Solve {l}{x+2/6-y+4/3+z/2=0}{x+1/2+frac{y-1}{2}-frac{z}{4}=frac{9}{2}}{frac{x-5}{4}+frac{y+1}{3}+frac{z-2}{2}=frac{19}{4}}

$$\left. \begin{array} { l } { \frac { x + 2 } { 6 } - \frac { y + 4 } { 3 } + \frac { z } { 2 } = 0 } \\ { \frac { x + 1 } { 2 } + \frac { y - 1 } { 2 } - \frac { z } { 4 } = \frac { 9 } { 2 } } \\ { \frac { x - 5 } { 4 } + \frac { y + 1 } { 3 } + \frac { z - 2 } { 2 } = \frac { 19 } { 4 } } \end{array} \right.$$

Solve for x, y, z

$$x=4$$

$$y=8$$

$$z=6$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$-6+x-2y+3z=0 2x+2y-z=18 -23+3x+4y+6z=57$$

Solve -6+x-2y+3z=0 for x.

$$x=6+2y-3z$$

Substitute 6+2y-3z for x in the second and third equation.

$$2\left(6+2y-3z\right)+2y-z=18 -23+3\left(6+2y-3z\right)+4y+6z=57$$

Solve these equations for y and z respectively.

$$y=1+\frac{7}{6}z z=-\frac{62}{3}+\frac{10}{3}y$$

Substitute 1+\frac{7}{6}z for y in the equation z=-\frac{62}{3}+\frac{10}{3}y.

$$z=-\frac{62}{3}+\frac{10}{3}\left(1+\frac{7}{6}z\right)$$

Solve z=-\frac{62}{3}+\frac{10}{3}\left(1+\frac{7}{6}z\right) for z.

$$z=6$$

Substitute 6 for z in the equation y=1+\frac{7}{6}z.

$$y=1+\frac{7}{6}\times 6$$

Calculate y from y=1+\frac{7}{6}\times 6.

$$y=8$$

Substitute 8 for y and 6 for z in the equation x=6+2y-3z.

$$x=6+2\times 8-3\times 6$$

Calculate x from x=6+2\times 8-3\times 6.

$$x=4$$

The system is now solved.

$$x=4 y=8 z=6$$