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)![]() ![]() You can put this solution on YOUR website! 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 \n" ); document.write( " |