Solving Simultaneous Equations by Substitution (1)
• Introduction

Simultaneous equations involve two variables, each of whose value must be found.

The following method, in which one equation is substituted into the other, works best when one equation contains a single letter on its own.

Example

Solve simultaneously

3x + 2y = 7

y = 4x - 2

3x + 2y = 7     (a)

y = 4x - 2        (b)

We substitute (b) into (a).

Wherever we see a y in (a) we will replace it with (4x - 2), remembering the brackets.

(a) becomes:

3x + 2(4x - 2) = 7

Multiply out the brackets and solve.

Notice that we now have an equation just in x.

3x + 8x - 4 = 7

11x - 4 = 7

11x = 11

x = 1

Now use (b) to determine y.

y = 4x - 2

When x = 1, this gives:

y = 4×1 - 2 = 4 - 2 = 2

y = 2

Check by putting this into the other equation (a) to get 3 × 1 + 2 × 2 = 3 + 4 = 7

Solution is x = 1, y = 2

• Question 1

Solve simultaneously:

3x + 2y = 13

y = x - 1

x = 3, y = 2

x = 12, y = 3

x = 4, y = 1

• Question 2

Solve simultaneously:

x = y - 2

13x - 5y = 6

x = 6, y = -2

x = 2, y = 4

x = 4, y = 2

• Question 3

Solve simultaneously:

4x + 3y = 11

x = y + 1

x = 4, y = 2

x = 3, y = 2

x = 2, y = 1

• Question 4

Solve simultaneously:

2x + 5y = 31

y = 2x - 1

x = 4, y = -2

x = 13, y = 1

x = 3, y = 5

• Question 5

Solve simultaneously:

7x + 2y = 20

y = 4x - 5

x = 5, y = 5

x = 6, y = -11

x = 2, y = 3

• Question 6

Solve simultaneously:

5x + 6y = 17

y = 5x - 3

x = 1, y = 2

x = 3, y = 4

x = 2, y = 4

• Question 7

Solve simultaneously:

4x + 3y = 22

x = 2y

x = 10, y = -6

x = 6, y = 3

x = 4, y = 2

• Question 8

Solve simultaneously:

3x + 7y = 0

y = x - 10

x = 21, y = 1

x = 7, y = -3

x = 3, y = 7

• Question 9

Solve simultaneously:

5x + 3y = -33

y = -7 - x

x = -6, y = -1

x = 1, y = 4

x = 1, y = 3

• Question 10

Solve simultaneously:

6x + 5y = 29

x = y - 8

x = -1, y = 7

x = 1½, y = 4

x = 1, y = 10

Progress