document.write( "Question 6642: mr. west drove his car from home to chicago at the rate of 40 mph. amd returned at a rate of 45mph. if his time going exceed his time returning by 30 minutes, find his time going and his time returning \n" ); document.write( "
Algebra.Com's Answer #3620 by prince_abubu(198)\"\" \"About 
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We're going to use rate * time = distance in this situation. Before we go on, we must keep in mind that the distance for both to and from trips is the same.\r
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\n" ); document.write( "\n" ); document.write( "Let's nail down the trip home first.\r
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\n" ); document.write( "\n" ); document.write( "\"+r%5Bh%5D+\" = Rate going back home = 45 mph.
\n" ); document.write( "\"+t%5Bh%5D+\" = We don't know yet.
\n" ); document.write( "Distance from Chicago to home would then be \"+r%5Bh%5D%2At%5Bh%5D+=+45t%5Bh%5D+\".\r
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\n" ); document.write( "\n" ); document.write( "Let's nail down the trip to Chicago then.\r
\n" ); document.write( "\n" ); document.write( "\"+r%5Bc%5D+\" = Rate to Chicago = 40 mph
\n" ); document.write( "\"+t%5Bc%5D+\" = time it took to go to Chicago = \"+t%5Bh%5D+%2B+0.5+\" <----- We said 0.5 hours (instead of saying + 30) because the speed is in miles per hour.
\n" ); document.write( "Distance from home to Chicago would have to be \"+r%5Bc%5D%2At%5Bc%5D+=+40%28t%5Bh%5D%2B0.5%29\".\r
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\n" ); document.write( "\n" ); document.write( "Since the rates and times HAVE TO equal the same distance, we can set them equal to each other.\r
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\n" ); document.write( "\n" ); document.write( "\"+40%28t%5Bh%5D+%2B+0.5%29+=+45t%5Bh%5D+\"\r
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\n" ); document.write( "\n" ); document.write( "\"+40t%5Bh%5D+%2B+20+=+45t%5Bh%5D+\" <---- used distributive property\r
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\n" ); document.write( "\n" ); document.write( "\"+20+=+5t%5Bh%5D+\" <---- subtracted \"+40t%5Bh%5D+\" from both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"+t%5Bh%5D+=+4+\" <---- Took him four hours to get home from Chicago. If his trip to Chicago took 30 minutes longer, then his trip to Chicago took 4.5 hours.
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