document.write( "Question 166131: Two planes travel toward each other from cities that are 1615 kilometers apart. The two planes passed each other in the air after 3 hours and 48 minutes. The slower plane was traveling at a rate of speed that was 80 k/h more than half the rate of speed of the faster plane. Assuming they left at the same time, at what rate of speed was each plane traveling? \n" ); document.write( "
Algebra.Com's Answer #122462 by Mathtut(3670)\"\" \"About 
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lets call the distance traveled d(d1) and 1615-d(d2)
\n" ); document.write( "fast plane rate of speed lets call f
\n" ); document.write( "slow planes rate of speed will call s which equals 1/2f+80--->(2f+160)/2
\n" ); document.write( "time = 3 4/5hours or 19/5 hours. Time for plane 1 and plane 2 are equal\r
\n" ); document.write( "\n" ); document.write( "so d1=(rate of speed)(time)--->
\n" ); document.write( " d=(f+160/2)(19/5)=(19f+3040)/10\r
\n" ); document.write( "\n" ); document.write( "and d2=f(19/5)---->1615-d=19f/5\r
\n" ); document.write( "\n" ); document.write( "placing d's value into the d2 equation we have 1615-(19f+3040)/10=19f/5\r
\n" ); document.write( "\n" ); document.write( "multiply each term by 10 16150-19f-3040=38f--->57f=13110=\"230\"\r
\n" ); document.write( "\n" ); document.write( "so the speed of the fast plane was \"230mph\"
\n" ); document.write( "hence the speed of the slow plane was s=1/2(230)+80=\"195mph\"
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