Solve {c}{2x_{1}-x_{2}+x_{3}=8}{x_{1}+2x_{2}+2x_{3}=6}{x_{1}-2x_{2}-x_{3}=1}

$$\left.\begin{array} { c } { 2 x _ { 1 } - x _ { 2 } + x _ { 3 } = 8 } \\ { x _ { 1 } + 2 x _ { 2 } + 2 x _ { 3 } = 6 } \\ { x _ { 1 } - 2 x _ { 2 } - x _ { 3 } = 1 } \end{array} \right\}$$

Solve for x_1, x_2, x_3

$$x_{1}=2$$

$$x_{2}=-1$$

$$x_{3}=3$$

Short Steps Using Substitution

Solve 2x_{1}-x_{2}+x_{3}=8 for x_{2}.

$$x_{2}=2x_{1}+x_{3}-8$$

Substitute 2x_{1}+x_{3}-8 for x_{2} in the second and third equation.

$$x_{1}+2\left(2x_{1}+x_{3}-8\right)+2x_{3}=6 x_{1}-2\left(2x_{1}+x_{3}-8\right)-x_{3}=1$$

Solve these equations for x_{1} and x_{3} respectively.

$$x_{1}=\frac{22}{5}-\frac{4}{5}x_{3} x_{3}=5-x_{1}$$

Substitute \frac{22}{5}-\frac{4}{5}x_{3} for x_{1} in the equation x_{3}=5-x_{1}.

$$x_{3}=5-\left(\frac{22}{5}-\frac{4}{5}x_{3}\right)$$

Solve x_{3}=5-\left(\frac{22}{5}-\frac{4}{5}x_{3}\right) for x_{3}.

$$x_{3}=3$$

Substitute 3 for x_{3} in the equation x_{1}=\frac{22}{5}-\frac{4}{5}x_{3}.

$$x_{1}=\frac{22}{5}-\frac{4}{5}\times 3$$

Calculate x_{1} from x_{1}=\frac{22}{5}-\frac{4}{5}\times 3.

$$x_{1}=2$$

Substitute 2 for x_{1} and 3 for x_{3} in the equation x_{2}=2x_{1}+x_{3}-8.

$$x_{2}=2\times 2+3-8$$

Calculate x_{2} from x_{2}=2\times 2+3-8.

$$x_{2}=-1$$

The system is now solved.

$$x_{1}=2 x_{2}=-1 x_{3}=3$$