d) | 2x - $\dfrac{1}{5}$| + $\dfrac{1}{3}$ = $\dfrac{1}{2}$
⇒ | 2x - $\dfrac{1}{5}$| = $\dfrac{1}{2}$ - $\dfrac{1}{3}$
⇒ | 2x - $\dfrac{1}{5}$| = $\dfrac{3}{6}$ - $\dfrac{2}{6}$
⇒ | 2x - $\dfrac{1}{5}$| = $\dfrac{1}{6}$
⇒ $\left[\begin{matrix} 2x - \dfrac{1}{5} = \dfrac{1}{6}\\2x - \dfrac{1}{5} = -\dfrac{1}{6}\end{matrix}\right.$
⇒ $\left[\begin{matrix} 2x = \dfrac{1}{6} + \dfrac{1}{5}\\2x = -\dfrac{1}{6} + \dfrac{1}{5}\end{matrix}\right.$
⇒ $\left[\begin{matrix} 2x = \dfrac{11}{30}\\2x = \dfrac{1}{30}\end{matrix}\right.$
⇒ $\left[\begin{matrix} x = \dfrac{11}{30} ÷ 2\\2x = \dfrac{1}{30} ÷ 2\end{matrix}\right.$
⇒ $\left[\begin{matrix} x = \dfrac{11}{60}\\x = -\dfrac{1}{60}\end{matrix}\right.$
