"Simultaneous equations" are two or more equations which have two or more unknowns to be found. At GCSE, it is unlikely that you will have more than two equations with 2 values (x and y) which need to be found.

Example

A man buys 3 fish and 2 chips for £2.80
A woman buys 1 fish and 4 chips for £2.60
How much are the fish and how much are the chips?

First form the equations. Let fish be f and chips be c.
We know that:
3f + 2c = 280    (1)
f + 4c = 260      (2)

These two equations are both true at the same time, hence the name 'simultaneous'. 
There are two methods of solving simultaneous equations. Use the method which you prefer:

Elimination

The method of elimination involves manipulating the two equations so that one can be added/ subtracted from the other to leave us with an equation with only one unknown.

In our above example:

Doubling (1) gives:
6f + 4c = 560 (3)
(3)-(2) gives 5f = 300
\ f = 60
Therefore the price of fish is 60p

Substitute this value into (1):
3(60) + 2c = 280
\ 2c = 100
c = 50
Therefore the price of chips is 50p

Substitution

The method of substitution involves transforming one equation into x = something or y = something and then substituting this into the other equation.

So,

Rearrange one of the original equations to isolate a variable.
Rearranging (2): f = 260 - 4c
Substitute this into the other equation:
3(260 - 4c) + 2c = 280
\ 780 - 12c + 2c = 280
\ 10c = 500
\ c = 50
Substitute this into one of the original equations to get f = 60 .

Harder simultaneous equations

To solve a pair of equations, one of which contains x², y² or xy, we need to use the method of substitution.

Example

2xy + y = 10  (1)
x + y = 4        (2)
Take the simpler equation and get y = .... or x = ....
from (2), y = 4 - x    (3)
this can be substituted in the first equation. Since y = 4 - x, where there is a y in the first equation, it can be replaced by 4 - x .
sub (3) in (1), 2x(4 - x) + (4 - x) = 10
\ 8x - 2x² + 4 - x - 10 = 0
\ 7x - 2x² - 6 = 0
\ 2x² - 7x + 6 = 0  (taking everything to the other side of the equals sign)
\ (2x - 3)(x - 2) = 0
\ either 2x - 3 = 0 or x - 2 = 0
therefore x = 1.5 or 2 .

Substitute these x values into one of the original equations.
When x = 1.5,  y = 2.5
when x = 2, y = 2

Simultaneous equation can also be solved by graphical methods.

Copyright © Matthew Pinkney 2003