Question inolving algebra 1 with distance = rate * time

Question

Justin rode his bicycle at a rate of 15 mph and then slowed to 10 mph. He rode 30 min longer at 15 mph then he did at 10 mph. If he traveled a total of 25 mi how, how long did he ride at the faster rate?
question involves algeba 1 and i'm aware the forumula has something to do with distance = rate * time can anyone help?

Answer 1

distance = rate₁*time₁ + rate₂*time₂

 

Justin rode for t minutes at 10 mph, and for 30 minutes longer at 15 mph.  Since the rate is miles per HOUR, convert 30 minutes to 0.5 hours.  He rode t+ 0.5 at 15 mph.

 

25 miles = 15*(t+0.5) + 10*t

25 = 15t + 7.5 + 10t

17.5 = 25t

17.5/25 = t

0.7 hours = t  This is the time he rode at the slower rate (10 mph).  The time he rode at the faster rate is 0.7 + 0.5 = 1.2 hours

 

Check:

25 miles = 15*(t+0.5) + 10*t

25 = 15*(0.7+0.5) + 10*0.7

25 = 15*1.2 + 7

25 = 18 + 7

25 = 25

Check!