Solve {c}{x+y+z=10}{3y+2z=100}{x+y=9}

$$\left. \begin{array} { c } { x + y + z = 10 } \\ { 3 y + 2 z = 100 } \\ { x + y = 9 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{71}{3} = -23\frac{2}{3} \approx -23.666666667$$

$$y = \frac{98}{3} = 32\frac{2}{3} \approx 32.666666667$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=10 for x.

$$x=-y-z+10$$

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

$$-y-z+10+y=9$$

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

$$y=\frac{100}{3}-\frac{2}{3}z z=1$$

Substitute 1 for z in the equation y=\frac{100}{3}-\frac{2}{3}z.

$$y=\frac{100}{3}-\frac{2}{3}$$

Calculate y from y=\frac{100}{3}-\frac{2}{3}.

$$y=\frac{98}{3}$$

Substitute \frac{98}{3} for y and 1 for z in the equation x=-y-z+10.

$$x=-\frac{98}{3}-1+10$$

Calculate x from x=-\frac{98}{3}-1+10.

$$x=-\frac{71}{3}$$

The system is now solved.

$$x=-\frac{71}{3} y=\frac{98}{3} z=1$$