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Notice that 

$\dfrac1p = \dfrac{q}{pq}\\ \dfrac1q = \dfrac{p}{pq}$

If we rewrote the fractions like this, then we could add the two fractions.

$\dfrac{1}p + \dfrac1q = \dfrac{q}{pq} + \dfrac{p}{pq} = \dfrac{p + q}{pq}$

Substituting pq = 9/2,

$p + q = \dfrac92$

Subtracting q on both sides,

$p = \dfrac92 - q$

Substituting this into pq = 9/2,

$q\left(\dfrac92 - q\right) = \dfrac92\\ q^2 - \dfrac92 q + \dfrac92 = 0\\ 2q^2 - 9q + 9 = 0$

You can solve this equation by using the formula $q = \dfrac{-b\pm\sqrt{b^2- 4ac}}{2a}$, where a is the coefficient of x^2, b is the coefficient of x, and c is the constant term.