Question

Please help solve algebra question

Answer 1

Let be:Speed of the wind: WSpeed of the plane in still air: P
Against the wind the plane flew:Distance: d=175 milesTime: ta=1 hour 10 minutesta=1 hour (10 minutes)*(1 hour/60 minutes)ta=1 hour + 1/6 hourta=(6+1)/6 hourta=7/6 hourSpeed against the wind: Sa=d/taSa=(175 miles) / (7/6 hour)Sa=175*(6/7) miles/hourSa=1,050/7 miles per hourSa=150 mph
(1) P-W=Sa(1) P-W=150
The return trip only took 50 minutesDistance: d=175 milesTime: tr=50 minutestr=(50 minutes)*(1 hour/60 minutes)tr=5/6 hour
Speed retur trip: Sr=d/trSr=(175 miles) / (5/6 hour)Sr=175*(6/5) miles/hourSr=1,050/5 miles per hourSr=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+2102P=360Solving for P:2P/2=360/2P=180
Replacing P by 180 in equation (2):(2) P+W=210180+W=210
Solving for W:180+W-180=210-180W=30
Answers:The speed of the plane in still air was 180 mphThe speed of the wind was 30 mph