document.write( "Question 276065: At 2:00 p.m. two cars start toward each other from towns 240 miles apart. If the rate of one car is 10 mph faster than the other, how fast does each car go if they meet at 5:00 p.m.? \n" ); document.write( "
Algebra.Com's Answer #201272 by stanbon(75887)\"\" \"About 
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At 2:00 p.m. two cars start toward each other from towns 240 miles apart. If the rate of one car is 10 mph faster than the other, how fast does each car go if they meet at 5:00 p.m.?
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\n" ); document.write( "Slower Car DATA:
\n" ); document.write( "distance = x ; time = 3 hrs ; rate = d/t = x/3 hrs
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\n" ); document.write( "Faster Car DATA:
\n" ); document.write( "distance = (240-x) ; time = 3 hrs ; rate = d/t = (240-x)/3
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\n" ); document.write( "Equation:
\n" ); document.write( "faster rate - slower rate = 10 mph
\n" ); document.write( "(240-x)/3 - x/3 = 10 mph
\n" ); document.write( "240-x - x = 30
\n" ); document.write( "-2x = -210
\n" ); document.write( "x = 105 miles
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\n" ); document.write( "slower car rate = 105/3 = 35 mph
\n" ); document.write( "faster car rate = 35+10 = 45 mph
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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