Solve {l}{x+y+z=17}{z+y-x=3}{99z-99x=198}

$$\left\{ \begin{array} { l } { x + y + z = 17 } \\ { z + y - x = 3 } \\ { 99 z - 99 x = 198 } \end{array} \right.$$

Solve for x, y, z

$$x=7$$

$$y=1$$

$$z=9$$

Short Steps Using Substitution

Solve x+y+z=17 for x.

$$x=-y-z+17$$

Substitute -y-z+17 for x in the second and third equation.

$$z+y-\left(-y-z+17\right)=3 99z-99\left(-y-z+17\right)=198$$

Solve these equations for y and z respectively.

$$y=10-z z=\frac{19}{2}-\frac{1}{2}y$$

Substitute 10-z for y in the equation z=\frac{19}{2}-\frac{1}{2}y.

$$z=\frac{19}{2}-\frac{1}{2}\left(10-z\right)$$

Solve z=\frac{19}{2}-\frac{1}{2}\left(10-z\right) for z.

$$z=9$$

Substitute 9 for z in the equation y=10-z.

$$y=10-9$$

Calculate y from y=10-9.

$$y=1$$

Substitute 1 for y and 9 for z in the equation x=-y-z+17.

$$x=-1-9+17$$

Calculate x from x=-1-9+17.

$$x=7$$

The system is now solved.

$$x=7 y=1 z=9$$