SOLUTION: Mary can jog, on average, 2 miles per hour faster than her husband, James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. Mary can jog __________

Algebra ->  Square-cubic-other-roots -> SOLUTION: Mary can jog, on average, 2 miles per hour faster than her husband, James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. Mary can jog __________       Log On


   

Question 942488: Mary can jog, on average, 2 miles per hour faster than her husband, James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. Mary can jog __________ miles per hour?
Another word problem I can not comprehend
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
M=Mary's rate J=Jame's rate
M=J+2 Mary is 2mph faster than James
6.6 mi/J=9 mi/M James runs 6.6 miles in the time it takes Mary to run 9 miles
Substitute M=J+2 in the second equation to solve for J
6.6/J=9/J+2 Multiply each side by J
6.6J/J=9J/J+2
6.6=9J/J+2 Multiply each side by J+2
6.6(J+2)=9J(J+2)/(J+2)
6.6J+13.2=9J subtract 6.6J from each side
6.6J+13.2-6.6J=9J-6.6J
13.2=2.4J Divide each side by 2.4
13.2/2.4=2.4J/2.4
5.5=J So James can run at 5.5 mph and Mary runs at 7.5 mph (J+2)
CHECK
James runs 6.6 miles in 6.6 mi/5.5 mph=1.2 hrs
Mary runs 9 miles in 9mi/7.5 mph=1.2 hrs