Solve {l}{text{(1)}2x_{1}+5x_{2}-x_{3}=8}{3x_{1}+4x_{2}+2x_{3}=8}{8x_{3}+5x_{2}-3x_{2}=6}

$$\left. \begin{array} { l } { \text { (1) } 2 x _ { 1 } + 5 x _ { 2 } - x _ { 3 } = 8 } \\ { 3 x _ { 1 } + 4 x _ { 2 } + 2 x _ { 3 } = 8 } \\ { 8 x _ { 3 } + 5 x _ { 2 } - 3 x _ { 2 } = 6 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=\frac{2}{5}=0.4$$

$$x_{2} = \frac{53}{35} = 1\frac{18}{35} \approx 1.514285714$$

$$x_{3}=\frac{13}{35}\approx 0.371428571$$

Short Steps Using Substitution

Solve 2x_{1}+5x_{2}-x_{3}=8 for x_{3}.

$$x_{3}=2x_{1}+5x_{2}-8$$

Substitute 2x_{1}+5x_{2}-8 for x_{3} in the second and third equation.

$$3x_{1}+4x_{2}+2\left(2x_{1}+5x_{2}-8\right)=8 8\left(2x_{1}+5x_{2}-8\right)+5x_{2}-3x_{2}=6$$

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

$$x_{2}=-\frac{1}{2}x_{1}+\frac{12}{7} x_{1}=\frac{35}{8}-\frac{21}{8}x_{2}$$

Substitute -\frac{1}{2}x_{1}+\frac{12}{7} for x_{2} in the equation x_{1}=\frac{35}{8}-\frac{21}{8}x_{2}.

$$x_{1}=\frac{35}{8}-\frac{21}{8}\left(-\frac{1}{2}x_{1}+\frac{12}{7}\right)$$

Solve x_{1}=\frac{35}{8}-\frac{21}{8}\left(-\frac{1}{2}x_{1}+\frac{12}{7}\right) for x_{1}.

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

Substitute \frac{2}{5} for x_{1} in the equation x_{2}=-\frac{1}{2}x_{1}+\frac{12}{7}.

$$x_{2}=-\frac{1}{2}\times \frac{2}{5}+\frac{12}{7}$$

Calculate x_{2} from x_{2}=-\frac{1}{2}\times \frac{2}{5}+\frac{12}{7}.

$$x_{2}=\frac{53}{35}$$

Substitute \frac{53}{35} for x_{2} and \frac{2}{5} for x_{1} in the equation x_{3}=2x_{1}+5x_{2}-8.

$$x_{3}=2\times \frac{2}{5}+5\times \frac{53}{35}-8$$

Calculate x_{3} from x_{3}=2\times \frac{2}{5}+5\times \frac{53}{35}-8.

$$x_{3}=\frac{13}{35}$$

The system is now solved.

$$x_{1}=\frac{2}{5} x_{2}=\frac{53}{35} x_{3}=\frac{13}{35}$$