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:

$x=2$

$y=2$