Solve {l}{x+y+z=2}{x+2y-z=6}{2x+y-z=4}

$$\left. \begin{array} { l } { x + y + z = 2 } \\ { x + 2 y - z = 6 } \\ { 2 x + y - z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{2}{5}=0.4$$

$$y = \frac{12}{5} = 2\frac{2}{5} = 2.4$$

$$z=-\frac{4}{5}=-0.8$$

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.

$$-y-z+2+2y-z=6 2\left(-y-z+2\right)+y-z=4$$

Solve these equations for y and z respectively.

$$y=4+2z z=-\frac{1}{3}y$$

Substitute 4+2z for y in the equation z=-\frac{1}{3}y.

$$z=-\frac{1}{3}\left(4+2z\right)$$

Solve z=-\frac{1}{3}\left(4+2z\right) for z.

$$z=-\frac{4}{5}$$

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

$$y=4+2\left(-\frac{4}{5}\right)$$

Calculate y from y=4+2\left(-\frac{4}{5}\right).

$$y=\frac{12}{5}$$

Substitute \frac{12}{5} for y and -\frac{4}{5} for z in the equation x=-y-z+2.

$$x=-\frac{12}{5}-\left(-\frac{4}{5}\right)+2$$

Calculate x from x=-\frac{12}{5}-\left(-\frac{4}{5}\right)+2.

$$x=\frac{2}{5}$$

The system is now solved.

$$x=\frac{2}{5} y=\frac{12}{5} z=-\frac{4}{5}$$