Intermediate Algebra for College Students (7th Edition)

$30$ miles
distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$ We are given:$\displaystyle \quad t_{2}=t_{1}+\frac{1}{6}$ (in hours), where the indices 1 and 2 represent boats 1 and 2. The time $t_{1}$ is shorter for the faster boat (boat 1) Let d be the distance (length of the course). $\displaystyle \frac{d}{18}=\frac{d}{20}+\frac{1}{6}\qquad$.../$\times$LCD= $180$ $10d=9d+30\qquad.../-9d$ $d=30$ miles