Question 212741
{{{system(2x + y-z=5,x+3y+z=14,3x-y+2z=1)}}}
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Solve for any letter in any equation.

I'll arbitrarily choose to solve the middle equation for z.

{{{x+3y+z=14}}}
{{{z=14-x-3y}}}

Now substitute {{{(14-x-3y)}}} for {{{z}}} in both
the other two equations:

{{{system(2x + y-z=5,3x-y+2z=1)}}}

so we get

{{{system(2x + y-(14-x-3y)=5,3x-y+2(14-x-3y)=1)}}}

Then we simplify:

{{{system(2x + y-14+x+3y=5,3x-y+28-2x-6y)=1)}}}

{{{system(3x +4y=19,x-7y=-27)}}}

Now we have it down to 2 equations and 2 unknowns

We solve the 2nd one here for x:

{{{x-7y=-27)}}}
{{{x=7y-27)}}}

Then we substitute {{{(7y-27)}}} for {{{x}}} in the
other equation:

{{{3x +4y=19}}}

becomes:

{{{3(7y-27)+4y=19}}}

Then we solve for y:

{{{21y-81+4y=19)}}}

{{{25y=100)}}}

{{{y=4}}}

Now we substitute {{{(4)}}} for y in
the equation above solved for x:
 
{{{x-7y=-27)}}}
{{{x-7(4)=-27)}}}
{{{x-28=-27)}}}
{{{x=1}}}

Finally we substitute {{{(4)}}} for y and
{{{(1)}}} for x in the equation above solved for z:

{{{z=14-x-3y}}}

{{{z=14-(1)-3(4)}}}

{{{z=14-1-12}}}

{{{z=1}}}

So the final solution is {{{"(x,y,z)"="(1,4,1)"}}}

Edwin</pre>