Solve {l}{x+y=1/8}{y+z=1/12}{x+y+z=1/10}

$$\left. \begin{array} { l } { x + y = \frac { 1 } { 8 } } \\ { y + z = \frac { 1 } { 12 } } \\ { x + y + z = \frac { 1 } { 10 } } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{1}{60}\approx 0.016666667$$

$$y=\frac{13}{120}\approx 0.108333333$$

$$z=-\frac{1}{40}=-0.025$$

Short Steps Using Substitution

Solve x+y=\frac{1}{8} for x.

$$x=-y+\frac{1}{8}$$

Substitute -y+\frac{1}{8} for x in the equation x+y+z=\frac{1}{10}.

$$-y+\frac{1}{8}+y+z=\frac{1}{10}$$

Solve the second equation for y and the third equation for z.

$$y=-z+\frac{1}{12} z=-\frac{1}{40}$$

Substitute -\frac{1}{40} for z in the equation y=-z+\frac{1}{12}.

$$y=-\left(-\frac{1}{40}\right)+\frac{1}{12}$$

Calculate y from y=-\left(-\frac{1}{40}\right)+\frac{1}{12}.

$$y=\frac{13}{120}$$

Substitute \frac{13}{120} for y in the equation x=-y+\frac{1}{8}.

$$x=-\frac{13}{120}+\frac{1}{8}$$

Calculate x from x=-\frac{13}{120}+\frac{1}{8}.

$$x=\frac{1}{60}$$

The system is now solved.

$$x=\frac{1}{60} y=\frac{13}{120} z=-\frac{1}{40}$$