Solve {l}{x/3+y/2+z/4=2}{x/4+frac{y}{3}-frac{z}{2}=frac{1}{6}}{frac{x}{2}-frac{y}{4}+frac{z}{3}=frac{23}{6}}

$$\left. \begin{array} { l } { \frac { x } { 3 } + \frac { y } { 2 } + \frac { z } { 4 } = 2 } \\ { \frac { x } { 4 } + \frac { y } { 3 } - \frac { z } { 2 } = \frac { 1 } { 6 } } \\ { \frac { x } { 2 } - \frac { y } { 4 } + \frac { z } { 3 } = \frac { 23 } { 6 } } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{2322}{395} = 5\frac{347}{395} \approx 5.878481013$$

$$y=-\frac{362}{395}\approx -0.916455696$$

$$z = \frac{788}{395} = 1\frac{393}{395} \approx 1.994936709$$

Short Steps Using Substitution

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

$$4x+6y+3z=24 3x+4y-6z=2 6x-3y+4z=46$$

Solve 4x+6y+3z=24 for x.

$$x=6-\frac{3}{2}y-\frac{3}{4}z$$

Substitute 6-\frac{3}{2}y-\frac{3}{4}z for x in the second and third equation.

$$3\left(6-\frac{3}{2}y-\frac{3}{4}z\right)+4y-6z=2 6\left(6-\frac{3}{2}y-\frac{3}{4}z\right)-3y+4z=46$$

Solve these equations for y and z respectively.

$$y=32-\frac{33}{2}z z=-20-24y$$

Substitute 32-\frac{33}{2}z for y in the equation z=-20-24y.

$$z=-20-24\left(32-\frac{33}{2}z\right)$$

Solve z=-20-24\left(32-\frac{33}{2}z\right) for z.

$$z=\frac{788}{395}$$

Substitute \frac{788}{395} for z in the equation y=32-\frac{33}{2}z.

$$y=32-\frac{33}{2}\times \frac{788}{395}$$

Calculate y from y=32-\frac{33}{2}\times \frac{788}{395}.

$$y=-\frac{362}{395}$$

Substitute -\frac{362}{395} for y and \frac{788}{395} for z in the equation x=6-\frac{3}{2}y-\frac{3}{4}z.

$$x=6-\frac{3}{2}\left(-\frac{362}{395}\right)-\frac{3}{4}\times \frac{788}{395}$$

Calculate x from x=6-\frac{3}{2}\left(-\frac{362}{395}\right)-\frac{3}{4}\times \frac{788}{395}.

$$x=\frac{2322}{395}$$

The system is now solved.

$$x=\frac{2322}{395} y=-\frac{362}{395} z=\frac{788}{395}$$