# Question #bf791

##### 1 Answer

#### Answer:

The answer is **(B)**

#### Explanation:

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

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

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

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

Now you can add these fractions by adding their numerators

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

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**

#color(blue)(a/(b/c) = a * c/b)#

In your case,

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

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

#(x+2)(x+3) = x^2 + 2x + 3x + 6 = x^2 + 5x + 6#

This means that the expression will be equal to

#((x+2)(x+3))/(2x+5) = (x^2 + 5x + 6)/(2x+5)#

Therefore, you have

#1/(1/(x+2) + 1/(x+3)) = color(green)((x^2 + 5x + 6)/(2x+5))#