math problem

Question

A cruise ship sailed through a 30-mile inland passageway at a constant rate before its speed was increased by 15 mph. Another 75 mi was traveled at the increased rate. The total time for the 105-mile trip was 6 h. Find the rate of the ship during the last 75 mi.

Answer 1

distance=rate*time

d=rt

30=rt

30/r=t

75=(r+15)(6-t)

75=(r+15)(6-[30/r])

75=(r+15)([6r-30]/r)

75r=(r+15)(6r-30)

75r=6r^2+90r-30r-450

6r^2+90r-30r-75r-450=0

6r^2-15r-450=0

2r^2-5r-150=0

2*150=2*2*3*5*5=(2*2*5)(3*5)=20*15 and 20-15=5, the coefficient of the middle term

factor

(2r+15)(r-10)=0

r-10=0

r=10

r+15=10+15=25 mph for the last 75 miles  (answer)