# Thread: Rational Equations- Multiplying and dividing

1. ## Rational Equations- Multiplying and dividing

I really need help with this.

First I had to do a problem that is :
Write an algebraic expression for the total time, in hours, that it takes a jogger to run 2 miles by running 1 mile uphill and then going downhill one mile if the jogger runs uphill at an average speed that is c miles per hour slower than ground level speed of 6 mph and runs downhill at an average speed c more than 6 miles per hour. Simplify your answer to a single algebraic fraction.

12
______
36- c squared

I then had to write an expression for the average speed of a jogger running 1 mile uphill and one mile downhill (2 miles total), which turned out to be (and I am POSITIVE these first two are right):

36-c squared
____________
6

Then, here is the one I am having trouble on:

For what value of c would the jogger's average speed for the two mile trip (one mile up and one mile down) be 4.5 miles per hour? For this value of c, what would be the average rate uphill and downhill?

2. Hello kpapetti
Originally Posted by kpapetti
I really need help with this.

First I had to do a problem that is :
Write an algebraic expression for the total time, in hours, that it takes a jogger to run 2 miles by running 1 mile uphill and then going downhill one mile if the jogger runs uphill at an average speed that is c miles per hour slower than ground level speed of 6 mph and runs downhill at an average speed c more than 6 miles per hour. Simplify your answer to a single algebraic fraction.

12
______
36- c squared

I then had to write an expression for the average speed of a jogger running 1 mile uphill and one mile downhill (2 miles total), which turned out to be (and I am POSITIVE these first two are right):

36-c squared
____________
6

Then, here is the one I am having trouble on:

For what value of c would the jogger's average speed for the two mile trip (one mile up and one mile down) be 4.5 miles per hour? For this value of c, what would be the average rate uphill and downhill?

Welcome to Math Help Forum!

You're right so far, and I think you've done the hardest bits! All you need to do for the final part is to put the expression you have for part 2 equal to $\displaystyle 4.5$, and solve the equation for $\displaystyle c$. So you get:

$\displaystyle \frac{36-c^2}{6} = 4.5$

$\displaystyle \Rightarrow 36-c^2 = 4.5 \times 6 = 27$

$\displaystyle \Rightarrow 36 - 27 = c^2$

$\displaystyle \Rightarrow c^2 = 9$

$\displaystyle \Rightarrow c = 3$ (not $\displaystyle -3$, because $\displaystyle c$ must be positive)

So the jogger runs at $\displaystyle 3$ mpg uphill and $\displaystyle 9$ mph downhill.

Just in case you think that answer is wrong, and would give an average speed of $\displaystyle 6$ mph, just try it out.

The time taken to run 1 mile at 3 mph = $\displaystyle \frac13$ hours, and at 9 mph is $\displaystyle \frac19$ hours. So the total time taken to run the 2 miles is $\displaystyle \frac13 + \frac19 = \frac49$ hours. So the average speed is $\displaystyle 2 \div \frac49 = 2 \times \frac94 = 4.5$ mph