GCSE Mathematics/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:
 * $$x = 2$$
 * $$y = 2$$