Please explain #13, test 1, Section 3.

Yeah, this is a great question, right? 🙂

You’re given this:

\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}

You’re going to have to deal with it in two parts. First, get the bottom simplified. Step 1 is to multiply by some clever forms of 1 to get a common denominator:

\dfrac{1}{\left(\dfrac{1}{x+2}\times\dfrac{x+3}{x+3}\right)+\left(\dfrac{1}{x+3}\times\dfrac{x+2}{x+2}\right)}

Now multiply those fractions on the bottom. Your common denominator will be (x+2)(x+3)=x^2+5x+6.

\dfrac{1}{\left(\dfrac{x+3}{x^2+5x+6}\right)+\left(\dfrac{x+2}{x^2+5x+6}\right)}

Now that you’ve got a common denominator, add to get one big fraction on the bottom.

\dfrac{1}{\left(\dfrac{(x+3)+(x+2)}{x^2+5x+6}\right)}

\dfrac{1}{\left(\dfrac{2x+5}{x^2+5x+6}\right)}

From there, all you need to remember is that when you divide by a fraction, you flip the divisor and multiply. In other words:

\dfrac{1}{\left(\dfrac{2x+5}{x^2+5x+6}\right)}=1\times\dfrac{x^2+5x+6}{2x+5}

 

So there you have it—the answer is B.

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