Question 763884
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Your problem

3x  +  2y  +   z  =  -5  --- eqn (1)
x   -  3y   + 2z  =  28/3 ---- eqn (2)
-6x +   y   + 2z  =  -5  ----- eqn (3)

Solution.

Step 1) We first try to eliminate 1 variable (say z) from the 3 equations, to get 2 equations in x and y, which we can solve.

Step 2) Subtract eqn 3 from eqn 2 to get

{{{(x - (-6*x)) - 3*y - y = 28/3 + 5}}}   (note that the two 2z terms cancel out)

{{{7*x - 4*y = 43/3}}}  ---- eqn (4)

Step 3) Multiply eqn (1) by 2 to get another 2z term that can be cancelled.

{{{6*x + 4*y + 2*z = -10}}}  ---- eqn (5)

Step 4) Subtract eqn 5 from eqn 3 (both have 2z terms which will cancel out)

{{{-12*x  - 3*y = 5}}} ----- eqn (6)

Step 5) Now we have 2 eqns (4) and (6) in x and y. We will solve it by eliminating y.

Multiply eqn (4) by 3 to get

{{{21*x - 12*y = 43}}}   ---- eqn (7)

Multiply eqn (6) by 4 to get

{{{-48*x - 12*y = 20}}}   ---- eqn (8)

Step 6)
Subtract eqn 8 from eqn 7 to get

{{{69*x = 23}}} or {{{x = 1/3}}}

Step 7) Substitute for x in eqn (7)
{{{21*1/3 - 12*y = 43}}}
{{{7 - 12*y = 43}}}
{{{12*y = -36}}} or {{{y = -3}}}

Step 8)
Now substitute for x and y in eqn(1)
{{{3*1/3 + 2*(-3) + z = -5}}}
{{{1 - 6 + z = -5}}} or {{{z = 0}}}

So the final solution is 
{{{highlight(x = 1/3)}}}
{{{highlight(y = -3)}}}
{{{highlight(z = 0)}}}

Check: Substitute in original 3 equations.

Eqn 1: 3*1/3 + 2*(-3) + 0 = 1 - 6 = -5 - correct
Eqn 2: 1/3 -3*(-3) + 2*0 = 1/3 + 9 = 28/3 - correct
Eqn 3: -6*1/3 -3 + 2*0 = -2 - 3 = -5 --- correct

:)

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