Solve {l}{1/8!+1/9!=x/10!}{y=8}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{1}{8 !} + \frac{1}{9 !} = \frac{x}{10 !} }\\ { y = 8 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=8$$

Solution Steps

Consider the first equation. The factorial of 8 is 40320.

$$\frac{1}{40320}+\frac{1}{9!}=\frac{x}{10!}$$

The factorial of 9 is 362880.

$$\frac{1}{40320}+\frac{1}{362880}=\frac{x}{10!}$$

Add \frac{1}{40320} and \frac{1}{362880} to get \frac{1}{36288}.

$$\frac{1}{36288}=\frac{x}{10!}$$

The factorial of 10 is 3628800.

$$\frac{1}{36288}=\frac{x}{3628800}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{x}{3628800}=\frac{1}{36288}$$

Multiply both sides by 3628800.

$$x=\frac{1}{36288}\times 3628800$$

Multiply \frac{1}{36288} and 3628800 to get 100.

$$x=100$$

The system is now solved.

$$x=100 y=8 z=8$$