# GCSE Mathematics/Simultaneous Equations

## Solving simultaneous equations

### By elimination

One way of solving a simultaneous equation is by canceling out either the x or y values so that you are left with a linear equation.

#### First example

$20x+15y=135$ $20x-8y=20$ In this example, we could subtract the second equation from the first to get this:

$23y=115$ $y=5$ Once we know this, we can go back to one of the original equations, and replace y with 5, then solve it, like this:

$20x+15(5)=135$ $20x=135-75$ $x={\frac {60}{20}}=3$ So, the final solution is:

$x=3$ $y=5$ #### Second example

$4x+2y=12$ $x+y=4$ We can see that in this example the equations will not cancel each other out. To make them cancel each other out, we multiply the second equation by two and get:

$2x+2y=8$ We can now subtract this from the original equation in order to get a linear equation that we can solve:

$2x=4$ $x=2$ Now that we know the value of x, we can substitute it in the first equation in order to solve it:

$4(2)+2y=12$ $2y=12-8$ $y={\frac {4}{2}}=2$ So, the final solution is:

$x=2$ $y=2$ 