Question 1000185
Knowing how to do it, you will solve the problem as you want and without the mistake you are making.  


The equivalent system is  {{{system(x=4-2y,2y=-x-4)}}} and the substitution was not yet done.


{{{2y=-(4-2y)-4}}}

{{{2y=-4+2y-4}}}

{{{2y=2y-8}}}, a false statement.
NO SOLUTION.



Notice the two equations solved for y in terms of x, starting with the way each was given:
{{{x+2y=4}}}
{{{2y=-x+4}}}
{{{y=-x/2+2}}}
{{{y=-(1/2)x+2}}}
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{{{4y=-2x-8}}}
{{{y=-2x/4-2}}}
{{{y=-(1/2)x-2}}}
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BOTH SLOPES are  {{{-1/2}}}, the lines for each equation are parallel, therefore they never intersect, and this means NO POINT COMMON TO BOTH LINES. 


Meaning, solving the given system using whatever method will yield NO SOLUTION.