document.write( "Question 369512: Christine and Jessica leave from the same spot at the same time and begin jogging in opposite directions. At the end of an 1 1/2 hours they are 15 miles apart. Find Christine's average speed if her average speed is 2 miles per hour more than Jessica's average speed. \n" ); document.write( "
Algebra.Com's Answer #263330 by CharlesG2(834)\"\" \"About 
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Christine and Jessica leave from the same spot at the same time and begin jogging in opposite directions. At the end of an 1 1/2 hours they are 15 miles apart. Find Christine's average speed if her average speed is 2 miles per hour more than Jessica's average speed.\r
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\n" ); document.write( "\n" ); document.write( "distance D = rate R * time T
\n" ); document.write( "let R = Jessica's average speed
\n" ); document.write( "let R + 2 = Christine's average speed
\n" ); document.write( "let T = 1 1/2 = 1.5 hours
\n" ); document.write( "let D = total distance of 15 miles
\n" ); document.write( "15 = R * 1.5 + (R + 2)1.5
\n" ); document.write( "15 = 1.5R + 1.5R + 3
\n" ); document.write( "12 = 3R
\n" ); document.write( "4 = R
\n" ); document.write( "Jessica's average speed is 4 mph
\n" ); document.write( "Christine's average speed is 6 mph
\n" ); document.write( "Jessica's distance = 4 * 1.5 = 6 miles
\n" ); document.write( "Christine's distance = 6 * 1.5 = 9 miles
\n" ); document.write( "6 miles + 9 miles = 15 miles
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