SOLUTION: Hi, Thank you for answering my question in advance. An airplane flies 355 miles to city A. Then, with better winds it continues onto city B, 448 miles from A, at a speed 15.8

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hi, Thank you for answering my question in advance. An airplane flies 355 miles to city A. Then, with better winds it continues onto city B, 448 miles from A, at a speed 15.8      Log On


   

Question 1010956: Hi,
Thank you for answering my question in advance.
An airplane flies 355 miles to city A. Then, with better winds it continues onto city B, 448 miles from A, at a speed 15.8 mi/h greater than the first leg of the trip. The total flying time was 5.20 hrs. Find the speed at which the plane traveled to city A.
Thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = plane's speed in mi/hr going to A
Let +t+ = plane's time in hrs going to A
+s+%2B+15.8+ = plane's speed in mi/hr going to B
+5.2+-+t+ = plane's time in hrs going to B
-----------------------------------------
Plane's equation going to A:
(1) +355+=+s%2At+
Plane's equation going to B:
(2) +448+=+%28+s+%2B+15.8+%29%2A%28+5.2+-+t+%29+
-------------------------------
(1) +t+=+355%2Fs+
Substitute this into (2)
(2) +448+=+%28+s+%2B+15.8+%29%2A%28+5.2+-355%2Fs+%29+
(2) +448+=+5.2s+%2B+82.16+-+355+-+5609%2Fs+
(2) +720.84+=+5.2s+-+5609%2Fs+
(2) +720.84s+=+5.2s%5E2+-+5609+
(2) +5.2s%5E2+-+720.84+-+5609+=+0+
(2) +s%5E2+-+138.623s+-+1078.654+=+0+
You can solve with quadratic formula. The numbers
are really strange, so check my math. I think my
method is good. Another opinion would be good too.