3 so x = -2

When y = 2  x = 1 + 2 so x = 3

So the two pairs of solutions are x = 3, y = 2 or x = -2, y = -3.

These are sometimes written as co-ordinates (3 , 2) or (-2 , -3).

At this point, we could check our solutions by substituting them into equation (2) but to save time we will not do this in this activity.

It will always be the case with these types of simultaneous equation that there will be two pairs of solutions and you need to find them both.

Now, it's time to try some yourself.

Below you will see all the steps to solving the equations

xy + x + y = -1  (1)

and y = 3x + 1  (2)

but in the wrong order. Put the steps into the correct order.

## Column B

1st
3x² + x + x + 3x + 1 = -1 [expand]
2nd
x(3x + 1) + x + (3x + 1) = -1 [sub (2) into (1)]
3rd
(x + 1)(3x + 2) = 0 [factorise]
4th
Solutions: x = -1, y = -2 or x = -2/3, y = -1
5th
When x = -1 y = 3(-1) + 1 = -2 [sub into (2)]
6th
xy + x + y = -1 (1) and y = 3x + 1 (2)
7th
Either x + 1 = 0 or 3x + 2 = 0
8th
So x = -1 or x = -2/3
9th
3x² + 5x + 2 = 0 [simplify]
10th
When x = -3/2 y = 3(-2/3) + 1 = -1 [sub into (2)]

Below you will see all the steps to solving the equations

y² - x² = 60   (1)

y = 3x + 2   (2)

but in the wrong order. Put the steps into the correct order.

## Column B

1st
2x² + 3x - 14 = 0 [simplify]
2nd
(2x + 7)(x - 2) = 0 [factorise]
3rd
y = 3x(-7/2) + 2 = -17/2 [sub into (2)]
4th
So x = -7/2 or x = 2
5th
y² - x² = 60 (1) and y = 3x + 2 (2)
6th
Solutions x = -7/2, y = -17/2 or x = 2, y = 8
7th
Either 2x + 7 = 0 or x - 2 = 0
8th
9x² + 12x + 4 - x² = 60 [expand]
9th
(3x + 2)² - x² = 60 [sub (2) into (1)]
10th
y = 3x2 + 2 = 8 [sub into (2)]

Using the substitution method, solve the simultaneous equations

xy = 2

and

y = x + 1

by filling in the blanks below.

## Column B

1st
2x² + 3x - 14 = 0 [simplify]
2nd
(2x + 7)(x - 2) = 0 [factorise]
3rd
y = 3x(-7/2) + 2 = -17/2 [sub into (2)]
4th
So x = -7/2 or x = 2
5th
y² - x² = 60 (1) and y = 3x + 2 (2)
6th
Solutions x = -7/2, y = -17/2 or x = 2, y = 8
7th
Either 2x + 7 = 0 or x - 2 = 0
8th
9x² + 12x + 4 - x² = 60 [expand]
9th
(3x + 2)² - x² = 60 [sub (2) into (1)]
10th
y = 3x2 + 2 = 8 [sub into (2)]

Using the substitution method, solve the simultaneous equations

y = 2x + 3   (1)

and

y(x + 1) = 10  (2)

by filling in the blanks below.

## Column B

1st
2x² + 3x - 14 = 0 [simplify]
2nd
(2x + 7)(x - 2) = 0 [factorise]
3rd
y = 3x(-7/2) + 2 = -17/2 [sub into (2)]
4th
So x = -7/2 or x = 2
5th
y² - x² = 60 (1) and y = 3x + 2 (2)
6th
Solutions x = -7/2, y = -17/2 or x = 2, y = 8
7th
Either 2x + 7 = 0 or x - 2 = 0
8th
9x² + 12x + 4 - x² = 60 [expand]
9th
(3x + 2)² - x² = 60 [sub (2) into (1)]
10th
y = 3x2 + 2 = 8 [sub into (2)]

For the pair of simultaneous equations below, select the correct solutions from the options given.

y = 3x²

and

y - 3x = 6

 -12 -3 -2 -1 1 2 3 12 x y

For the pair of simultaneous equations below, select the correct solutions from the options given.

x + y = 9

and

x² - 2xy + y² = 1

 -5 -4 4 5 x y

On a piece of paper solve the following simultaneous equations using the substitution method. Fill in your solutions in the blanks.

xy - 2y - x = 2

and

x + y = 7

 -5 -4 4 5 x y

The simultaneous equations

3x + y = 7

and

xy + x² = 6

are to be solved using the method of substitution.

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

 -5 -4 4 5 x y

The diagram shows the circle x² + y² = 25 and the line y = x + 1. To find the points of intersection, solve the simultaneous equations

x² + y² = 25

and

y = x + 1.

Give your solutions as coordinates in the spaces below.

 -5 -4 4 5 x y

The function x² + 3xy + y² = 11 and the straight line x + y = 3 intersect twice.

On paper, find the points of intersection and fill in your solutions below.

 -5 -4 4 5 x y
• Question 1

Below you will see all the steps to solving the equations

xy + x + y = -1  (1)

and y = 3x + 1  (2)

but in the wrong order. Put the steps into the correct order.

## Column B

1st
xy + x + y = -1 (1) and y = 3x +...
2nd
x(3x + 1) + x + (3x + 1) = -1 [s...
3rd
3x² + x + x + 3x + 1 = -1 [expa...
4th
3x² + 5x + 2 = 0 [simplify]
5th
(x + 1)(3x + 2) = 0 [factorise]
6th
Either x + 1 = 0 or 3x + 2 = 0
7th
So x = -1 or x = -2/3
8th
When x = -1 y = 3(-1) + 1 = -2 [s...
9th
When x = -3/2 y = 3(-2/3) + 1 = -...
10th
Solutions: x = -1, y = -2 or x ...
EDDIE SAYS
First, we substitute equation (2) into (1) and expand the brackets to obtain an equation in x. Then we simplify to obtain a quadratic equation. Next, we factorise this quadratic and solve to find the two x-values. We substitute each of the x-values into equation (2) to find the two y-values. Finally, we give two pairs of solutions.
• Question 2

Below you will see all the steps to solving the equations

y² - x² = 60   (1)

y = 3x + 2   (2)

but in the wrong order. Put the steps into the correct order.

## Column B

1st
y² - x² = 60 (1) and y = 3x +...
2nd
(3x + 2)² - x² = 60 [sub (2) i...
3rd
9x² + 12x + 4 - x² = 60 [expan...
4th
2x² + 3x - 14 = 0 [simplify]
5th
(2x + 7)(x - 2) = 0 [factorise]
6th
Either 2x + 7 = 0 or x - 2 = 0
7th
So x = -7/2 or x = 2
8th
y = 3x(-7/2) + 2 = -17/2 [sub in...
9th
y = 3x2 + 2 = 8 [sub into (2)]
10th
Solutions x = -7/2, y = -17/2 or...
EDDIE SAYS
First, we substitute equation (2) into (1) and expand the brackets to obtain an equation in x. Then we simplify to obtain a quadratic equation. Next, we factorise this quadratic and solve to find the two x-values. We substitute each of the x-values into equation (2) to find the two y-values. Finally, we give two pairs of solutions.
• Question 3

Using the substitution method, solve the simultaneous equations

xy = 2

and

y = x + 1

by filling in the blanks below.

EDDIE SAYS
Here's the full solution. xy = 2 (1) y = x + 1 (2) x(x + 1) = 2 [sub (2) into (1)] x² + x = 2 [expand] x² + x - 2 = 0 [rearrange] (x +2)(x - 1) = 0 [factorise] Either x + 2 = 0 or x - 1 = 0 So x = -2 or x = 1 y = -2 + 1 = -1 [sub into (2)] y = 1 + 1 = 2 [sub into (2)] Solutions x = -2, y = -1 or x = 1, y = 2
• Question 4

Using the substitution method, solve the simultaneous equations

y = 2x + 3   (1)

and

y(x + 1) = 10  (2)

by filling in the blanks below.

EDDIE SAYS
here is the full solution: y(x + 1) = 10 (1) y = 2x + 3 (2) (2x + 3)(x + 1) = 10 [sub (2) into (1)] 2x² + 5x + 3 = 10 [expand] 2x² + 5x - 7 = 0 [rearrange] (2x + 7)(x - 1) = 0 [factorise] Either 2x + 7 = 0 or x - 1 = 0 So x = -7/2 or x = 1 y = 2x(-7/2) +3 = -4 [sub into (2)] y = 2x1 +3 = 5 [sub into (2)] Solutions: x = -7/2, y = -4 or x = 1, y = 5 How are you getting on?
• Question 5

For the pair of simultaneous equations below, select the correct solutions from the options given.

y = 3x²

and

y - 3x = 6

 -12 -3 -2 -1 1 2 3 12 x y
EDDIE SAYS
y = 3x² (1) y - 3x = 6 (2) 3x² - 3x = 6 [sub (1) into (2)] 3x² - 3x - 6 = 0 [rearrange] x² - x - 2 = 0 [divide by 2] (x - 2)(x + 1) = 0 [factorise] Either x -2 = 0 or x + 1 = 0 So x = 2 or x = -1 y = 3 x 2² = 12 [sub into (1)] y = 3 x (-1)² = 3 [sub into (1)] Solutions x = 2, y = 12 or x = -1, y = 3
• Question 6

For the pair of simultaneous equations below, select the correct solutions from the options given.

x + y = 9

and

x² - 2xy + y² = 1

 -5 -4 4 5 x y
EDDIE SAYS
Here is the full solution x + y = 9 (1) x² - 2xy + y² = 1 (2) y = 9 - x [rearrange (1)] x² - 2x(9 - x) + (9 - x)² = 1 [sub (1) into (2)] x² - 18x + 2x² + 81 - 18x + x² = 1 [expand] 4x² - 36x + 81 = 1 [simplify] 4x² - 36x + 80 = 0 [rearrange] x² - 9x + 20 = 0 [divide by 4] (x - 5)(x - 4) = 0 [factorise] Either x - 5 = 0 or x - 4 = 0 So x = 5 or x = 4 y = 9 - 5 = 4 [sub into (1)] y = 9 - 4 = 5 [sub into (1)] Solutions: x = 5, y= 4 or x = 4, y = 5
• Question 7

On a piece of paper solve the following simultaneous equations using the substitution method. Fill in your solutions in the blanks.

xy - 2y - x = 2

and

x + y = 7

EDDIE SAYS
So the full solution is as follows. xy - 2y - x = 2 (1) x + y = 7 (2) x = 7 - y [rearrange (2)] (7 - y)y - 2y - (7 - y) = 2 [sub (2) into (1)] 7y - y² - 2y - 7 + y = 2 [expand] 6y - y² - 7 = 2 [simplify] y² - 6y + 9 = 0 [rearrange] (y - 3)(y - 3) = 0 [factorise] y - 3 = 0 y = 3 x = 7 - 3 = 4 Solution x = 4, y = 3 Notice there is only one pair of solutions here as the quadratic in y has a repeated factor. This is unusual but you must be prepared for it.
• Question 8

The simultaneous equations

3x + y = 7

and

xy + x² = 6

are to be solved using the method of substitution.

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

EDDIE SAYS
Hear's the full solution 3x + y = 7 (1) xy + x² = 6 (2) y = 7 - 3x [rearrange (1)] x(7 - 3x) + x² = 6 [sub (1) into (2)] 7x - 3x² + x² = 6 [expand] -2x² + 7x - 6 = 0 2x² - 7x + 6 = 0 (2x - 3)(x - 2) = 0 Either 2x - 3 = 0 or x - 2 = 0 So x = 3/2 or x = 2 y = 7 - 3x(3/2) = 5/2 [sub into (1)] y = 7 - 3x2 = 1 [sub into (1)] Solutions: x = 3/2, y = 5/2 or x = 2, y = 1
• Question 9

The diagram shows the circle x² + y² = 25 and the line y = x + 1. To find the points of intersection, solve the simultaneous equations

x² + y² = 25

and

y = x + 1.

Give your solutions as coordinates in the spaces below.

EDDIE SAYS
So here is the solution: x² + y² = 25 (1) y = x + 1 (2) x² + (x + 1)² = 25 x² + x² + 2x + 1 = 25 2x² + 2x - 24 = 0 x² + x - 12 = 0 (x - 3)(x + 4) = 0 Either x = 3 or x = -4 y = 3 + 1 = 4 [sub into (2)] y = -4 + 1 = -3 [sub into (2)] Solutions (3 , 4) or (-4, -3)
• Question 10

The function x² + 3xy + y² = 11 and the straight line x + y = 3 intersect twice.

On paper, find the points of intersection and fill in your solutions below.

EDDIE SAYS
So we need to solve the simultaneous equations x² + 3xy + y² = 11 (1) and x + y = 3 (2) y = 3 - x [rearrange (2)] x² + 3x(3 - x) + (3 - x)² = 11 [sub (2) into (1)] x² + 9x - 3x² + 9 - 6x + x² = 11 -x² + 3x - 2 = 0 x² - 3x + 2 = 0 (x - 2)(x - 1) = 0 x = 2 or x = 1 y = 3 - 2 = 1 [sub into (2) y = 3 - 1 = 2 [sub into (2)] Solutions (2 , 1) or (1 , 2) That's it. Done for now.
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