How do you solve the following linear system: #-2x+y=1 , -4x+2y=-8 #?

There is two ways you can try to solve the linear system, either graphing it or algebraically.
If you graph both linear equations on the graph, they are parallel to each other and will never meet.
graph{-2x+y=1 [-10, 10, -5, 5]}
graph{-4x+2y=-8 [-10, 10, -5, 5]}
To solve it algebraically, the substitution method would be the easiest.
For the first equation, #-2x+y=1#, isolate #y# by adding #-2x# to both sides of the equation. This will give you #y=1+2x#.
Substitute #1+2x# for #y# into the other equation.
Solve for x: #-4x+2(1+2x)=-8#
#-4x+2+4x=-8#
#2!=-8#

First, we need to isolate #y# in the first equation so that we have an equation to plug in:

#y=1+2x#

Next, let's plug it in and solve for #x#:

#-4x+2(1+2x)=-8#
#-4x+2+4x=-8#
#2=-8#

Wait a second. What happened here? Let's look at the equations we started out with. If we multiply the first equation by 2, we get #-4x+2y=2#. And if we combine the two equations...

#-4x+2y=2#
#-4x+2y=-8#
#2=-4x+2y=-8#
#2!=-8#

There aren't any solutions to this! If you graph these two lines, you'll see that they are parallel lines; they have the same slope, but different y-intercepts.

This is something that's very important to keep in mind. Make sure to double check, but not every system has a solution.