 # Solve Linear and Quadratic Simultaneous Equations

In this worksheet, students will learn how to solve a pair of simultaneous equations (one linear and one quadratic) 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

You may have already learnt how to use the substitution method to solve two linear simultaneous equations.

This activity contains challenging problems, so make sure you are confident to solve simultaneous equations using a variety of methods to solve a pair of linear equations.

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 stay the same, but the problems will become more complicated and lengthy.

Let's look at an example to explore this now.

Solve these 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 both and y², it does not matter which we choose, so let's make x the subject.

Adding 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 involving y, so needs to be solved using any of the methods we have in our toolkit for solving quadratics.

Firstly we can divide the whole equation by 2 to make life easier:

y² + y - 6 = 0

Always try to factorise before you turn to any other method for solving quadratics, as this is usually the easiest and quickest method.

This equation will factorise into two brackets:

(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 coordinates: (3 , 2) or (-2 , -3).

At this point, we could check our solutions by substituting them into equation (2).

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.

The working in some of these questions is long and complex, so make sure you have a pen and paper handy to record your calculations.

You can then compare your working to the working written by our maths teacher.

Consider these two equations:

xy + x + y = -1   (1)

y = 3x + 1   (2)

Written below you will see all the steps required to solve these equations, but in the wrong order.

Put the steps into the correct order by matching each to the correct position in the sequence.

## Column B

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

Review these two equations:

y² - x² = 60   (1)

y = 3x + 2   (2)

Written below you will see all the steps required to solve these equations, but in the wrong order.

Put the steps into the correct order by matching each to the correct position in the sequence.

## Column B

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

Using the substitution method, solve these simultaneous equations:

xy = 2   (1)

y = x + 1   (2)

Complete the working below by filling in the blanks with missing numbers.

## Column B

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

Using the substitution method, solve the simultaneous equations

Using the substitution method, solve these simultaneous equations:

y = 2x + 3   (1)

y(x + 1) = 10   (2)

Complete the working below by filling in the blanks with missing numbers.

## Column B

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

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

y = 3x²

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

x² - 2xy + y² = 1

 -5 -4 4 5 x y

Solve the following simultaneous equations using the substitution method:

xy - 2y - x = 2

x + y = 7

Then type your solutions into the correct spaces below.

 -5 -4 4 5 x y

Consider these two equations:

3x + y = 7

xy + x² = 6

Solve these equations on paper using substitution then type in your two possible pairs of 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 these simultaneous equations:

x² + y² = 25

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.

Find the points of intersection then type in the values of your solutions below.

 -5 -4 4 5 x y
• Question 1

Consider these two equations:

xy + x + y = -1   (1)

y = 3x + 1   (2)

Written below you will see all the steps required to solve these equations, but in the wrong order.

Put the steps into the correct order by matching each to the correct position in the sequence.

## 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
Here we have one linear equation and one quadratic equation. First, we need to substitute equation (2) into (1) and expand the brackets to obtain an equation in the form: x = ... Then we simplify to obtain a quadratic equation. Next, we factorise this quadratic and solve to find the two possible x-values. We then substitute each of the x-values into equation (2) to find the two related y-values. Finally, we write our two pairs of solutions in the approved format. Did you order those steps correctly? If you need to, review the Introduction before you move on to tackle the rest of this activity.
• Question 2

Review these two equations:

y² - x² = 60   (1)

y = 3x + 2   (2)

Written below you will see all the steps required to solve these equations, but in the wrong order.

Put the steps into the correct order by matching each to the correct position in the sequence.

## Column B

1st
y² - x² = 60 (1) and y = 3x + 2...
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 = 3 × 2 + 2 = 8 [sub int...
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 the form: x = ... Then we simplify to obtain a quadratic equation. Next, we factorise this quadratic and solve to find the two possible x-values. Then we substitute each of the x-values into equation (2) to find the two related y-values. Finally, we give two pairs of solutions in the approved format.
• Question 3

Using the substitution method, solve these simultaneous equations:

xy = 2   (1)

y = x + 1   (2)

Complete the working below by filling in the blanks with missing numbers.

EDDIE SAYS
Here's the full solution for this pair of equations: xy = 2 (1) y = x + 1 (2) Let's substitute (2) into (1) now: x(x + 1) = 2 Let's expand our brackets: x² + x = 2 And rearrange our equation: x² + x - 2 = 0 Now we can factorise: (x + 2)(x - 1) = 0 So either x + 2 = 0 or x - 1 = 0 So x = -2 or x = 1 Now we can substitute these values of x into (2) to find the corresponding y values: y = -2 + 1 = -1 y = 1 + 1 = 2 Finally, our solutions are: x = -2, y = -1 or x = 1, y = 2
• Question 4

Using the substitution method, solve the simultaneous equations

Using the substitution method, solve these simultaneous equations:

y = 2x + 3   (1)

y(x + 1) = 10   (2)

Complete the working below by filling in the blanks with missing numbers.

EDDIE SAYS
Here is the full solution for this pair, without a detailed explanation: 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 = 2 × (-7/2) + 3 = -4 [sub into (2)] y = 2 × 1 + 3 = 5 Solutions: x = -7/2, y = -4 or x = 1, y = 5 How are you getting on with these? Are you starting to become more confident with this process?
• Question 5

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

y = 3x²

y - 3x = 6

 -12 -3 -2 -1 1 2 3 12 x y
EDDIE SAYS
This time, there is no extended written explanation to support you. Compare what you have written to our example working below: 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 × 2² = 12 [sub into (1)] y = 3 × (-1)² = 3 Solutions x = 2, y = 12 or x = -1, y = 3 How did your process compare to this example?
• Question 6

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

x + y = 9

x² - 2xy + y² = 1

 -5 -4 4 5 x y
EDDIE SAYS
Here is our full solution this time: 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 Solutions: x = 5, y= 4 or x = 4, y = 5 Don't worry if you made a slip along the way in your method. Examiners will give marks for your workings in these long and complex questions, so make sure you are recording these accurately as it is great practise.
• Question 7

Solve the following simultaneous equations using the substitution method:

xy - 2y - x = 2

x + y = 7

Then type your solutions into the correct spaces below.

EDDIE SAYS
So our 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 [sub into (2)] Solutions: x = 4, y = 3 Notice there is only one pair of solutions here, as the quadratic in y has a repeated factor. If the factor in both brackets is exactly the same, there will only be one outcome for y. This will lead to only one viable option for x too. This is unusual, but you must be prepared for it.
• Question 8

Consider these two equations:

3x + y = 7

xy + x² = 6

Solve these equations on paper using substitution then type in your two possible pairs of solutions below.

EDDIE SAYS
Hear's our full solution here: 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 [simplify] 2x² - 7x + 6 = 0 (2x - 3)(x - 2) = 0 [factorise] Either 2x - 3 = 0 or x - 2 = 0 So x = 3/2 or x = 2 y = 7 - 3 × (3/2) = 5/2 [sub into (1)] y = 7 - 3 × 2 = 1 Solutions: x = 3/2, y = 5/2 or x = 2, y = 1 In order to reach the correct answer, you must pair the correct x-value with its relevant y-value, as mis-matched pairs will not be accepted.
• Question 9

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

x² + y² = 25

y = x + 1

Give your solutions as coordinates in the spaces below.

EDDIE SAYS
Don't let the diagram throw you off here, we can just solve this pair of equations as usual. So here is our solution for this question: 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 y = -4 + 1 = -3 Solutions: (3, 4) or (-4, -3) Just one more challenge to go now!
• Question 10

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

Find the points of intersection then type in the values of your solutions below.

EDDIE SAYS
So we need to solve the simultaneous equations: x² + 3xy + y² = 11 (1) 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 [expand] -x² + 3x - 2 = 0 [simplify] x² - 3x + 2 = 0 (x - 2)(x - 1) = 0 [factorise] x = 2 or x = 1 y = 3 - 2 = 1 [sub into (2) y = 3 - 1 = 2 Solutions: (2 , 1) or (1 , 2) Great work! Some of these questions are as tough as it gets, so you should feel very proud of yourself. Don't forget to always record your working, as you can earn valuable marks for this in your exam, even if you do not reach the correct answer.
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