Solve {l}{20+y-z=0}{2x-y+3z=2}{x-4y-2z+6=0}

$$\left. \begin{array} { l } { 20 + y - z = 0 } \\ { 2 x - y + 3 z = 2 } \\ { x - 4 y - 2 z + 6 = 0 } \end{array} \right.$$

Solve for y, z, x

$$x=-20$$

$$y=-9$$

$$z=11$$

Short Steps Using Substitution

Solve 20+y-z=0 for y.

$$y=-20+z$$

Substitute -20+z for y in the second and third equation.

$$2x-\left(-20+z\right)+3z=2 x-4\left(-20+z\right)-2z+6=0$$

Solve these equations for z and x respectively.

$$z=-9-x x=-86+6z$$

Substitute -9-x for z in the equation x=-86+6z.

$$x=-86+6\left(-9-x\right)$$

Solve x=-86+6\left(-9-x\right) for x.

$$x=-20$$

Substitute -20 for x in the equation z=-9-x.

$$z=-9-\left(-20\right)$$

Calculate z from z=-9-\left(-20\right).

$$z=11$$

Substitute 11 for z in the equation y=-20+z.

$$y=-20+11$$

Calculate y from y=-20+11.

$$y=-9$$

The system is now solved.

$$y=-9 z=11 x=-20$$