Question 1196110
<pre>
{{{system(2x-4y+5z=-33,
4x-y=-5,
-2x+2y-3z=19)}}}

Since z is already eliminated from the 2nd equation,
let's eliminate z from the 1st and 3rd equations:

Multiply the first equation by 3 and the 3rd equation by 5,
so the coefficients of z will be alike with opposite signs:

{{{system(matrix(1,3,"","",6x-12y+15z=-99),

-10x+10y-15z=95)}}}

Add the equations term by term and the 15z and the -15z will cancel:

{{{-4x-2y=-4}}}

Put that with the original 2nd equation:

{{{system(-4x-2y=-4,matrix(1,3,"","",4x-y=-5)) }}}

Add those term by term, the 4x and the -4x will cancel:

{{{-3y=-9

{{{y=3}}}

Substitute 3 for y in 

{{{4x-y=-5}}}

{{{4x-3=-5}}}

{{{4x = -2}}}

{{{x=-1/2}}}

Substitute -1/2 for x and 3 for y in 2x-4y+5z=-33

{{{2x-4y+5z=-33}}}
{{{2(-1/2)-4(3)+5z=-33}}}
{{{-1-12+5z=-33}}}
{{{-13+5z=-33}}}
{{{5z=-20}}}
{{{z=-4}}}

(x,y,z) = (-1/2,3,-4}}}

Edwin</pre>