Solve {l}{10x=15y+40z=}{-30y=40z}{10x+15y+40z=300}

$$\left. \begin{array} { l } { 10 x = 15 y + 40 z = } \\ { - 30 y = 40 z } \\ { 10 x + 15 y + 40 z = 300 } \end{array} \right.$$

Solve for x, y, z

$$x=15$$

$$y=-10$$

$$z = \frac{15}{2} = 7\frac{1}{2} = 7.5$$

Short Steps Using Substitution

Solve 10x=15y+40z for x.

$$x=\frac{3}{2}y+4z$$

Substitute \frac{3}{2}y+4z for x in the equation 10x+15y+40z=300.

$$10\left(\frac{3}{2}y+4z\right)+15y+40z=300$$

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

$$y=-\frac{4}{3}z z=\frac{15}{4}-\frac{3}{8}y$$

Substitute -\frac{4}{3}z for y in the equation z=\frac{15}{4}-\frac{3}{8}y.

$$z=\frac{15}{4}-\frac{3}{8}\left(-\frac{4}{3}\right)z$$

Solve z=\frac{15}{4}-\frac{3}{8}\left(-\frac{4}{3}\right)z for z.

$$z=\frac{15}{2}$$

Substitute \frac{15}{2} for z in the equation y=-\frac{4}{3}z.

$$y=-\frac{4}{3}\times \frac{15}{2}$$

Calculate y from y=-\frac{4}{3}\times \frac{15}{2}.

$$y=-10$$

Substitute -10 for y and \frac{15}{2} for z in the equation x=\frac{3}{2}y+4z.

$$x=\frac{3}{2}\left(-10\right)+4\times \frac{15}{2}$$

Calculate x from x=\frac{3}{2}\left(-10\right)+4\times \frac{15}{2}.

$$x=15$$

The system is now solved.

$$x=15 y=-10 z=\frac{15}{2}$$