Solve {r}{x+y+z=8}{-x+3y+4z=15}{6x-2y-z=0}

$$\left. \begin{array} { r } { x + y + z = 8 } \\ { - x + 3 y + 4 z = 15 } \\ { 6 x - 2 y - z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{25}{12} = 2\frac{1}{12} \approx 2.083333333$$

$$y = \frac{79}{12} = 6\frac{7}{12} \approx 6.583333333$$

$$z=-\frac{2}{3}\approx -0.666666667$$

Short Steps Using Substitution

Solve x+y+z=8 for x.

$$x=-y-z+8$$

Substitute -y-z+8 for x in the second and third equation.

$$-\left(-y-z+8\right)+3y+4z=15 6\left(-y-z+8\right)-2y-z=0$$

Solve these equations for y and z respectively.

$$y=-\frac{5}{4}z+\frac{23}{4} z=\frac{48}{7}-\frac{8}{7}y$$

Substitute -\frac{5}{4}z+\frac{23}{4} for y in the equation z=\frac{48}{7}-\frac{8}{7}y.

$$z=\frac{48}{7}-\frac{8}{7}\left(-\frac{5}{4}z+\frac{23}{4}\right)$$

Solve z=\frac{48}{7}-\frac{8}{7}\left(-\frac{5}{4}z+\frac{23}{4}\right) for z.

$$z=-\frac{2}{3}$$

Substitute -\frac{2}{3} for z in the equation y=-\frac{5}{4}z+\frac{23}{4}.

$$y=-\frac{5}{4}\left(-\frac{2}{3}\right)+\frac{23}{4}$$

Calculate y from y=-\frac{5}{4}\left(-\frac{2}{3}\right)+\frac{23}{4}.

$$y=\frac{79}{12}$$

Substitute \frac{79}{12} for y and -\frac{2}{3} for z in the equation x=-y-z+8.

$$x=-\frac{79}{12}-\left(-\frac{2}{3}\right)+8$$

Calculate x from x=-\frac{79}{12}-\left(-\frac{2}{3}\right)+8.

$$x=\frac{25}{12}$$

The system is now solved.

$$x=\frac{25}{12} y=\frac{79}{12} z=-\frac{2}{3}$$