Solve {l}{2x-y+z=0}{x-2y-z=3}{3x+y+2z=1}

$$\left\{ \begin{array} { l } { 2 x - y + z = 0 } \\ { x - 2 y - z = 3 } \\ { 3 x + y + 2 z = 1 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=1$$

$$z=-3$$

Short Steps Using Substitution

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

$$y=2x+z$$

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

$$x-2\left(2x+z\right)-z=3 3x+2x+z+2z=1$$

Solve these equations for x and z respectively.

$$x=-1-z z=-\frac{5}{3}x+\frac{1}{3}$$

Substitute -1-z for x in the equation z=-\frac{5}{3}x+\frac{1}{3}.

$$z=-\frac{5}{3}\left(-1-z\right)+\frac{1}{3}$$

Solve z=-\frac{5}{3}\left(-1-z\right)+\frac{1}{3} for z.

$$z=-3$$

Substitute -3 for z in the equation x=-1-z.

$$x=-1-\left(-3\right)$$

Calculate x from x=-1-\left(-3\right).

$$x=2$$

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

$$y=2\times 2-3$$

Calculate y from y=2\times 2-3.

$$y=1$$

The system is now solved.

$$x=2 y=1 z=-3$$