document.write( "Question 939011: On a ground trip of 540 km in each direction, Mr. Armstrong averaged 9 km/h more returning than he did going. If the return trip took 2 hours less than the trip going, find the time for each direction.
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Algebra.Com's Answer #572089 by mananth(16946)\"\" \"About 
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on hisway there speed x
\n" ); document.write( "on his return speed x + 9
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\n" ); document.write( " Distance = same 540 km 1
\n" ); document.write( " original time – time with increased speed = 2 hour
\n" ); document.write( " t=d/t
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\n" ); document.write( " 540 / x - 540 / ( x + 9 ) = 2.00
\n" ); document.write( " LCD= x ( x + 9 )
\n" ); document.write( " multiply by LCD
\n" ); document.write( " 540 ( x + 9 ) - 540 x = 2 x (
\n" ); document.write( " 540 x + 4860 - 540 x = 2 X^2 + 18 x
\n" ); document.write( " 4860 = 2 X^2 + 18 x
\n" ); document.write( " 2 X^2 + 18 x- -4860 = 0 \r
\n" ); document.write( "\n" ); document.write( "/2
\n" ); document.write( "x^2+9x-2430=0\r
\n" ); document.write( "\n" ); document.write( "x^2+54x-45x-2430=0\r
\n" ); document.write( "\n" ); document.write( "x(x+54)-45(x+54)=0\r
\n" ); document.write( "\n" ); document.write( "(x-45)(x+54)=0\r
\n" ); document.write( "\n" ); document.write( "x=45 OR -54
\n" ); document.write( "ignore negative\r
\n" ); document.write( "\n" ); document.write( "speed = 45 km/h
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\n" ); document.write( "54 km/h \n" ); document.write( "
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