Solve {r}{3x+y+3z=0}{x-4y-z-10=0}{-4x+3y+4z+2=0}

$$\left. \begin{array} { r } { 3 x + y + 3 z = 0 } \\ { x - 4 y - z - 10 = 0 } \\ { - 4 x + 3 y + 4 z + 2 = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{14}{39}\approx -0.358974359$$

$$y = -\frac{38}{13} = -2\frac{12}{13} \approx -2.923076923$$

$$z = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Short Steps Using Substitution

Solve 3x+y+3z=0 for y.

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

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

$$x-4\left(-3x-3z\right)-z-10=0 -4x+3\left(-3x-3z\right)+4z+2=0$$

Solve these equations for x and z respectively.

$$x=-\frac{11}{13}z+\frac{10}{13} z=-\frac{13}{5}x+\frac{2}{5}$$

Substitute -\frac{11}{13}z+\frac{10}{13} for x in the equation z=-\frac{13}{5}x+\frac{2}{5}.

$$z=-\frac{13}{5}\left(-\frac{11}{13}z+\frac{10}{13}\right)+\frac{2}{5}$$

Solve z=-\frac{13}{5}\left(-\frac{11}{13}z+\frac{10}{13}\right)+\frac{2}{5} for z.

$$z=\frac{4}{3}$$

Substitute \frac{4}{3} for z in the equation x=-\frac{11}{13}z+\frac{10}{13}.

$$x=-\frac{11}{13}\times \frac{4}{3}+\frac{10}{13}$$

Calculate x from x=-\frac{11}{13}\times \frac{4}{3}+\frac{10}{13}.

$$x=-\frac{14}{39}$$

Substitute -\frac{14}{39} for x and \frac{4}{3} for z in the equation y=-3x-3z.

$$y=-3\left(-\frac{14}{39}\right)-3\times \frac{4}{3}$$

Calculate y from y=-3\left(-\frac{14}{39}\right)-3\times \frac{4}{3}.

$$y=-\frac{38}{13}$$

The system is now solved.

$$x=-\frac{14}{39} y=-\frac{38}{13} z=\frac{4}{3}$$