Solve {l}{2x+3y-8z=2}{4x+3y-2z=-3}{x+y+4z=0}

$$\left\{ \begin{array} { l } { 2 x + 3 y - 8 z = 2 } \\ { 4 x + 3 y - 2 z = - 3 } \\ { x + y + 4 z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{44}{17} = -2\frac{10}{17} \approx -2.588235294$$

$$y = \frac{42}{17} = 2\frac{8}{17} \approx 2.470588235$$

$$z=\frac{1}{34}\approx 0.029411765$$

Short Steps Using Substitution

Reorder the equations.

$$x+y+4z=0 4x+3y-2z=-3 2x+3y-8z=2$$

Solve x+y+4z=0 for x.

$$x=-y-4z$$

Substitute -y-4z for x in the second and third equation.

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

Solve these equations for y and z respectively.

$$y=-18z+3 z=-\frac{1}{8}+\frac{1}{16}y$$

Substitute -18z+3 for y in the equation z=-\frac{1}{8}+\frac{1}{16}y.

$$z=-\frac{1}{8}+\frac{1}{16}\left(-18z+3\right)$$

Solve z=-\frac{1}{8}+\frac{1}{16}\left(-18z+3\right) for z.

$$z=\frac{1}{34}$$

Substitute \frac{1}{34} for z in the equation y=-18z+3.

$$y=-18\times \frac{1}{34}+3$$

Calculate y from y=-18\times \frac{1}{34}+3.

$$y=\frac{42}{17}$$

Substitute \frac{42}{17} for y and \frac{1}{34} for z in the equation x=-y-4z.

$$x=-\frac{42}{17}-4\times \frac{1}{34}$$

Calculate x from x=-\frac{42}{17}-4\times \frac{1}{34}.

$$x=-\frac{44}{17}$$

The system is now solved.

$$x=-\frac{44}{17} y=\frac{42}{17} z=\frac{1}{34}$$