Question 258240
<pre><font size = 4 color = "indigo"><b>
{{{system(x+2y+5z= -1,
2x-y+z=2,
3x+4y-4z=14)}}}

Pick a letter and an equation to solve for it in.

I will pick z and the second equation to solve for z in:

{{{2x-y+z=2}}}
{{{z=2-2x+y}}}

Substitute {{{(2-2x+y)}}} for z in the other two
equations, (but not the same one you just used) 

Substituting it in

{{{x+2y+5z=-1}}}
{{{x+2y+5(2-2x+y)=-1}}}
{{{x+2y+10-10x+5y=-1}}}
{{{-9x+7y+10=-1}}}
{{{-9x+7y=-11}}}

Substituting it in

{{{3x+4y-4z=14}}}
{{{3x+4y-4(2-2x+y)=14}}}
{{{3x+4y-8+8x-4y=14}}}
{{{11x-8=14}}}
{{{11x=22}}}
{{{x=2}}}

Now you just have this simple system:

{{{system(-9x+7y=-11,x=2)}}}

and I'll bet you can finish that by yourself, just by
substituting back

Answer x=2, y=1, z=-1 

Edwin</pre>