Solve {c}{x-y-z=0}{-150+10x+10y=0}{-120+10x+4z=0}

$$\left. \begin{array} { c } { x - y - z = 0 } \\ { - 150 + 10 x + 10 y = 0 } \\ { - 120 + 10 x + 4 z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=10$$

$$y=5$$

$$z=5$$

Short Steps Using Substitution

Solve x-y-z=0 for x.

$$x=y+z$$

Substitute y+z for x in the second and third equation.

$$-150+10\left(y+z\right)+10y=0 -120+10\left(y+z\right)+4z=0$$

Solve these equations for y and z respectively.

$$y=\frac{15}{2}-\frac{1}{2}z z=\frac{60}{7}-\frac{5}{7}y$$

Substitute \frac{15}{2}-\frac{1}{2}z for y in the equation z=\frac{60}{7}-\frac{5}{7}y.

$$z=\frac{60}{7}-\frac{5}{7}\left(\frac{15}{2}-\frac{1}{2}z\right)$$

Solve z=\frac{60}{7}-\frac{5}{7}\left(\frac{15}{2}-\frac{1}{2}z\right) for z.

$$z=5$$

Substitute 5 for z in the equation y=\frac{15}{2}-\frac{1}{2}z.

$$y=\frac{15}{2}-\frac{1}{2}\times 5$$

Calculate y from y=\frac{15}{2}-\frac{1}{2}\times 5.

$$y=5$$

Substitute 5 for y and 5 for z in the equation x=y+z.

$$x=5+5$$

Calculate x from x=5+5.

$$x=10$$

The system is now solved.

$$x=10 y=5 z=5$$