Solve {c}{3x+2y-z=8}{-3x+4y+5z=-14}{x-3y+4z=-14}

$$\left. \begin{array} { c } { 3 x + 2 y - z = 8 } \\ { - 3 x + 4 y + 5 z = - 14 } \\ { x - 3 y + 4 z = - 14 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=1$$

$$z=-3$$

Short Steps Using Substitution

Solve 3x+2y-z=8 for z.

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

Substitute 3x+2y-8 for z in the second and third equation.

$$-3x+4y+5\left(3x+2y-8\right)=-14 x-3y+4\left(3x+2y-8\right)=-14$$

Solve these equations for y and x respectively.

$$y=\frac{13}{7}-\frac{6}{7}x x=-\frac{5}{13}y+\frac{18}{13}$$

Substitute \frac{13}{7}-\frac{6}{7}x for y in the equation x=-\frac{5}{13}y+\frac{18}{13}.

$$x=-\frac{5}{13}\left(\frac{13}{7}-\frac{6}{7}x\right)+\frac{18}{13}$$

Solve x=-\frac{5}{13}\left(\frac{13}{7}-\frac{6}{7}x\right)+\frac{18}{13} for x.

$$x=1$$

Substitute 1 for x in the equation y=\frac{13}{7}-\frac{6}{7}x.

$$y=\frac{13}{7}-\frac{6}{7}$$

Calculate y from y=\frac{13}{7}-\frac{6}{7}.

$$y=1$$

Substitute 1 for y and 1 for x in the equation z=3x+2y-8.

$$z=3\times 1+2\times 1-8$$

Calculate z from z=3\times 1+2\times 1-8.

$$z=-3$$

The system is now solved.

$$x=1 y=1 z=-3$$