College Algebra, Exercise P, Exercise P.1, Section Exercise P.1, Problem 22

Suppose at a certain car rental agency a compact car rents for $30 a day and 10 cents a mile.
a.) How much does it cost to rent a car for 3 days if the car is driven 280 miles?
Since the rental charge for a car per day is $30 and 10 cents per mile. The total cost would be (suppose that $T$ is the total cost).

$\begin{equation} \begin{aligned} T &= \$ 30 (3) + 10 \text{cents} (280 \text{mi})\\ \\ T &= \$ 30(3) + \$ \frac{10}{100} (280 \text{mi})\\ \\ T &= 90 + 0.1(280)\\ \\ T &= 90 + 28\\ \\ T &= \$ 118 && \text{Cost to rent a car for 3 days and 250 miles} \end{aligned} \end{equation}$


b.) Find a formula that models the cost $C$ of renting this car for $n$ days if it is driven $m$ miles.
$C = \$ 30(n) + 10 \text{cents} (m)$ model

c.) If the cost for a 3-day rental was $ 140, how many miles was the car driven?
From the model in part(b), we solve $C = \$ 30(n) + 10 \text{cents}(m)$ for $m$.

$\begin{equation} \begin{aligned} C - \$ 30(n) &= \$ 30(n) + 10 \text{cents} (m) - \$ 30(n) && \text{Subtract both sides by \$30 }(n)\\ \\ \frac{C- \$30 (n)}{10\text{cents}} &= \frac{\cancel{10\text{cents}}(m)}{\cancel{10\text{cents}}} && \text{Divide both sides by 10cents}\\ \\ m &= \frac{C - \$30 (n)}{10 \text{cents}} \text{ model}\\ \\ m &= \frac{\$ 140 - \$ 30(3)}{10 \text{cents}} && \text{Substitute } C = \$ 140 \text{ and } n = 3\\ \\ m &= \frac{\$ 140 - \$ 30(3)}{\$ 0.1} && \text{Recall that } \$ 1 = 100 \text{cents}\\ \\ m &= \frac{140-90}{0.1} && \text{Simplify}\\ \\ m &= 500 \text{miles} \end{aligned} \end{equation}$