Question

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

Answer 1

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

$⇒\frac{1}{6!}+\frac{1}{7\times6!}=\frac{x}{8\times7\times6!}$

$⇒\frac{1}{6!}(1+\frac{1}{7})=\frac{x}{8\times7\times6!}$

$⇒1+\frac{1}{7}=\frac{x}{8\times7}$

$⇒\frac{8}{7}=\frac{x}{8\times7}$

$⇒x=\frac{8\times8\times7}{7}$

$∴x=64$