\left\{ \begin{array} { l } { 3 a - b + c = 10 } \\ { a - 2 b - c = 6 } \\ { a + b + c = 12 } \end{array} \right.
Solve for a, b, c
a=19
b=20
c=-27
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b=3a+c-10
Solve 3a-b+c=10 for b.
a-2\left(3a+c-10\right)-c=6 a+3a+c-10+c=12
Substitute 3a+c-10 for b in the second and third equation.
a=-\frac{3}{5}c+\frac{14}{5} c=11-2a
Solve these equations for a and c respectively.
c=11-2\left(-\frac{3}{5}c+\frac{14}{5}\right)
Substitute -\frac{3}{5}c+\frac{14}{5} for a in the equation c=11-2a.
c=-27
Solve c=11-2\left(-\frac{3}{5}c+\frac{14}{5}\right) for c.
a=-\frac{3}{5}\left(-27\right)+\frac{14}{5}
Substitute -27 for c in the equation a=-\frac{3}{5}c+\frac{14}{5}.
a=19
Calculate a from a=-\frac{3}{5}\left(-27\right)+\frac{14}{5}.
b=3\times 19-27-10
Substitute 19 for a and -27 for c in the equation b=3a+c-10.
b=20
Calculate b from b=3\times 19-27-10.
a=19 b=20 c=-27
The system is now solved.
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