# Find x in each of the following:

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

Solution:

(i) $\frac{1}{4 !}+\frac{1}{5 !}=\frac{x}{6 !}$

$\Rightarrow \frac{1}{4 !}+\frac{1}{5(4 !)}=\frac{x}{6 !}$

$\Rightarrow \frac{5+1}{5(4 !)}=\frac{x}{6 !}$

$\Rightarrow \frac{6}{5 !}=\frac{x}{6 !}$

$\Rightarrow \frac{6}{5 !}=\frac{x}{6 \times 5 !}$

$\Rightarrow x=36$

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

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

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

$\Rightarrow \frac{x}{10 !}=\frac{10}{9 !}$

$\Rightarrow \frac{x}{10 \times 9 !}=\frac{10}{9 !}$

$\Rightarrow x=100$

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

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

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

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

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

$\Rightarrow x=64$