Solve {c}{x+y-z=0}{-140+20x+6z=0}{90-5y-6z=0}

$$\left. \begin{array} { c } { x + y - z = 0 } \\ { - 140 + 20 x + 6 z = 0 } \\ { 90 - 5 y - 6 z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=4$$

$$y=6$$

$$z=10$$

Short Steps Using Substitution

Solve x+y-z=0 for x.

$$x=-y+z$$

Substitute -y+z for x in the equation -140+20x+6z=0.

$$-140+20\left(-y+z\right)+6z=0$$

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

$$y=-7+\frac{13}{10}z z=15-\frac{5}{6}y$$

Substitute -7+\frac{13}{10}z for y in the equation z=15-\frac{5}{6}y.

$$z=15-\frac{5}{6}\left(-7+\frac{13}{10}z\right)$$

Solve z=15-\frac{5}{6}\left(-7+\frac{13}{10}z\right) for z.

$$z=10$$

Substitute 10 for z in the equation y=-7+\frac{13}{10}z.

$$y=-7+\frac{13}{10}\times 10$$

Calculate y from y=-7+\frac{13}{10}\times 10.

$$y=6$$

Substitute 6 for y and 10 for z in the equation x=-y+z.

$$x=-6+10$$

Calculate x from x=-6+10.

$$x=4$$

The system is now solved.

$$x=4 y=6 z=10$$