Solve {r}{x+y-z=-1}{2x-y+z=4}{x-4y+3z=-8}

$$\left. \begin{array} { r } { x + y - z = - 1 } \\ { 2 x - y + z = 4 } \\ { x - 4 y + 3 z = - 8 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=15$$

$$z=17$$

Short Steps Using Substitution

Solve x+y-z=-1 for x.

$$x=-y+z-1$$

Substitute -y+z-1 for x in the second and third equation.

$$2\left(-y+z-1\right)-y+z=4 -y+z-1-4y+3z=-8$$

Solve these equations for y and z respectively.

$$y=z-2 z=\frac{5}{4}y-\frac{7}{4}$$

Substitute z-2 for y in the equation z=\frac{5}{4}y-\frac{7}{4}.

$$z=\frac{5}{4}\left(z-2\right)-\frac{7}{4}$$

Solve z=\frac{5}{4}\left(z-2\right)-\frac{7}{4} for z.

$$z=17$$

Substitute 17 for z in the equation y=z-2.

$$y=17-2$$

Calculate y from y=17-2.

$$y=15$$

Substitute 15 for y and 17 for z in the equation x=-y+z-1.

$$x=-15+17-1$$

Calculate x from x=-15+17-1.

$$x=1$$

The system is now solved.

$$x=1 y=15 z=17$$