## Intermediate Algebra for College Students (7th Edition)

$12$ miles
distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$ We are given:$\quad t_{2}=t_{1}+0.5$ (in hours), where the indices 1 and 2 represent runners 1 and 2. The time $t_{1}$ is shorter for the faster runner (runner 1) Let d be the distance (length of the trail). $\displaystyle \frac{d}{6}=\frac{d}{8}+\frac{1}{2}\qquad$.../$\times$LCD= 24 $4d=3d+12\qquad.../-3d$ $d=12$ miles