Solve x-2y=10,quad2x-y-z=8,-2y+z=7

$$x - 2 y = 10 , \quad 2 x - y - z = 8 , - 2 y + z = 7$$

Solve for x, y, z

$$x=0$$

$$y=-5$$

$$z=-3$$

Short Steps Using Substitution

Solve x-2y=10 for x.

$$x=10+2y$$

Substitute 10+2y for x in the equation 2x-y-z=8.

$$2\left(10+2y\right)-y-z=8$$

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

$$y=-4+\frac{1}{3}z z=2y+7$$

Substitute -4+\frac{1}{3}z for y in the equation z=2y+7.

$$z=2\left(-4+\frac{1}{3}z\right)+7$$

Solve z=2\left(-4+\frac{1}{3}z\right)+7 for z.

$$z=-3$$

Substitute -3 for z in the equation y=-4+\frac{1}{3}z.

$$y=-4+\frac{1}{3}\left(-3\right)$$

Calculate y from y=-4+\frac{1}{3}\left(-3\right).

$$y=-5$$

Substitute -5 for y in the equation x=10+2y.

$$x=10+2\left(-5\right)$$

Calculate x from x=10+2\left(-5\right).

$$x=0$$

The system is now solved.

$$x=0 y=-5 z=-3$$