Question 253163
{{{system(y+2x=-1,
y-4=-2x)}}}
<pre>
Solve either either equation for either letter.  I'll pick
the easies letter to solve for, which is y because it only
has an understood coefficient of 1 in either equation, whereas
x has other coefficients.  I could pick either equation, 
so I'll just arbitrarily pick the first one to solve for y: 

    y + 2x = -1

Add -2x to both sides:

    y + 2x = -1
       -2x     -2x
   ---------------
    y      = -1-2x

Now substitute (-1-2x) for y in the OTHER equations.

Substitute {{{(-1-2x)}}} for y in


      y - 4 = -2x
 (1-2x) - 4 = -2x
 1 - 2x - 4 = -2x 
    -3 - 2x = -2x
        +2x   +2x
    -------------
    -3      = 0

Uh! Oh!  That's always false, -3 never equals 0, 
so there is no solution.

If you were to graph those two equations, you'd find
that they were parallel and never had any points in common.

Here they are graphed:

{{{drawing(400,400,-5,5,-5,5, graph(400,400,-5,5,-5,5, y=-2x-1),
graph(400,400,-5,5,-5,5, y=4-2x) )}}}

Edwin</pre>