document.write( "Question 168139: Two traveler were 125 Kilometers apart at 2:00pm and were headed toward each other. If they met at 3:15p.m. and one was traveling 20 kilometers per hour faster than the other, what was the speed of each traveler? \n" ); document.write( "

Algebra.Com's Answer #123944 by Mathtut(3670) $\"\"$ $\"About$

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lets call one travelers rate r and the other r+20. time is the same and lets call the distances x and 125-x\r
\n" ); document.write( "\n" ); document.write( "r(5/4)=x.........eq 1
\n" ); document.write( "(r+20)(5/4)=125-x..eq 2\r
\n" ); document.write( "\n" ); document.write( "lets use x's value from 1st equation and plug it into the 2nd eq\r
\n" ); document.write( "\n" ); document.write( "r+20(5/4)=125-(5/4)r\r
\n" ); document.write( "\n" ); document.write( "multiply both sides by 4\r
\n" ); document.write( "\n" ); document.write( "5r+100=500-5r\r
\n" ); document.write( "\n" ); document.write( "10r=400\r
\n" ); document.write( "\n" ); document.write( " $\"highlight%28r=40%29\"$ speed of the slower traveler\r
\n" ); document.write( "\n" ); document.write( " $\"highlight%28r%2B20=60%29\"$ speed of the faster traveler\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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