Solve {c}{3x+2y+2z=-2}{2x+y-z=-2}{x-3y+z=0}

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

Solve for x, y, z

$$x=-\frac{10}{13}\approx -0.769230769$$

$$y=-\frac{2}{13}\approx -0.153846154$$

$$z=\frac{4}{13}\approx 0.307692308$$

Short Steps Using Substitution

Reorder the equations.

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

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

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

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

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

Solve these equations for x and z respectively.

$$x=4z-2 z=\frac{7}{2}x+3$$

Substitute 4z-2 for x in the equation z=\frac{7}{2}x+3.

$$z=\frac{7}{2}\left(4z-2\right)+3$$

Solve z=\frac{7}{2}\left(4z-2\right)+3 for z.

$$z=\frac{4}{13}$$

Substitute \frac{4}{13} for z in the equation x=4z-2.

$$x=4\times \frac{4}{13}-2$$

Calculate x from x=4\times \frac{4}{13}-2.

$$x=-\frac{10}{13}$$

Substitute -\frac{10}{13} for x and \frac{4}{13} for z in the equation y=-2x+z-2.

$$y=-2\left(-\frac{10}{13}\right)+\frac{4}{13}-2$$

Calculate y from y=-2\left(-\frac{10}{13}\right)+\frac{4}{13}-2.

$$y=-\frac{2}{13}$$

The system is now solved.

$$x=-\frac{10}{13} y=-\frac{2}{13} z=\frac{4}{13}$$