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

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

Solve for x, y, z

$$x=1$$

$$y=0$$

$$z=-1$$

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.

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

Solve these equations for y and z respectively.

$$y=1+z z=-\frac{1}{2}y-1$$

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

$$z=-\frac{1}{2}\left(1+z\right)-1$$

Solve z=-\frac{1}{2}\left(1+z\right)-1 for z.

$$z=-1$$

Substitute -1 for z in the equation y=1+z.

$$y=1-1$$

Calculate y from y=1-1.

$$y=0$$

Substitute 0 for y and -1 for z in the equation x=-y-z.

$$x=-0-\left(-1\right)$$

Calculate x from x=-0-\left(-1\right).

$$x=1$$

The system is now solved.

$$x=1 y=0 z=-1$$