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

Car: $50$ mph Truck: $30$ mph.
distance = (rate)$\times$(time) $\Rightarrow$ time = $\displaystyle \frac{\text{distance}}{\text{rate}}$ Let x be the rate of the car. The rate of the truck is $x-20.$ We are given: time = $\displaystyle \frac{300}{\text{rate of car}}$ = $\displaystyle \frac{180}{\text{rate of truck}}$ That is, $\displaystyle \frac{300}{x}=\frac{180}{x-20} \qquad$ ... $/\times x(x-20)$ $300(x-20)=180\cdot x \qquad$ ... distribute... $300x-6000=180x \qquad$ ... add $6000-180x$ $120x=6000 \qquad$ ... $/\div 120$ $x=50$ mph ... is the rate of the car, and the truck's is $30$ mph.