Solve {l}{x+y+z=6}{3x-y-2z=12}{x-y-3z=-4}

$$\left\{ \begin{array} { l } { x + y + z = 6 } \\ { 3 x - y - 2 z = 12 } \\ { x - y - 3 z = - 4 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{17}{3} = 5\frac{2}{3} \approx 5.666666667$$

$$y = -\frac{13}{3} = -4\frac{1}{3} \approx -4.333333333$$

$$z = \frac{14}{3} = 4\frac{2}{3} \approx 4.666666667$$

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.

$$3\left(-y-z+6\right)-y-2z=12 -y-z+6-y-3z=-4$$

Solve these equations for y and z respectively.

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

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

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

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

$$z=\frac{14}{3}$$

Substitute \frac{14}{3} for z in the equation y=-\frac{5}{4}z+\frac{3}{2}.

$$y=-\frac{5}{4}\times \frac{14}{3}+\frac{3}{2}$$

Calculate y from y=-\frac{5}{4}\times \frac{14}{3}+\frac{3}{2}.

$$y=-\frac{13}{3}$$

Substitute -\frac{13}{3} for y and \frac{14}{3} for z in the equation x=-y-z+6.

$$x=-\left(-\frac{13}{3}\right)-\frac{14}{3}+6$$

Calculate x from x=-\left(-\frac{13}{3}\right)-\frac{14}{3}+6.

$$x=\frac{17}{3}$$

The system is now solved.

$$x=\frac{17}{3} y=-\frac{13}{3} z=\frac{14}{3}$$