Distance, rate, time problem

Art leaves his house at 10 o'clock traveling at 50 mph. Jennifer leaves her house at 10:30 traveling at 60 mph. At what time will Jennifer catch up to Art?

Distance = rate x time 

Art's distance
d = rt   {distance = rate x time}
rate = 50
time = t
d = 50t 

Jennifer's distance
d = rt   {distance = rate x time}
rate = 60
time = t - 0.5   {left 1/2 hour later than Art}
d = 60(t - 0.5) = 60t - 30   {used distributive property} 

50t = 60t - 30   {when Jennifer catches up with Art, the distances will be equal}
-10t = -30   {subtracted 60t from each side}
t = 3   {divided each side by -10} 

In 3 hours, Jennifer will catch up with Art

- Algebra House