Solve {l}{x_{1}=2x_{2}-2x_{3}}{-x_{1}+5/2x_{2}-1/2x_{3}=0}{-x_{1}-1/2x_{2}+5/2x_{3}=-5}

$$\left. \begin{array} { l } { x _ { 1 } = 2 x _ { 2 } - 2 x _ { 3 } } \\ { - x _ { 1 } + \frac { 5 } { 2 } x _ { 2 } - \frac { 1 } { 2 } x _ { 3 } = 0 } \\ { - x _ { 1 } - \frac { 1 } { 2 } x _ { 2 } + \frac { 5 } { 2 } x _ { 3 } = - 5 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

$$x_{2} = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x_{3}=-\frac{5}{12}\approx -0.416666667$$

Short Steps Using Substitution

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

$$-\left(2x_{2}-2x_{3}\right)+\frac{5}{2}x_{2}-\frac{1}{2}x_{3}=0 -\left(2x_{2}-2x_{3}\right)-\frac{1}{2}x_{2}+\frac{5}{2}x_{3}=-5$$

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

$$x_{2}=-3x_{3} x_{3}=-\frac{10}{9}+\frac{5}{9}x_{2}$$

Substitute -3x_{3} for x_{2} in the equation x_{3}=-\frac{10}{9}+\frac{5}{9}x_{2}.

$$x_{3}=-\frac{10}{9}+\frac{5}{9}\left(-3\right)x_{3}$$

Solve x_{3}=-\frac{10}{9}+\frac{5}{9}\left(-3\right)x_{3} for x_{3}.

$$x_{3}=-\frac{5}{12}$$

Substitute -\frac{5}{12} for x_{3} in the equation x_{2}=-3x_{3}.

$$x_{2}=-3\left(-\frac{5}{12}\right)$$

Calculate x_{2} from x_{2}=-3\left(-\frac{5}{12}\right).

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

Substitute \frac{5}{4} for x_{2} and -\frac{5}{12} for x_{3} in the equation x_{1}=2x_{2}-2x_{3}.

$$x_{1}=2\times \frac{5}{4}-2\left(-\frac{5}{12}\right)$$

Calculate x_{1} from x_{1}=2\times \frac{5}{4}-2\left(-\frac{5}{12}\right).

$$x_{1}=\frac{10}{3}$$

The system is now solved.

$$x_{1}=\frac{10}{3} x_{2}=\frac{5}{4} x_{3}=-\frac{5}{12}$$