Solve {l}{2x_{1}-3x_{2}+x_{3}=10}{x_{1}+4x_{2}-2x_{3}=8}{3x_{1}+2x_{2}-x_{3}=1}

$$\left\{ \begin{array} { l } { 2 x _ { 1 } - 3 x _ { 2 } + x _ { 3 } = 10 } \\ { x _ { 1 } + 4 x _ { 2 } - 2 x _ { 3 } = 8 } \\ { 3 x _ { 1 } + 2 x _ { 2 } - x _ { 3 } = 1 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

$$x_{2}=-17$$

$$x_{3} = -\frac{193}{5} = -38\frac{3}{5} = -38.6$$

Short Steps Using Substitution

Solve 2x_{1}-3x_{2}+x_{3}=10 for x_{3}.

$$x_{3}=-2x_{1}+3x_{2}+10$$

Substitute -2x_{1}+3x_{2}+10 for x_{3} in the second and third equation.

$$x_{1}+4x_{2}-2\left(-2x_{1}+3x_{2}+10\right)=8 3x_{1}+2x_{2}-\left(-2x_{1}+3x_{2}+10\right)=1$$

Solve these equations for x_{2} and x_{1} respectively.

$$x_{2}=\frac{5}{2}x_{1}-14 x_{1}=\frac{11}{5}+\frac{1}{5}x_{2}$$

Substitute \frac{5}{2}x_{1}-14 for x_{2} in the equation x_{1}=\frac{11}{5}+\frac{1}{5}x_{2}.

$$x_{1}=\frac{11}{5}+\frac{1}{5}\left(\frac{5}{2}x_{1}-14\right)$$

Solve x_{1}=\frac{11}{5}+\frac{1}{5}\left(\frac{5}{2}x_{1}-14\right) for x_{1}.

$$x_{1}=-\frac{6}{5}$$

Substitute -\frac{6}{5} for x_{1} in the equation x_{2}=\frac{5}{2}x_{1}-14.

$$x_{2}=\frac{5}{2}\left(-\frac{6}{5}\right)-14$$

Calculate x_{2} from x_{2}=\frac{5}{2}\left(-\frac{6}{5}\right)-14.

$$x_{2}=-17$$

Substitute -17 for x_{2} and -\frac{6}{5} for x_{1} in the equation x_{3}=-2x_{1}+3x_{2}+10.

$$x_{3}=-2\left(-\frac{6}{5}\right)+3\left(-17\right)+10$$

Calculate x_{3} from x_{3}=-2\left(-\frac{6}{5}\right)+3\left(-17\right)+10.

$$x_{3}=-\frac{193}{5}$$

The system is now solved.

$$x_{1}=-\frac{6}{5} x_{2}=-17 x_{3}=-\frac{193}{5}$$