document.write( "Question 322579: Airplane A travels 1000 mi at a certain speed. Airplaine B travels the same distance in 1 hr less time but travels 50 mph faster. Find the time that Airplane B travels. \n" ); document.write( "
Algebra.Com's Answer #230943 by mananth(16946)\"\" \"About 
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Airplane Time = 1000/x where x is the speed of the plane
\n" ); document.write( "Airplane B Time = 1000/x+50 ( 50 mph faster)
\n" ); document.write( "..
\n" ); document.write( "Time taken by airplane A - Time by Airplane B =1
\n" ); document.write( "1000/x - 1000/ x+50=1
\n" ); document.write( "LCM = x(x+50)
\n" ); document.write( "1000(x+50)- 1000x= x(x+50)
\n" ); document.write( "1000x+50,000 - 1000x=x^2 +50x
\n" ); document.write( "x^2+50x-50000=0
\n" ); document.write( "x^2+250x-200x-50000=0
\n" ); document.write( "x(x+250)-200(x+2 b50)=0
\n" ); document.write( "(x+250)(x-200)=0
\n" ); document.write( "x= 200 mph speed of plane A
\n" ); document.write( "speed of plane B = 250 mph
\n" ); document.write( "distance traveled = 1000miles
\n" ); document.write( "time it traveled = 1000/250 = 4 hours
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