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

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

Solve for x_1, x_3, x_2

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

$$x_{3}=-\frac{3}{5}=-0.6$$

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

Short Steps Using Substitution

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

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

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

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

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

$$x_{1}=\frac{2}{3}+\frac{2}{3}x_{2} x_{2}=8-6x_{1}$$

Substitute \frac{2}{3}+\frac{2}{3}x_{2} for x_{1} in the equation x_{2}=8-6x_{1}.

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

Solve x_{2}=8-6\left(\frac{2}{3}+\frac{2}{3}x_{2}\right) for x_{2}.

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

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

$$x_{1}=\frac{2}{3}+\frac{2}{3}\times \frac{4}{5}$$

Calculate x_{1} from x_{1}=\frac{2}{3}+\frac{2}{3}\times \frac{4}{5}.

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

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

$$x_{3}=2\times \frac{6}{5}-3$$

Calculate x_{3} from x_{3}=2\times \frac{6}{5}-3.

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

The system is now solved.

$$x_{1}=\frac{6}{5} x_{3}=-\frac{3}{5} x_{2}=\frac{4}{5}$$