Question 160537
<font size = 7 color = "red"><b>Edwin's solution:</font></b>
<pre><font size = 4 color = "indigo"><b>

That would work, but it would be the best way.

First you should observe that z is already eliminated
from the third equation, so you should eliminate z 
from the first two equations:

To eliminate z from the first two equations:

Multiply the first equation by 3 and the second equation 
by 7, then add them:

3[-10x - 11y + 7z = 145]
7[  7x -  4y - 3z =  53]

  -30x - 33y + 21z = 435
  <u> 49x - 28y - 21z = 371</u>
   19x - 61y       = 806

Now we take the third original equation with
this equation and solve this system:

{{{system(-5x-y= 146, 19x - 61y       = 806)}}}

We can do this by substitution:

Solve the first equation for y:

{{{matrix(3,1,-5x-y=146,-y=146+5x,y=-146-5x)}}}

Substitute {{{-146-5x}}} for {{{y}}} in

{{{19x-61y=806}}}
{{{19x-61(-146-5x)=806}}}
{{{19x+8906+305x=806}}}
{{{324x+8906=806}}}
{{{324x=-8100}}}
{{{x=-25}}}

Substitute that into {{{y=-146-5x}}}
{{{y=-146-5x}}}
{{{y=-146-5(-25)}}}
{{{y=-146+125)}}}
{{{y=-21}}}

Now substitute {{{x=-25}}} and {{{y=-21}}} into
either one of the first two original equations.

{{{7x -  4y - 3z =  53}}}
{{{7(-25) -  4(-21) - 3z =  53}}}
{{{-175 +84 - 3z =  53}}}
{{{-91-3z=53}}}
{{{-3z=144}}}
{{{z=-48}}}

Edwin</pre>