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?

There are two methods of solving simultaneous equations. Use the method which you prefer.

Method 1: elimination

First form 2 equations. Let fish be f and chips be c.

We know that:

3f + 2c = 280 (1)

f + 4c = 260 (2)

Doubling (1) gives:

6f + 4c = 560 (3)

(3)-(2) is 5f = 300

so f = 60

Therefore the price of fish is 60p

Substitute this value into (1):

3(60) + 2c = 280

so 2c = 100

c = 50

Therefore the price of chips is 50p

Method 2: Substitution

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

so 780 - 12c + 2c = 280

so 10c = 500

so 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

so 8x - 2x² + 4 - x - 10 = 0

so 2x² - 7x + 6 = 0

so (2x - 3)(x - 2) = 0

so 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 equations can also be solved by graphical methods.