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

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

Solve for x, y, z

$$x=\frac{1}{4}=0.25$$

$$y=-\frac{5}{8}=-0.625$$

$$z = \frac{19}{8} = 2\frac{3}{8} = 2.375$$

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.

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

Solve these equations for y and z respectively.

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

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

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

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

$$z=\frac{19}{8}$$

Substitute \frac{19}{8} for z in the equation y=z-3.

$$y=\frac{19}{8}-3$$

Calculate y from y=\frac{19}{8}-3.

$$y=-\frac{5}{8}$$

Substitute -\frac{5}{8} for y and \frac{19}{8} for z in the equation x=-y-z+2.

$$x=-\left(-\frac{5}{8}\right)-\frac{19}{8}+2$$

Calculate x from x=-\left(-\frac{5}{8}\right)-\frac{19}{8}+2.

$$x=\frac{1}{4}$$

The system is now solved.

$$x=\frac{1}{4} y=-\frac{5}{8} z=\frac{19}{8}$$