To solve the system of equations we need to use elimination. But to do that we are required to rewrite the equations in implicit form.
![ax+by+c=0](/tpl/images/0154/7819/bd04f.png)
So we have,
![3x-y+3=0 \\ -2x-y+3=0](/tpl/images/0154/7819/e511d.png)
Then multiply the first equation by 2 on both sides and second equation by 3 on both sides. Resulting with,
![6x-2y+6=0 \\ -6x-3y+9=0](/tpl/images/0154/7819/7a939.png)
Adding these two equations eliminates the first term in both since 6x - 6x = 0. Hence,
![-5y+15=0\Longrightarrow y=3](/tpl/images/0154/7819/bb6a4.png)
Now that we know the value of y we can insert it in either one of the equations in the system. I'll pick first one to get x.
![3=3x+3\Longrightarrow 3x=0\Longrightarrow x=0](/tpl/images/0154/7819/a30c4.png)
So the solution to this system of equation are ![\boxed{x=0},\boxed{y=3}](/tpl/images/0154/7819/cb443.png)
Hope this helps.
r3t40