## Elementary Algebra

Let X represent the speed of the second car. The second car overtook the first car 5 hours and 20 minutes after the second car left the town. Since the second car left the town 2 hours after the first car left, the first car traveled for 2 hours more. So, it traveled for 7 hours and 20 minutes. 5 hours and 20 minutes = 5$\frac{1}{3}$ hours. 7 hours and 20 minutes = 7$\frac{1}{3}$ hours. Since the second car overtook the first car, the distance they both traveled is the same. So, we can set up the following equation: 40 $\times$ 7$\frac{1}{3}$ = X $\times$ 5$\frac{1}{3}$ 40 $\times$ $\frac{22}{3}$ = X $\times$ $\frac{16}{3}$ Divide both sides by $\frac{16}{3}$. 40 $\times$ $\frac{22}{3}$ $\div$ $\frac{16}{3}$ = X 40 $\times$ $\frac{22}{3}$ $\times$ $\frac{3}{16}$ = X $\frac{40 \times 22}{16}$ = X X = 55 The speed of the second car was 55 miles per hour.