\left\{ \begin{array} { r } { x _ { 1 } + 2 x _ { 2 } + x _ { 3 } = 3 } \\ { 3 x _ { 1 } + 2 x _ { 2 } + 2 x _ { 3 } = 4 } \\ { 2 x _ { 1 } + x _ { 2 } + x _ { 3 } = 2 } \end{array} \right.
Solve for x_1, x_2, x_3
x_{1}=0
x_{2}=1
x_{3}=1
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x_{1}=-2x_{2}-x_{3}+3
Solve x_{1}+2x_{2}+x_{3}=3 for x_{1}.
3\left(-2x_{2}-x_{3}+3\right)+2x_{2}+2x_{3}=4 2\left(-2x_{2}-x_{3}+3\right)+x_{2}+x_{3}=2
Substitute -2x_{2}-x_{3}+3 for x_{1} in the second and third equation.
x_{2}=-\frac{1}{4}x_{3}+\frac{5}{4} x_{3}=-3x_{2}+4
Solve these equations for x_{2} and x_{3} respectively.
x_{3}=-3\left(-\frac{1}{4}x_{3}+\frac{5}{4}\right)+4
Substitute -\frac{1}{4}x_{3}+\frac{5}{4} for x_{2} in the equation x_{3}=-3x_{2}+4.
x_{3}=1
Solve x_{3}=-3\left(-\frac{1}{4}x_{3}+\frac{5}{4}\right)+4 for x_{3}.
x_{2}=-\frac{1}{4}+\frac{5}{4}
Substitute 1 for x_{3} in the equation x_{2}=-\frac{1}{4}x_{3}+\frac{5}{4}.
x_{2}=1
Calculate x_{2} from x_{2}=-\frac{1}{4}+\frac{5}{4}.
x_{1}=-2-1+3
Substitute 1 for x_{2} and 1 for x_{3} in the equation x_{1}=-2x_{2}-x_{3}+3.
x_{1}=0
Calculate x_{1} from x_{1}=-2-1+3.
x_{1}=0 x_{2}=1 x_{3}=1
The system is now solved.
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