Solve {l}{x_{1}-x_{2}+x_{3}=2}{x_{1}+2x_{2}=1}{x_{1}-x_{3}=4}

$$\left\{ \begin{array}{l}{ x _ { 1 } - x _ { 2 } + x _ { 3 } = 2 }\\{ x _ { 1 } + 2 x _ { 2 } = 1 }\\{ x _ { 1 } - x _ { 3 } = 4 }\end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{13}{5} = 2\frac{3}{5} = 2.6$$

$$x_{2}=-\frac{4}{5}=-0.8$$

$$x_{3} = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

Short Steps Using Substitution

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

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

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

$$x_{2}-x_{3}+2+2x_{2}=1 x_{2}-x_{3}+2-x_{3}=4$$

Solve these equations for x_{2} and x_{3} respectively.

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

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

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

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

$$x_{3}=-\frac{7}{5}$$

Substitute -\frac{7}{5} for x_{3} in the equation x_{2}=-\frac{1}{3}+\frac{1}{3}x_{3}.

$$x_{2}=-\frac{1}{3}+\frac{1}{3}\left(-\frac{7}{5}\right)$$

Calculate x_{2} from x_{2}=-\frac{1}{3}+\frac{1}{3}\left(-\frac{7}{5}\right).

$$x_{2}=-\frac{4}{5}$$

Substitute -\frac{4}{5} for x_{2} and -\frac{7}{5} for x_{3} in the equation x_{1}=x_{2}-x_{3}+2.

$$x_{1}=-\frac{4}{5}-\left(-\frac{7}{5}\right)+2$$

Calculate x_{1} from x_{1}=-\frac{4}{5}-\left(-\frac{7}{5}\right)+2.

$$x_{1}=\frac{13}{5}$$

The system is now solved.

$$x_{1}=\frac{13}{5} x_{2}=-\frac{4}{5} x_{3}=-\frac{7}{5}$$