How fast did the boat travel under power, under sail, and under both power and sail?

Question

This is from Page 226, #24 of Fundamentals of Technical Mathematics (2nd edition): Solve this system for I1, I2, and I3.
A sailboat leaves a harbor at 0600 hours (6 a.m.) and motors under power with no wind. At 0800 hours (8 a.m.) the wind picks up, and the skipper shuts off the boat's engine and sails until 1200 hours (noon), covering 22 miles. At this time the skipper decides to use both power and sail to make more progress. The boat arrives in port at 1800 hours (6 p.m.) after covering a total of 64 miles. If the speed under power and sail equals the speed under power plus the speed under sail, how fast did the boat travel under power, under sail, and under both power and sail?

Answer 1

We know that from 0800 hours to 1200 hours the boat was traveling under sail power and it covered 22 miles. So the speed of sail power = 22 / 4 = 5.5 mph

Let E represent the speed of the boat with only Engine power and S represent the speed of the boat with only Sail power.
So now we have:
2E + 22 + 6 (E + S) = 64
2E + 22 + 6 (E + 5.5) = 64
2E + 22 + 6E+ 33 = 64
2E + 6E = 64 - 22 - 33
8E = 9
E = 11/8
E = 1.125 mph

E+S = 1.125 + 5.5 = 6.625 mph

Let's double check the answers
2(1.125) + 4(5.5) + 6(6.625)
= 2.25 + 22 + 39.75
= 64

I hope this helped

Kia