Simultaneous Equations by Substitution
Simultaneous equations are two equations with two unknown variables, and we want to find the same solution
There are 5 steps
In steps b and d we use a combination of Solving Equations − Add and Subtract and Solving Equations − Multiply and Divide
Please look at the above before continuing
Example 1. Solve the following two equations by substitution
2x + y = 5
3x − y = 15Example 2. Solve the following two equations by substitution
4a + b = 20
5a + 3b = 32
i. Label each equation
| Equation | Label | ||||
| 2x + y | = | 5 | (1) | ||
| 3x − y | = | 15 | (2) | ||
ii. Rearrange equation (1) to make y the subject
| 2x + y | = | 5 | Note (3) | |||||
| − | 2x | = | − 2x |
| y = 5 − 2x |
Note: (3) inverse of + 2x is − 2x
iii. Substitute y = 5 − 2x into equation (2):
| 3x | − | y | = | 15 | |||
| 3x | − | (5 − 2x) | = | 15 | Note (4) | ||
| 3x | − | 5 + 2x | = | 15 |
Note: (4) − ( − 2) = + 2
iv. Rearrange to find the value of x
| 3x | − | 5 + 2x | = | 15 | Note (5) | ||
| + | 5 | = | + 5 |
| 5x | = | 20 | Note (6) | |||||
| ÷ | 5 | = | ÷ | 5 |
| x | = | 4 |
Note: (5) inverse of − 5 is + 5
Note: (6) inverse of × 5 is ÷ 5
v. put x = 4 into equation:
So x = 4 and y = −3
i. Label each equation
| Equation | Label | ||||
| 4a + b | = | 20 | (1) | ||
| 5a + 3b | = | 32 | (2) | ||
ii. Rearrange equation (1) to make b the subject
| 4a + b | = | 20 | Note (3) | |||||
| − | 4a | = | − 4a |
| b = 20 − 4a |
Note: (3) inverse of + 4a is − 4a
iii. Substitute b = 20 − 4a into equation (2):
| 5a | + | 3(b) | = | 32 | |
| 5a | + | 3(20 − 4a) | = | 32 | |
| 5a | + | 60 − 12a | = | 32 |
iv. Rearrange to find the value of a
| 5a | + | 60 − 12a | = | 32 | Note | ||
| − | 60 | = | − 60 | (4) |
| −7a | = | − | 28 | Note | ||||
| ÷ | −7 | = | ÷ | −7 | (5) |
| a | = | 4 |
Note: (4) inverse of + 60 is − 60
Note: (5) inverse of × −7 is ÷ −7
v. put a = 4 into equation:
So a = 4 and b = 4
to: 
