Solve {l}{2x+3y-4z=12}{x+3y+z=6}{-x-2y+3z=-8}

$$\left\{ \begin{array} { l } { 2 x + 3 y - 4 z = 12 } \\ { x + 3 y + z = 6 } \\ { - x - 2 y + 3 z = - 8 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=2$$

$$z=-1$$

Short Steps Using Substitution

Reorder the equations.

$$x+3y+z=6 2x+3y-4z=12 -x-2y+3z=-8$$

Solve x+3y+z=6 for x.

$$x=-3y-z+6$$

Substitute -3y-z+6 for x in the second and third equation.

$$2\left(-3y-z+6\right)+3y-4z=12 -\left(-3y-z+6\right)-2y+3z=-8$$

Solve these equations for y and z respectively.

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

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

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

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

$$z=-1$$

Substitute -1 for z in the equation y=-2z.

$$y=-2\left(-1\right)$$

Calculate y from y=-2\left(-1\right).

$$y=2$$

Substitute 2 for y and -1 for z in the equation x=-3y-z+6.

$$x=-3\times 2-\left(-1\right)+6$$

Calculate x from x=-3\times 2-\left(-1\right)+6.

$$x=1$$

The system is now solved.

$$x=1 y=2 z=-1$$