Question 331613
solve the system: 

{{{system(x-y+2z=7,
2x+z=4,
x+5y+z=9)}}}
<pre><b>
Since y is already missing in the middle equation, we will
eliminate y from the other two equations:

{{{system(x-y+2z=7,
x+5y+z=9)}}}

To make the y's cancel we multiply the first one through by 5

{{{system(5x-5y+10z=35,
x+5y+z=9)}}}

Adding corresponding terms causes the y's to cancel:

{{{6x+11z=44}}}

Now we put that with the original middle equation that
already had y eliminated:

{{{system(6x+11z=44,
2x+z=4)}}}

To make the x's cancel we multiply the second one through by -3:

{{{system(6x+11z=44,
-6x-3z=-12)}}}

Adding corresponding terms causes the x's to cancel:

{{{8z=32}}}
{{{z=4}}}

Substitute that in 

{{{2x+z=4)}}}
{{{2x+4=4}}}
{{{2x=0}}}
{{{x=0}}}

Substitute that and {{{z=4}}} in one of the original
three equations, that contains a y:

{{{x-y+2z=7}}}
{{{(0)-y+2(4)=7}}}
{{{-y+8=7}}}
{{{-y=-1}}}
{{{y=1}}}

So the solution is (x,y,z) = (0,1,4)

Edwin</pre>