Find x from the following:$\sqrt{1+\frac{27}{169}} = 1+ \frac{x}{13}$
Given :
The given expression is $\sqrt{1+\frac{27}{169}} = 1+ \frac{x}{13}$.
To do :
We have to find the value of x.
Solution :
$\sqrt{1+\frac{27}{169}} = 1+ \frac{x}{13}$
$\Longrightarrow \sqrt{\frac{169 +27}{169}} = 1+ \frac{x}{13}$
$\Longrightarrow \sqrt{\frac{196}{169}} = 1+ \frac{x}{13}$
$\Longrightarrow \frac{14}{13} = 1+ \frac{x}{13}$
$\Longrightarrow \frac{14}{13} -1= \frac{x}{13}$
$\Longrightarrow \frac{14-13}{13} = \frac{x}{13}$
$\Longrightarrow \frac{1}{13} = \frac{x}{13}$
Rewrite,
$\Longrightarrow \frac{x}{13} = \frac{1}{13}$
$\Longrightarrow x = 1$.
The value of x is 1.
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