Solve {l}{(1/20+1/10+1/50)x_{1}-1/10x_{2}-frac{1}{50}x_{3}=frac{50}{20}}{-frac{1}{10}x_{1}+(frac{1}{10}+frac{1}{10}+frac{1}{40})x_{2}-frac{1}{40}x_{3}=frac{20}{10}}{-frac{1}{50}x_{1}-frac{1}{40}x_{2}+(frac{1}{5}+frac{1}{40}+frac{1}{50})x_{3}=frac{10}{5}}

$$\left\{ \begin{array} { l } { ( \frac { 1 } { 20 } + \frac { 1 } { 10 } + \frac { 1 } { 50 } ) x _ { 1 } - \frac { 1 } { 10 } x _ { 2 } - \frac { 1 } { 50 } x _ { 3 } = \frac { 50 } { 20 } } \\ { - \frac { 1 } { 10 } x _ { 1 } + ( \frac { 1 } { 10 } + \frac { 1 } { 10 } + \frac { 1 } { 40 } ) x _ { 2 } - \frac { 1 } { 40 } x _ { 3 } = \frac { 20 } { 10 } } \\ { - \frac { 1 } { 50 } x _ { 1 } - \frac { 1 } { 40 } x _ { 2 } + ( \frac { 1 } { 5 } + \frac { 1 } { 40 } + \frac { 1 } { 50 } ) x _ { 3 } = \frac { 10 } { 5 } } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{1602}{53} = 30\frac{12}{53} \approx 30.226415094$$

$$x_{2} = \frac{1260}{53} = 23\frac{41}{53} \approx 23.773584906$$

$$x_{3} = \frac{692}{53} = 13\frac{3}{53} \approx 13.056603774$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$17x_{1}-10x_{2}-2x_{3}=250 -4x_{1}+9x_{2}-x_{3}=80 -4x_{1}-5x_{2}+49x_{3}=400$$

Reorder the equations.

$$-4x_{1}+9x_{2}-x_{3}=80 17x_{1}-10x_{2}-2x_{3}=250 -4x_{1}-5x_{2}+49x_{3}=400$$

Solve -4x_{1}+9x_{2}-x_{3}=80 for x_{3}.

$$x_{3}=-4x_{1}+9x_{2}-80$$

Substitute -4x_{1}+9x_{2}-80 for x_{3} in the second and third equation.

$$17x_{1}-10x_{2}-2\left(-4x_{1}+9x_{2}-80\right)=250 -4x_{1}-5x_{2}+49\left(-4x_{1}+9x_{2}-80\right)=400$$

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

$$x_{2}=-\frac{45}{14}+\frac{25}{28}x_{1} x_{1}=-\frac{108}{5}+\frac{109}{50}x_{2}$$

Substitute -\frac{45}{14}+\frac{25}{28}x_{1} for x_{2} in the equation x_{1}=-\frac{108}{5}+\frac{109}{50}x_{2}.

$$x_{1}=-\frac{108}{5}+\frac{109}{50}\left(-\frac{45}{14}+\frac{25}{28}x_{1}\right)$$

Solve x_{1}=-\frac{108}{5}+\frac{109}{50}\left(-\frac{45}{14}+\frac{25}{28}x_{1}\right) for x_{1}.

$$x_{1}=\frac{1602}{53}$$

Substitute \frac{1602}{53} for x_{1} in the equation x_{2}=-\frac{45}{14}+\frac{25}{28}x_{1}.

$$x_{2}=-\frac{45}{14}+\frac{25}{28}\times \frac{1602}{53}$$

Calculate x_{2} from x_{2}=-\frac{45}{14}+\frac{25}{28}\times \frac{1602}{53}.

$$x_{2}=\frac{1260}{53}$$

Substitute \frac{1260}{53} for x_{2} and \frac{1602}{53} for x_{1} in the equation x_{3}=-4x_{1}+9x_{2}-80.

$$x_{3}=-4\times \frac{1602}{53}+9\times \frac{1260}{53}-80$$

Calculate x_{3} from x_{3}=-4\times \frac{1602}{53}+9\times \frac{1260}{53}-80.

$$x_{3}=\frac{692}{53}$$

The system is now solved.

$$x_{1}=\frac{1602}{53} x_{2}=\frac{1260}{53} x_{3}=\frac{692}{53}$$