\left\{ \begin{array} { l } { 3 a - b + c = 2 } \\ { 3 a + b - c = - 2 } \\ { 3 a + b + c = 0 } \end{array} \right.
Solve for a, b, c
a=0
b=-1
c=1
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b=3a+c-2
Solve 3a-b+c=2 for b.
3a+3a+c-2-c=-2 3a+3a+c-2+c=0
Substitute 3a+c-2 for b in the second and third equation.
a=0 c=1-3a
Solve these equations for a and c respectively.
c=1-3\times 0
Substitute 0 for a in the equation c=1-3a.
c=1
Calculate c from c=1-3\times 0.
b=3\times 0+1-2
Substitute 0 for a and 1 for c in the equation b=3a+c-2.
b=-1
Calculate b from b=3\times 0+1-2.
a=0 b=-1 c=1
The system is now solved.
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