First we pick the equation. Let's say we pick first one. From it we express y.
[tex]-5x+y=-2\Longrightarrow y=-2+5x[/tex]
Then we use this y and plug it in instead of y in the second equation.
[tex]-3x+6(-2+5x)=-12[/tex]
Now just solve for x.
[tex]
-3x-12+30x=-12 \\
-3x=-30x \\
\underline{x=10}
[/tex]
We plug this x in the first equation.
[tex]-5\cdot10+y=-2[/tex]
And solve for y.
[tex]
-50+y=-2 \\
\underline{y=48}
[/tex]
So the solution to the equation is geometrically a point P which lies on the intersection of the two lines.
[tex]\boxed{P(10,48)}[/tex]
Hope this helps.
r3t40
