Ask your own question, for FREE!
Mathematics 73 Online
OpenStudy (anonymous):

Set up an equation and solve the following problem. To travel 132 miles, it takes Sue, riding a moped, 2 hours less time than it takes Doreen to travel 72 miles riding a bicycle. Sue travels 13 miles per hour faster than Doreen. Find the times and rates of both girls.

OpenStudy (calculusfunctions):

Sorry, my computer lost connection as I was typing the solution. I'll star over. Give me a few minutes. Thanks.

OpenStudy (anonymous):

Thank you! Its no problem!!

OpenStudy (calculusfunctions):

Have you learned linear systems and how to work with two variables? Substitution and elimination methods?

OpenStudy (anonymous):

I learned linear systems with only 1 variable, but I'm not very good at these types of problems.

OpenStudy (anonymous):

Yes to substitution and elimination methods.

OpenStudy (calculusfunctions):

Let x represent the time it takes Doreen to travel 72 mi., then x - 2 represents the time it takes Sue to travel 132 mi. because it takes Sue two hours less time according to the question. Now make a chart as follows: |dw:1349493423857:dw| Since Sue travels 13 mi/h faster than Doreen, Sue's speed = Doreen's speed + 13 Hence\[\frac{ 132 }{ x -2 }=\frac{ 72 }{ x }+13\]Now multiply both sides of the equation by x(x - 2) (the least common denominator) to eliminate the denominators. Thus by doing so, we obtain\[132x =72(x -2)+13x(x -2)\]Now\[132x =72x -144+13x ^{2}-26x\]Thus\[0=13x ^{2}-86x -144\]Now factor this expression to obtain\[(13x +18)(x -8)=0\]Thus\[x =-\frac{ 18 }{ 3 },x =8\]However\[x =-\frac{ 18 }{ 3 }\]is inadmissible so the only possible solution is \[x =8\] ∴ it takes Sue 6 hours to travel 132 miles at 22 mi/h, and it takes Doreen 8 hours to travel 72 miles at 9 mi/h. Do you understand?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!