# Question bf791

Nov 3, 2015

The answer is (B) $\frac{{x}^{2} + 5 x + 6}{2 x + 5}$

#### Explanation:

The idea here is that you need to take your starting eexpressions

$\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$

and do some algebraic manipulations to get it to math one of the four expressions given.

Focus on the denominator of the original fraction

$\frac{1}{x + 2} + \frac{1}{x + 3}$

In order to be able to add these two fractions, you need them to have the same denominator. This means that you're going to have to multiply the first one by $1 = \frac{x + 3}{x + 3}$ and the second one by $1 = \frac{x + 2}{x + 2}$ to get their common denominator, $\left(x + 2\right) \left(x + 3\right)$

$\frac{1}{x + 2} \cdot \frac{x + 3}{x + 3} + \frac{1}{x + 3} \cdot \frac{x + 2}{x + 2}$

$\frac{x + 3}{\left(x + 2\right) \left(x + 3\right)} + \frac{x + 2}{\left(x + 2\right) \left(x + 3\right)}$

$\frac{x + 3 + x + 2}{\left(x + 2\right) \left(x + 3\right)} = \frac{2 x + 5}{\left(x + 2\right) \left(x + 3\right)}$

The original fraction can thus be written as

1/((2x+5)/((x+2)(x+3))#

Now, you know that dividing a number by a fraction is equivalent to multiplying the number by the inverse of the fraction

$\textcolor{b l u e}{\frac{a}{\frac{b}{c}} = a \cdot \frac{c}{b}}$

In your case, $a = 1$, $b = \left(2 x + 5\right)$ and $c = \left(x + 2\right) \left(x + 3\right)$. This means that you have

$\frac{1}{\frac{2 x + 5}{\left(x + 2\right) \left(x + 3\right)}} = 1 \cdot \frac{\left(x + 2\right) \left(x + 3\right)}{2 x + 5} = \frac{\left(x + 2\right) \left(x + 3\right)}{2 x + 5}$

Finally, you can expand the parantheses of the numerator to get

$\left(x + 2\right) \left(x + 3\right) = {x}^{2} + 2 x + 3 x + 6 = {x}^{2} + 5 x + 6$

This means that the expression will be equal to

$\frac{\left(x + 2\right) \left(x + 3\right)}{2 x + 5} = \frac{{x}^{2} + 5 x + 6}{2 x + 5}$

Therefore, you have

$\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}} = \textcolor{g r e e n}{\frac{{x}^{2} + 5 x + 6}{2 x + 5}}$