# GCSE Mathematics/Simultaneous Equations

## Solving simultaneous equationsEdit

### By eliminationEdit

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 exampleEdit

$20x + 15y = 135$
$20x - 8y = 20$

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 exampleEdit

$4x + 2y = 12$
$x + y = 4$

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:

$x = 2$
$y = 2$