Solve {l}{(y+z+2x=33}{x-y+2/5=z+4}{y-z+4/2=x-6}{z-x-7/3=y-5}

$$\left. \begin{array} { l } { ( y + z + 2 x = 33 } \\ { x - \frac { y + 2 } { 5 } = z + 4 } \\ { y - \frac { z + 4 } { 2 } = x - 6 } \\ { z - \frac { x - 7 } { 3 } = y - 5 } \end{array} \right.$$

Solve for x, y, z

$$x=10$$

$$y=8$$

$$z=4$$

Short Steps Using Substitution

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

$$-2+5x-y=5z+20 -4+2y-z=2x-12 7+3z-x=3y-15$$

Solve -2+5x-y=5z+20 for y.

$$y=-22+5x-5z$$

Substitute -22+5x-5z for y in the second and third equation.

$$-4+2\left(-22+5x-5z\right)-z=2x-12 7+3z-x=3\left(-22+5x-5z\right)-15$$

Solve these equations for x and z respectively.

$$x=\frac{9}{2}+\frac{11}{8}z z=-\frac{44}{9}+\frac{8}{9}x$$

Substitute \frac{9}{2}+\frac{11}{8}z for x in the equation z=-\frac{44}{9}+\frac{8}{9}x.

$$z=-\frac{44}{9}+\frac{8}{9}\left(\frac{9}{2}+\frac{11}{8}z\right)$$

Solve z=-\frac{44}{9}+\frac{8}{9}\left(\frac{9}{2}+\frac{11}{8}z\right) for z.

$$z=4$$

Substitute 4 for z in the equation x=\frac{9}{2}+\frac{11}{8}z.

$$x=\frac{9}{2}+\frac{11}{8}\times 4$$

Calculate x from x=\frac{9}{2}+\frac{11}{8}\times 4.

$$x=10$$

Substitute 10 for x and 4 for z in the equation y=-22+5x-5z.

$$y=-22+5\times 10-5\times 4$$

Calculate y from y=-22+5\times 10-5\times 4.

$$y=8$$

The system is now solved.

$$x=10 y=8 z=4$$