First we pick the equation. Let's say we pick first one. From it we express y.
![-5x+y=-2\Longrightarrow y=-2+5x](https://img.qammunity.org/2020/formulas/mathematics/high-school/qy5s91u8r0aireszrqtga75l1sk36yqxfz.png)
Then we use this y and plug it in instead of y in the second equation.
![-3x+6(-2+5x)=-12](https://img.qammunity.org/2020/formulas/mathematics/high-school/y1tqm8zujyy6zjwxtl8nc8jqdhwx6dgye0.png)
Now just solve for x.
![</p><p>-3x-12+30x=-12 \\</p><p>-3x=-30x \\</p><p>\underline{x=10}</p><p>](https://img.qammunity.org/2020/formulas/mathematics/high-school/i7ngsfdlxjzxvl9ilpuwdhoedb51gdykcs.png)
We plug this x in the first equation.
![-5\cdot10+y=-2](https://img.qammunity.org/2020/formulas/mathematics/high-school/13mwwe0sp7ds5yw5ofs6v7cg77r0rusbbu.png)
And solve for y.
![</p><p>-50+y=-2 \\</p><p>\underline{y=48}</p><p>](https://img.qammunity.org/2020/formulas/mathematics/high-school/64aogvvdw37i4104dvfnjgr0f9ivrq07ew.png)
So the solution to the equation is geometrically a point P which lies on the intersection of the two lines.
![\boxed{P(10,48)}](https://img.qammunity.org/2020/formulas/mathematics/high-school/s6zi4sjjkqlr0vwsxxo64v6cfem014c8ux.png)
Hope this helps.
r3t40