Question 333703
{{{system(2x+3z=2,
3x+6y=6,
x-2z=8)}}}
<pre><font size = 4 color = "indigo"><b>
The first and third equations are in x and z only, 
that is, neither has a y terms, so we start by taking them
as a system of 2 equations in 2 unknowns.

{{{system(2x+3z=2,
x-2z=8)}}} 

Multiply the second equation through by -2:

{{{system(2x+3z=2,
-2x+4z=-16)}}} 

We get 

{{{7z=-14}}} or

{{{z=-2}}}

Substituting in 

{{{x-2z=8}}}

{{{x-2(-2)=8}}}

{{{x+4=8}}}

{{{x=4}}}

Substitute that in the original equation which contained y:

{{{3x+6y=6}}}

{{{3(4)+6y=6}}}

{{{12+6y=6}}}

{{{6y=-6}}}

{{{y=-1}}}

So the solution is (x,y,z,) = (4,-1,-2)

Edwin</pre>