Solve {l}{x+y-z=6}{x-3y+2z=1}{3x+2y-z=4}

$$\left\{ \begin{array} { l } { x + y - z = 6 } \\ { x - 3 y + 2 z = 1 } \\ { 3 x + 2 y - z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{11}{5} = 2\frac{1}{5} = 2.2$$

$$y = -\frac{32}{5} = -6\frac{2}{5} = -6.4$$

$$z = -\frac{51}{5} = -10\frac{1}{5} = -10.2$$

Short Steps Using Substitution

Solve x+y-z=6 for x.

$$x=-y+z+6$$

Substitute -y+z+6 for x in the second and third equation.

$$-y+z+6-3y+2z=1 3\left(-y+z+6\right)+2y-z=4$$

Solve these equations for y and z respectively.

$$y=\frac{3}{4}z+\frac{5}{4} z=\frac{1}{2}y-7$$

Substitute \frac{3}{4}z+\frac{5}{4} for y in the equation z=\frac{1}{2}y-7.

$$z=\frac{1}{2}\left(\frac{3}{4}z+\frac{5}{4}\right)-7$$

Solve z=\frac{1}{2}\left(\frac{3}{4}z+\frac{5}{4}\right)-7 for z.

$$z=-\frac{51}{5}$$

Substitute -\frac{51}{5} for z in the equation y=\frac{3}{4}z+\frac{5}{4}.

$$y=\frac{3}{4}\left(-\frac{51}{5}\right)+\frac{5}{4}$$

Calculate y from y=\frac{3}{4}\left(-\frac{51}{5}\right)+\frac{5}{4}.

$$y=-\frac{32}{5}$$

Substitute -\frac{32}{5} for y and -\frac{51}{5} for z in the equation x=-y+z+6.

$$x=-\left(-\frac{32}{5}\right)-\frac{51}{5}+6$$

Calculate x from x=-\left(-\frac{32}{5}\right)-\frac{51}{5}+6.

$$x=\frac{11}{5}$$

The system is now solved.

$$x=\frac{11}{5} y=-\frac{32}{5} z=-\frac{51}{5}$$