First example

${\displaystyle 20x+15y=135}$

${\displaystyle 20x-8y=20}$

In this example, we could subtract the second equation from the first to get this:

${\displaystyle 23y=115}$

${\displaystyle y=5}$

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

${\displaystyle 20x+15(5)=135}$

${\displaystyle 20x=135-75}$

${\displaystyle x={\frac {60}{20}}=3}$

So, the final solution is:

${\displaystyle x=3}$

${\displaystyle y=5}$

Second example

${\displaystyle 4x+2y=12}$

${\displaystyle x+y=4}$

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:

${\displaystyle 2x+2y=8}$

We can now subtract this from the original equation in order to get a linear equation that we can solve:

${\displaystyle 2x=4}$

${\displaystyle x=2}$

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

${\displaystyle 4(2)+2y=12}$

${\displaystyle 2y=12-8}$

${\displaystyle y={\frac {4}{2}}=2}$

So, the final solution is:

${\displaystyle x=2}$

${\displaystyle y=2}$