Solve {l}{x_{1}+x_{2}-x_{3}=-10}{0.4x_{2}-0.8x_{1}=-4}{4x_{3}+0.4x_{2}=116}

$$\left. \begin{array} { l } { x _ { 1 } + x _ { 2 } - x _ { 3 } = - 10 } \\ { 0.4 x _ { 2 } - 0.8 x _ { 1 } = - 4 } \\ { 4 x _ { 3 } + 0.4 x _ { 2 } = 116 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=9.375$$

$$x_{2}=8.75$$

$$x_{3}=28.125$$

Short Steps Using Substitution

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

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

Substitute -x_{2}+x_{3}-10 for x_{1} in the equation 0.4x_{2}-0.8x_{1}=-4.

$$0.4x_{2}-0.8\left(-x_{2}+x_{3}-10\right)=-4$$

Solve the second equation for x_{2} and the third equation for x_{3}.

$$x_{2}=-10+\frac{2}{3}x_{3} x_{3}=-\frac{1}{10}x_{2}+29$$

Substitute -10+\frac{2}{3}x_{3} for x_{2} in the equation x_{3}=-\frac{1}{10}x_{2}+29.

$$x_{3}=-\frac{1}{10}\left(-10+\frac{2}{3}x_{3}\right)+29$$

Solve x_{3}=-\frac{1}{10}\left(-10+\frac{2}{3}x_{3}\right)+29 for x_{3}.

$$x_{3}=28.125$$

Substitute 28.125 for x_{3} in the equation x_{2}=-10+\frac{2}{3}x_{3}.

$$x_{2}=-10+\frac{2}{3}\times 28.125$$

Calculate x_{2} from x_{2}=-10+\frac{2}{3}\times 28.125.

$$x_{2}=8.75$$

Substitute 8.75 for x_{2} and 28.125 for x_{3} in the equation x_{1}=-x_{2}+x_{3}-10.

$$x_{1}=-8.75+28.125-10$$

Calculate x_{1} from x_{1}=-8.75+28.125-10.

$$x_{1}=9.375$$

The system is now solved.

$$x_{1}=9.375 x_{2}=8.75 x_{3}=28.125$$