Question:
If $\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$, find $x$
Solution:
$\frac{1}{6 !}+\frac{1}{7 !}=\frac{x}{8 !}$
$\Rightarrow \frac{1}{6 !}+\frac{1}{7 \times 6 !}=\frac{x}{8 \times 7 \times 6 !}$
$\Rightarrow \frac{1}{6 !}\left(1+\frac{1}{7}\right)=\frac{x}{8 \times 7 \times 6 !}$
$\Rightarrow 1+\frac{1}{7}=\frac{x}{8 \times 7}$
$\Rightarrow \frac{8}{7}=\frac{x}{8 \times 7}$
$\Rightarrow x=\frac{8 \times 8 \times 7}{7}$
$\therefore x=64$
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