To remove the $\frac{3}{7}$ from the left side, we need to add $\frac{3}{7}$ to each side of the equation. $$\begin{aligned} q-\frac{3}{7}\amp=\frac{3}{2}\\ q-\frac{3}{7}\addright{\frac{3}{7}}\amp=\frac{3}{2}\addright{\frac{3}{7}}\\ q\amp=\frac{27}{14} \end{aligned}$$ We will check the solution by substituting $q$ in the original equation with $\frac{27}{14}\text{:}$ $$\begin{aligned} q-\frac{3}{7}\amp=\frac{3}{2}\\ \substitute{\frac{27}{14}}-\frac{3}{7}\amp\stackrel{?}{=}\frac{3}{2}\\ \frac{27}{14}-\frac{6}{14}\amp\stackrel{?}{=}\frac{3}{2}\\ \frac{21}{14}\amp\stackrel{?}{=}\frac{3}{2}\\ \frac{3}{2}\amp\stackrel{\checkmark}{=}\frac{3}{2} \end{aligned}$$ The solution $\frac{27}{14}$ is checked and the solution set is $\left\{\frac{27}{14}\right\}\text{.}$