You can solve this system by finding what one variable equals from one of the equations, then put this into the other equation.

I went to find #y# here in the start. Because I saw that locking #x# by itself would be fair enough. It gave a clean #x=13-4y#, instead of fractions or such.

I then put what #x# equals to into the other #y# equation. So that I can find the integer value of #y# without having any #x# variables. Which gave the result of #y=3#.

From there, we can place the #y=3# into the other equation and find the #x# value, #x=13-4(3)# instead of #x=13-4y#. Which gave the result of #x=1#.

From that, we now know that:

#y=3# and #x=1#