Solve {l}{1/4!+1/5!=x/6!}{y=5*6*7*8*9*10}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{1}{4 !} + \frac{1}{5 !} = \frac{x}{6 !} }\\ { y = 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=151200$$

Solution Steps

Consider the first equation. The factorial of 4 is 24.

$$\frac{1}{24}+\frac{1}{5!}=\frac{x}{6!}$$

The factorial of 5 is 120.

$$\frac{1}{24}+\frac{1}{120}=\frac{x}{6!}$$

Add \frac{1}{24} and \frac{1}{120} to get \frac{1}{20}.

$$\frac{1}{20}=\frac{x}{6!}$$

The factorial of 6 is 720.

$$\frac{1}{20}=\frac{x}{720}$$

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

$$\frac{x}{720}=\frac{1}{20}$$

Multiply both sides by 720.

$$x=\frac{1}{20}\times 720$$

Multiply \frac{1}{20} and 720 to get 36.

$$x=36$$

Consider the second equation. Multiply 5 and 6 to get 30.

$$y=30\times 7\times 8\times 9\times 10$$

Multiply 30 and 7 to get 210.

$$y=210\times 8\times 9\times 10$$

Multiply 210 and 8 to get 1680.

$$y=1680\times 9\times 10$$

Multiply 1680 and 9 to get 15120.

$$y=15120\times 10$$

Multiply 15120 and 10 to get 151200.

$$y=151200$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=151200$$

The system is now solved.

$$x=36 y=151200 z=151200$$