Solve {c}{2y+z=-8}{x-2y-3z=0}{-x+y+2z=3}

$$\left\{ \begin{array} { c } { 2 y + z = - 8 } \\ { x - 2 y - 3 z = 0 } \\ { - x + y + 2 z = 3 } \end{array} \right.$$

Solve for y, z, x

$$x=-4$$

$$y=-5$$

$$z=2$$

Short Steps Using Substitution

Solve 2y+z=-8 for z.

$$z=-8-2y$$

Substitute -8-2y for z in the second and third equation.

$$x-2y-3\left(-8-2y\right)=0 -x+y+2\left(-8-2y\right)=3$$

Solve these equations for y and x respectively.

$$y=-\frac{1}{4}x-6 x=-3y-19$$

Substitute -\frac{1}{4}x-6 for y in the equation x=-3y-19.

$$x=-3\left(-\frac{1}{4}x-6\right)-19$$

Solve x=-3\left(-\frac{1}{4}x-6\right)-19 for x.

$$x=-4$$

Substitute -4 for x in the equation y=-\frac{1}{4}x-6.

$$y=-\frac{1}{4}\left(-4\right)-6$$

Calculate y from y=-\frac{1}{4}\left(-4\right)-6.

$$y=-5$$

Substitute -5 for y in the equation z=-8-2y.

$$z=-8-2\left(-5\right)$$

Calculate z from z=-8-2\left(-5\right).

$$z=2$$

The system is now solved.

$$y=-5 z=2 x=-4$$