Solve {l}{2x+y+z=0}{x+2y-z=-6}{x+50+2z=0}

$$\left. \begin{array} { l } { 2 x + y + z = 0 } \\ { x + 2 y - z = - 6 } \\ { x + 50 + 2 z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=54$$

$$y=-56$$

$$z=-52$$

Short Steps Using Substitution

Solve 2x+y+z=0 for y.

$$y=-2x-z$$

Substitute -2x-z for y in the equation x+2y-z=-6.

$$x+2\left(-2x-z\right)-z=-6$$

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

$$x=2-z z=-\frac{1}{2}x-25$$

Substitute 2-z for x in the equation z=-\frac{1}{2}x-25.

$$z=-\frac{1}{2}\left(2-z\right)-25$$

Solve z=-\frac{1}{2}\left(2-z\right)-25 for z.

$$z=-52$$

Substitute -52 for z in the equation x=2-z.

$$x=2-\left(-52\right)$$

Calculate x from x=2-\left(-52\right).

$$x=54$$

Substitute 54 for x and -52 for z in the equation y=-2x-z.

$$y=-2\times 54-\left(-52\right)$$

Calculate y from y=-2\times 54-\left(-52\right).

$$y=-56$$

The system is now solved.

$$x=54 y=-56 z=-52$$