SAT original mathematics module 1 question 13

Kabadosh, rewriting the expression would be confusing, so I'll say 'reciprocal' instead of writing it. We are to find the reciprocal of [1/(x+2)] + [1/(x+3)], I'll call that 'A', so we are to find the reciprocal of 'A'.

Let's find what 'A' is:
A = [1/(x+2)] + [1/(x+3)];
Just like any other addition of fraction, we make the two fractions have common denominators.
The LCM of (x+2) and (x+3) is (x+2)(x+3).
A = [(x+3)/(x+2)(x+3)] + [(x+2)/(x+2)(x+3)];
Now that we have common denominators, we can add.
A = [(x+3)/(x+2)(x+3)] + [(x+2)/(x+2)(x+3)];
A = [(x+3) +(x+2)]/(x+2)(x+3);
A = (x +3 +x +2)/(x+2)(x+3);
A = (2x +5)/(x+2)(x+3);

Let's find (x+2)(x+3)
(x+2)(x+3) = x (x+3) +2(x+3);
(x+2)(x+3) = x(x) +3(x) +2(x) +2(+3);
(x+2)(x+3) = x² +3x +2x +6;
(x+2)(x+3) = x² +5x +6;

Therefore A = (2x +5)/(x² +5x +6)

We are to find the reciprocal of 'A', i.e. either you flip the fraction; or you make the numerator go down, and denominator come up (same thing).
The reciprocal of 'A' is (x² +5x +6)/(2x +5)

Option B

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