Question

If $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$, find $x$.

Answer 1

Given, $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$

$\Rightarrow \frac{1}{6!}+\frac{1}{7\times 6!}=\frac{x}{8\times 7\times 6!}$

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

$\Rightarrow 1+\frac{1}{7}=\frac{x}{8\times 7}$

$\Rightarrow \frac{8}{7}=\frac{x}{8\times 7}$

$\Rightarrow x=\frac{8\times 8\times 7}{7}$

$\therefore x=64$