$\begin{array}{l} \dfrac{1}{f} = \frac{1}{{\dfrac{{3f}}{4} + 15}} + \dfrac{1}{{120 - 3f}}\\ \Leftrightarrow \dfrac{1}{f} = \dfrac{4}{{3f + 60}} + \dfrac{1}{{120 - 3f}}\\ \Leftrightarrow \dfrac{1}{f} = \dfrac{4}{{3\left( {f + 20} \right)}} + \dfrac{1}{{3\left( {40 - f} \right)}}\\ \Leftrightarrow 3\left( {f + 20} \right)\left( {40 - f} \right) = 4f\left( {40 - f} \right) + f\left( {f + 20} \right)\\ \Leftrightarrow 60f + 2400 = 160f + 20f\\ \Leftrightarrow 120f = 2400 \Leftrightarrow f = 20 \end{array}$
