Solve {c}{x+y+z=2}{3x-2y-z=4}{-2x+y+2z=2}

$$\left\{ \begin{array} { c } { x + y + z = 2 } \\ { 3 x - 2 y - z = 4 } \\ { - 2 x + y + 2 z = 2 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=-2$$

$$z=3$$

Short Steps Using Substitution

Solve x+y+z=2 for x.

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

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

$$3\left(-y-z+2\right)-2y-z=4 -2\left(-y-z+2\right)+y+2z=2$$

Solve these equations for y and z respectively.

$$y=\frac{2}{5}-\frac{4}{5}z z=\frac{3}{2}-\frac{3}{4}y$$

Substitute \frac{2}{5}-\frac{4}{5}z for y in the equation z=\frac{3}{2}-\frac{3}{4}y.

$$z=\frac{3}{2}-\frac{3}{4}\left(\frac{2}{5}-\frac{4}{5}z\right)$$

Solve z=\frac{3}{2}-\frac{3}{4}\left(\frac{2}{5}-\frac{4}{5}z\right) for z.

$$z=3$$

Substitute 3 for z in the equation y=\frac{2}{5}-\frac{4}{5}z.

$$y=\frac{2}{5}-\frac{4}{5}\times 3$$

Calculate y from y=\frac{2}{5}-\frac{4}{5}\times 3.

$$y=-2$$

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

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

Calculate x from x=-\left(-2\right)-3+2.

$$x=1$$

The system is now solved.

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