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

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

Solve for x, y, z

$$x=-2$$

$$y=8$$

$$z=6$$

Short Steps Using Substitution

Solve x+y-z=0 for x.

$$x=-y+z$$

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

$$2\left(-y+z\right)-y+2z=0 3\left(-y+z\right)+y+z=8$$

Solve these equations for y and z respectively.

$$y=\frac{4}{3}z z=2+\frac{1}{2}y$$

Substitute \frac{4}{3}z for y in the equation z=2+\frac{1}{2}y.

$$z=2+\frac{1}{2}\times \frac{4}{3}z$$

Solve z=2+\frac{1}{2}\times \frac{4}{3}z for z.

$$z=6$$

Substitute 6 for z in the equation y=\frac{4}{3}z.

$$y=\frac{4}{3}\times 6$$

Calculate y from y=\frac{4}{3}\times 6.

$$y=8$$

Substitute 8 for y and 6 for z in the equation x=-y+z.

$$x=-8+6$$

Calculate x from x=-8+6.

$$x=-2$$

The system is now solved.

$$x=-2 y=8 z=6$$