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

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

Solve for x, y, z

$$x=0$$

$$y=0$$

$$z=0$$

Short Steps Using Substitution

Solve 2x-3y+z=0 for z.

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

Substitute -2x+3y for z in the second and third equation.

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

Solve these equations for y and x respectively.

$$y=\frac{5}{7}x x=\frac{7}{2}y$$

Substitute \frac{5}{7}x for y in the equation x=\frac{7}{2}y.

$$x=\frac{7}{2}\times \frac{5}{7}x$$

Solve x=\frac{7}{2}\times \frac{5}{7}x for x.

$$x=0$$

Substitute 0 for x in the equation y=\frac{5}{7}x.

$$y=\frac{5}{7}\times 0$$

Calculate y from y=\frac{5}{7}\times 0.

$$y=0$$

Substitute 0 for y and 0 for x in the equation z=-2x+3y.

$$z=-2\times 0+3\times 0$$

Calculate z from z=-2\times 0+3\times 0.

$$z=0$$

The system is now solved.

$$x=0 y=0 z=0$$