Hi Mike
For question 13 test 1 no calculator, I used plugging in. I made x = 4 and solved to get 42/13. then I plugged 4 into my answer choices and B gave me 42/13.
I am curious as to why you did not use plug in for your answer and explanation.

(Link to the referenced solution.)

Good question. I definitely have this one categorized as plug-innable in my book, but because there’s some algebraic manipulation required whether you plug in or not, I guess I chose to take it as an opportunity to push that skill. Your solution (elaborated below) is also totally valid, and would definitely be my preferred method on a section where calculators were allowed.

    \begin{align*}\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}\\\\=\dfrac{1}{\dfrac{1}{4+2}+\dfrac{1}{4+3}}\\\\=\dfrac{1}{\dfrac{1}{6}+\dfrac{1}{7}}\\\\=\dfrac{1}{\dfrac{7}{42}+\dfrac{6}{42}}\\\\=\dfrac{1}{\left(\dfrac{13}{42}\right)}\\\\=\dfrac{42}{13}\end{align*}

Now set x=4 in each answer choice to see which one simplifies to \dfrac{42}{13}:

A) \dfrac{2(4)+5}{4^2+5(4)+6}=\dfrac{13}{42}

B) \dfrac{4^2+5(4)+6}{2(4)+5}=\dfrac{42}{13}

C) 2(4)+5=13

D) 4^2+5(4)+6=42