Solve {r}{x+y+z=142}{y=4z}{z=8x}

$$\left. \begin{array} { r } { x + y + z = 142 } \\ { y = 4 z } \\ { z = 8 x } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{142}{41} = 3\frac{19}{41} \approx 3.463414634$$

$$y = \frac{4544}{41} = 110\frac{34}{41} \approx 110.829268293$$

$$z = \frac{1136}{41} = 27\frac{29}{41} \approx 27.707317073$$

Short Steps Using Substitution

Reorder the equations.

$$y=4z x+y+z=142 z=8x$$

Substitute 4z for y in the equation x+y+z=142.

$$x+4z+z=142$$

Solve x+4z+z=142 for x.

$$x=-5z+142$$

Substitute -5z+142 for x in the equation z=8x.

$$z=8\left(-5z+142\right)$$

Solve z=8\left(-5z+142\right) for z.

$$z=\frac{1136}{41}$$

Substitute \frac{1136}{41} for z in the equation x=-5z+142.

$$x=-5\times \frac{1136}{41}+142$$

Calculate x from x=-5\times \frac{1136}{41}+142.

$$x=\frac{142}{41}$$

Substitute \frac{1136}{41} for z in the equation y=4z.

$$y=4\times \frac{1136}{41}$$

Calculate y from y=4\times \frac{1136}{41}.

$$y=\frac{4544}{41}$$

The system is now solved.

$$x=\frac{142}{41} y=\frac{4544}{41} z=\frac{1136}{41}$$