Question

Please help solve algebra question

Answer 1

Let be:
Speed of the wind: W
Speed of the plane in still air: P

Against the wind the plane flew:
Distance: d=175 miles
Time: ta=1 hour 10 minutes
ta=1 hour (10 minutes)*(1 hour/60 minutes)
ta=1 hour + 1/6 hour
ta=(6+1)/6 hour
ta=7/6 hour
Speed against the wind: Sa=d/ta
Sa=(175 miles) / (7/6 hour)
Sa=175*(6/7) miles/hour
Sa=1,050/7 miles per hour
Sa=150 mph

(1) P-W=Sa
(1) P-W=150

The return trip only took 50 minutes
Distance: d=175 miles
Time: tr=50 minutes
tr=(50 minutes)*(1 hour/60 minutes)
tr=5/6 hour

Speed retur trip: Sr=d/tr
Sr=(175 miles) / (5/6 hour)
Sr=175*(6/5) miles/hour
Sr=1,050/5 miles per hour
Sr=210 mph

(2) P+W=Sr
(2) P+W=210

We have a system of 2 equations and 2 unknows:
(1) P-W=150
(2) P+W=210

Adding the equations:
P-W+P+W=150+210
2P=360
Solving for P:
2P/2=360/2
P=180

Replacing P by 180 in equation (2):
(2) P+W=210
180+W=210

Solving for W:
180+W-180=210-180
W=30

Answers:
The speed of the plane in still air was 180 mph
The speed of the wind was 30 mph

Answer 2

The first trip had an average speed of
(175 mi)/(7/6 h) = 150 mi/h

The second trip had an average speed of
(175 mi)/(5/6 h) = 210 mi/h

The speed of the plane in still air is (150 +210)/2 = 180 mi/h.
The speed of the wind is 180 -150 = 30 mi/h.

_____
The two trip speeds are the sum and difference of the speed of the plane and the speed of the wind. Let p and w represent the speeds of plane and wind, respectively.
p -w = 150 mph
p +w = 210 mph
Adding these two equations gives
2p = (150 +210) mph
p = (150 +210)/2 mph = 180 mph
Then, from the first equation,
p -150 mph = w
(180 -150) mph = w = 30 mph