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

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

Solve for x, y, z

$$x=0$$

$$y=0$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=1 for x.

$$x=-y-z+1$$

Substitute -y-z+1 for x in the second and third equation.

$$-y-z+1-y+2z=2 3\left(-y-z+1\right)+y=0$$

Solve these equations for y and z respectively.

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

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

$$z=1-\frac{2}{3}\left(-\frac{1}{2}+\frac{1}{2}z\right)$$

Solve z=1-\frac{2}{3}\left(-\frac{1}{2}+\frac{1}{2}z\right) for z.

$$z=1$$

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

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

Calculate y from y=-\frac{1}{2}+\frac{1}{2}\times 1.

$$y=0$$

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

$$x=-0-1+1$$

Calculate x from x=-0-1+1.

$$x=0$$

The system is now solved.

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