document.write( "Question 1010956: Hi,\r
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\n" ); document.write( "\n" ); document.write( "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. \r
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Algebra.Com's Answer #626457 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+s+\" = plane's speed in mi/hr going to A
\n" ); document.write( "Let \"+t+\" = plane's time in hrs going to A
\n" ); document.write( "\"+s+%2B+15.8+\" = plane's speed in mi/hr going to B
\n" ); document.write( "\"+5.2+-+t+\" = plane's time in hrs going to B
\n" ); document.write( "-----------------------------------------
\n" ); document.write( "Plane's equation going to A:
\n" ); document.write( "(1) \"+355+=+s%2At+\"
\n" ); document.write( "Plane's equation going to B:
\n" ); document.write( "(2) \"+448+=+%28+s+%2B+15.8+%29%2A%28+5.2+-+t+%29+\"
\n" ); document.write( "-------------------------------
\n" ); document.write( "(1) \"+t+=+355%2Fs+\"
\n" ); document.write( "Substitute this into (2)
\n" ); document.write( "(2) \"+448+=+%28+s+%2B+15.8+%29%2A%28+5.2+-355%2Fs+%29+\"
\n" ); document.write( "(2) \"+448+=+5.2s+%2B+82.16+-+355+-+5609%2Fs+\"
\n" ); document.write( "(2) \"+720.84+=+5.2s+-+5609%2Fs+\"
\n" ); document.write( "(2) \"+720.84s+=+5.2s%5E2+-+5609+\"
\n" ); document.write( "(2) \"+5.2s%5E2+-+720.84+-+5609+=+0+\"
\n" ); document.write( "(2) \"+s%5E2+-+138.623s+-+1078.654+=+0+\"
\n" ); document.write( "You can solve with quadratic formula. The numbers
\n" ); document.write( "are really strange, so check my math. I think my
\n" ); document.write( "method is good. Another opinion would be good too.
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