Solve {l}{x-2y+z=-4}{y+2z=-4}{x+y+3z=-8}

$$\left. \begin{array} { l } { x - 2 y + z = - 4 } \\ { y + 2 z = - 4 } \\ { x + y + 3 z = - 8 } \end{array} \right.$$

Solve for x, y, z

$$x=-2$$

$$y=0$$

$$z=-2$$

Short Steps Using Substitution

Solve x-2y+z=-4 for x.

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

Substitute 2y-z-4 for x in the equation x+y+3z=-8.

$$2y-z-4+y+3z=-8$$

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

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

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

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

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

$$z=-2$$

Substitute -2 for z in the equation y=-4-2z.

$$y=-4-2\left(-2\right)$$

Calculate y from y=-4-2\left(-2\right).

$$y=0$$

Substitute 0 for y and -2 for z in the equation x=2y-z-4.

$$x=2\times 0-\left(-2\right)-4$$

Calculate x from x=2\times 0-\left(-2\right)-4.

$$x=-2$$

The system is now solved.

$$x=-2 y=0 z=-2$$