Solve {l}{x+y=27}{y+z=33}{x+z=30}

$$\left\{ \begin{array} { l } { x + y = 27 } \\ { y + z = 33 } \\ { x + z = 30 } \end{array} \right.$$

Solve for x, y, z

$$x=12$$

$$y=15$$

$$z=18$$

Short Steps Using Substitution

Solve x+y=27 for x.

$$x=-y+27$$

Substitute -y+27 for x in the equation x+z=30.

$$-y+27+z=30$$

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

$$y=-z+33 z=y+3$$

Substitute -z+33 for y in the equation z=y+3.

$$z=-z+33+3$$

Solve z=-z+33+3 for z.

$$z=18$$

Substitute 18 for z in the equation y=-z+33.

$$y=-18+33$$

Calculate y from y=-18+33.

$$y=15$$

Substitute 15 for y in the equation x=-y+27.

$$x=-15+27$$

Calculate x from x=-15+27.

$$x=12$$

The system is now solved.

$$x=12 y=15 z=18$$