The left side is effectively the same things as $-\frac{7}{12}H\text{,}$ so multiplying by $-\frac{12}{7}$ will isolate $H\text{.}$ $$\begin{aligned} \frac{-7H}{12}\amp=\frac{2}{3}\\ -\frac{7}{12}H\amp=\frac{2}{3}\\ \multiplyleft{\left(-\frac{12}{7}\right)}\left(-\frac{7}{12}H\right)\amp=\multiplyleft{\left(-\frac{12}{7}\right)}\frac{2}{3}\\ H\amp=-\frac{4}{7}\cdot\frac{2}{1}\\ H\amp=-\frac{8}{7} \end{aligned}$$ We will check the solution by substituting $H$ in the original equation with $-\frac{8}{7}\text{:}$ $$\begin{aligned} \frac{-7H}{12}\amp=\frac{2}{3}\\ \frac{-7\left(\substitute{-\frac{8}{7}}\right)}{12}\amp\stackrel{?}{=}\frac{2}{3}\\ \frac{8}{12}\amp\stackrel{?}{=}\frac{2}{3}\\ \frac{2}{3}\amp\stackrel{\checkmark}{=}\frac{2}{3} \end{aligned}$$ The solution $-\frac{8}{7}$ is checked and the solution set is $\left\{-\frac{8}{7}\right\}\text{.}$