Solve {l}{50x-40y=10}{-40x+110y-50z=20}{-50y+80z=-30}

$$\left. \begin{array}{l}{ 50 x - 40 y = 10 }\\{ - 40 x + 110 y - 50 z = 20 }\\{ - 50 y + 80 z = - 30 }\end{array} \right.$$

Solve for x, y, z

$$x=\frac{67}{187}\approx 0.35828877$$

$$y=\frac{37}{187}\approx 0.197860963$$

$$z=-\frac{47}{187}\approx -0.251336898$$

Short Steps Using Substitution

Solve 50x-40y=10 for x.

$$x=\frac{1}{5}+\frac{4}{5}y$$

Substitute \frac{1}{5}+\frac{4}{5}y for x in the equation -40x+110y-50z=20.

$$-40\left(\frac{1}{5}+\frac{4}{5}y\right)+110y-50z=20$$

Solve the second equation for y and the third equation for z.

$$y=\frac{14}{39}+\frac{25}{39}z z=-\frac{3}{8}+\frac{5}{8}y$$

Substitute \frac{14}{39}+\frac{25}{39}z for y in the equation z=-\frac{3}{8}+\frac{5}{8}y.

$$z=-\frac{3}{8}+\frac{5}{8}\left(\frac{14}{39}+\frac{25}{39}z\right)$$

Solve z=-\frac{3}{8}+\frac{5}{8}\left(\frac{14}{39}+\frac{25}{39}z\right) for z.

$$z=-\frac{47}{187}$$

Substitute -\frac{47}{187} for z in the equation y=\frac{14}{39}+\frac{25}{39}z.

$$y=\frac{14}{39}+\frac{25}{39}\left(-\frac{47}{187}\right)$$

Calculate y from y=\frac{14}{39}+\frac{25}{39}\left(-\frac{47}{187}\right).

$$y=\frac{37}{187}$$

Substitute \frac{37}{187} for y in the equation x=\frac{1}{5}+\frac{4}{5}y.

$$x=\frac{1}{5}+\frac{4}{5}\times \frac{37}{187}$$

Calculate x from x=\frac{1}{5}+\frac{4}{5}\times \frac{37}{187}.

$$x=\frac{67}{187}$$

The system is now solved.

$$x=\frac{67}{187} y=\frac{37}{187} z=-\frac{47}{187}$$