Solve {l}{2x+3y-z=2}{x-y+3z=-4}{3x+5y+z=4}

$$\left\{ \begin{array} { l } { 2 x + 3 y - z = 2 } \\ { x - y + 3 z = - 4 } \\ { 3 x + 5 y + z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x=-2$$

$$y=2$$

$$z=0$$

Short Steps Using Substitution

Solve 2x+3y-z=2 for z.

$$z=2x+3y-2$$

Substitute 2x+3y-2 for z in the second and third equation.

$$x-y+3\left(2x+3y-2\right)=-4 3x+5y+2x+3y-2=4$$

Solve these equations for y and x respectively.

$$y=-\frac{7}{8}x+\frac{1}{4} x=\frac{6}{5}-\frac{8}{5}y$$

Substitute -\frac{7}{8}x+\frac{1}{4} for y in the equation x=\frac{6}{5}-\frac{8}{5}y.

$$x=\frac{6}{5}-\frac{8}{5}\left(-\frac{7}{8}x+\frac{1}{4}\right)$$

Solve x=\frac{6}{5}-\frac{8}{5}\left(-\frac{7}{8}x+\frac{1}{4}\right) for x.

$$x=-2$$

Substitute -2 for x in the equation y=-\frac{7}{8}x+\frac{1}{4}.

$$y=-\frac{7}{8}\left(-2\right)+\frac{1}{4}$$

Calculate y from y=-\frac{7}{8}\left(-2\right)+\frac{1}{4}.

$$y=2$$

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

$$z=2\left(-2\right)+3\times 2-2$$

Calculate z from z=2\left(-2\right)+3\times 2-2.

$$z=0$$

The system is now solved.

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