document.write( "Question 90578: I posted this question with two more. I apologize for the confusion. I wasn't paying close attention to directions! Thanks so much!!!!\r
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\n" ); document.write( "\n" ); document.write( "1.) A bicyclist rode into the country for 5h. In returning, her speed was 5 mi/h faster and the trip took 4h. What was her speed each way?
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Algebra.Com's Answer #65761 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
You can use the formula:\"d+=+rt\" where d = distance, r = rate(speed) and t = time of travel.
\n" ); document.write( "For the outbound trip: \"t%5B1%5D+=+5\" hrs, so you can write:
\n" ); document.write( "\"d%5B1%5D+=+5r%5B1%5D\"
\n" ); document.write( "For the return trip, the rate was 5 mph faster than for the outbound trip, so \"r%5B2%5D+=+%28r%5B1%5D%2B5%29\" and \"t%5B2%5D+=+4\", so you can write:
\n" ); document.write( "\"d%5B2%5D+=+4%28r%5B1%5D%2B5%29\"
\n" ); document.write( "But the distance out is the same as the distance back, so...
\n" ); document.write( "\"d%5B1%5D+=+d%5B2%5D\" or...
\n" ); document.write( "\"5r%5B1%5D+=+4%28r%5B1%5D%2B5%29\" Simplify and solve for \"r%5B1%5D\"
\n" ); document.write( "\"5r%5B1%5D+=+4r%5B1%5D%2B20\"
\n" ); document.write( "\"r%5B1%5D+=+20\"mph This is the outbound speed.
\n" ); document.write( "\"r%5B2%5D+=+r%5B1%5D%2B5\"
\n" ); document.write( "\"r%5B2%5D+=+25\"mph This is the return speed.
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