document.write( "Question 1048766: Billy made a trip to California and back. The trip there took four hours and the trip back took 5 hours. He averaged 5mph faster on the trip there than on the return trip. Let r represent his average speed on the trip back. What was Billy's average speed on both trips? \n" ); document.write( "
Algebra.Com's Answer #664341 by solver91311(24713)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Distance equals rate times time, so the distance on the faster part of the trip is equal to , and the distance on the slower part of the trip is equal to . Presumably, neither California nor the place Billy started moved during the trip so we can assume that these two distances are the same. Therefore:\r
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\n" ); document.write( "\n" ); document.write( "Solve for and then calculate \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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