Solve {l}{2x_{1}+x_{2}-x_{3}=0}{x_{1}+2x_{2}-x_{3}=-1}{x_{1}+x_{2}+x_{3}=3}

$$\left. \begin{array} { l } { 2 x _ { 1 } + x _ { 2 } - x _ { 3 } = 0 } \\ { x _ { 1 } + 2 x _ { 2 } - x _ { 3 } = - 1 } \\ { x _ { 1 } + x _ { 2 } + x _ { 3 } = 3 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=1$$

$$x_{2}=0$$

$$x_{3}=2$$

Short Steps Using Substitution

Solve 2x_{1}+x_{2}-x_{3}=0 for x_{2}.

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

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

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

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

$$x_{1}=\frac{1}{3}+\frac{1}{3}x_{3} x_{3}=\frac{3}{2}+\frac{1}{2}x_{1}$$

Substitute \frac{1}{3}+\frac{1}{3}x_{3} for x_{1} in the equation x_{3}=\frac{3}{2}+\frac{1}{2}x_{1}.

$$x_{3}=\frac{3}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{3}x_{3}\right)$$

Solve x_{3}=\frac{3}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{1}{3}x_{3}\right) for x_{3}.

$$x_{3}=2$$

Substitute 2 for x_{3} in the equation x_{1}=\frac{1}{3}+\frac{1}{3}x_{3}.

$$x_{1}=\frac{1}{3}+\frac{1}{3}\times 2$$

Calculate x_{1} from x_{1}=\frac{1}{3}+\frac{1}{3}\times 2.

$$x_{1}=1$$

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

$$x_{2}=-2+2$$

Calculate x_{2} from x_{2}=-2+2.

$$x_{2}=0$$

The system is now solved.

$$x_{1}=1 x_{2}=0 x_{3}=2$$