document.write( "Question 147094: Brian and Jake left their homes, which are 500 miles apart, and drove straight toward each other. It took 4 hours for the two to meet. If Jake's speed was 15 mph slower than Brian's speed, what was Brian's speed? \n" ); document.write( "
Algebra.Com's Answer #107432 by nerdybill(7384)\"\" \"About 
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Brian and Jake left their homes, which are 500 miles apart, and drove straight toward each other. It took 4 hours for the two to meet. If Jake's speed was 15 mph slower than Brian's speed, what was Brian's speed?
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\n" ); document.write( "Use D = RT
\n" ); document.write( "where
\n" ); document.write( "D = distance
\n" ); document.write( "R = rate (mph)
\n" ); document.write( "T = time (hours)
\n" ); document.write( ".
\n" ); document.write( "Let x = Brian's speed
\n" ); document.write( "then
\n" ); document.write( "x-3 = Jake's speed
\n" ); document.write( ".
\n" ); document.write( "The idea is that the combined \"distance\" traveled by Brian and Jake should be 500 miles.
\n" ); document.write( "distance Brian traveled = RT = 4x
\n" ); document.write( "distance Jake traveled = RT = 4(x-3)
\n" ); document.write( ".
\n" ); document.write( "\"distance Brian traveled\" + \"distance Jake traveled\" = 500
\n" ); document.write( "4x + 4(x-3) = 500
\n" ); document.write( "4x + 4x - 12 = 500
\n" ); document.write( "8x = 512
\n" ); document.write( "x = 64 mph (Brian's speed)
\n" ); document.write( "x-3 = 64-3 = 61 mph (Jake's speed)\r
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