Question

Find x in each of the following:
(i) $\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!}$

(ii) $\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!}$

(iii) $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$

Answer 1

$$\begin{gathered} \left(i\right)\frac{1}{4!}+\frac{1}{5!}=\frac{x}{6!} \\ \Rightarrow \frac{1}{4!}+\frac{1}{5(4!)}=\frac{x}{6!} \\ \Rightarrow \frac{5+1}{5(4!)}=\frac{x}{6!} \\ \Rightarrow \frac{6}{5!}=\frac{x}{6!} \\ \Rightarrow \frac{6}{5!}=\frac{x}{6\times 5!} \\ \Rightarrow x=36 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{x}{10!}=\frac{1}{8!}+\frac{1}{9!} \\ \Rightarrow \frac{x}{10!}=\frac{1}{8!}+\frac{1}{9(8!)} \\ \Rightarrow \frac{x}{10!}=\frac{9+1}{9(8!)} \\ \Rightarrow \frac{x}{10!}=\frac{10}{9!} \\ \Rightarrow \frac{x}{10\times 9!}=\frac{10}{9!} \\ \Rightarrow x=100 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!} \\ \Rightarrow \frac{1}{6!}+\frac{1}{7(6!)}=\frac{x}{8!} \\ \Rightarrow \frac{7+1}{7(6!)}=\frac{x}{8!} \\ \Rightarrow \frac{8}{7!}=\frac{x}{8!} \\ \Rightarrow \frac{8}{7!}=\frac{x}{8\times 7!} \\ \Rightarrow x=64 \end{gathered}$$