Solve {c}{x+y-z=-3}{3x-y+z=3}{x-2y+4z=40}

$$\left. \begin{array} { c } { x + y - z = - 3 } \\ { 3 x - y + z = 3 } \\ { x - 2 y + 4 z = 40 } \end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=14$$

$$z=17$$

Short Steps Using Substitution

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

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

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

$$3\left(-y+z-3\right)-y+z=3 -y+z-3-2y+4z=40$$

Solve these equations for y and z respectively.

$$y=z-3 z=\frac{3}{5}y+\frac{43}{5}$$

Substitute z-3 for y in the equation z=\frac{3}{5}y+\frac{43}{5}.

$$z=\frac{3}{5}\left(z-3\right)+\frac{43}{5}$$

Solve z=\frac{3}{5}\left(z-3\right)+\frac{43}{5} for z.

$$z=17$$

Substitute 17 for z in the equation y=z-3.

$$y=17-3$$

Calculate y from y=17-3.

$$y=14$$

Substitute 14 for y and 17 for z in the equation x=-y+z-3.

$$x=-14+17-3$$

Calculate x from x=-14+17-3.

$$x=0$$

The system is now solved.

$$x=0 y=14 z=17$$