word problem involving distance formula: 50-mi bike ride

[serenity]

On a 50 mile bicycle ride, Irene averaged 4mph faster the first 36 miles that she did for the last 14 miles. The entire trip took 3 hours. Find her rate for the first 36 miles.

I was able to come up with equations. However I'm unable to come with the right answer. The equation for the first part using the distance formula 36=rt, and 14=(r-4)(3-t). Then, I can't go form here.

Are my equations put together incorrectly. What are the key words in solving this problem. should i add the miles. :? I'm really stuck on this problem. I'll be very thankful if you can help me solve the problem and give me some good suggestions in solving this kind of problems.

 

[galactus]

Re: word problem involving the distance formula.

You have two equations with two unknowns. It's just a simple system of equations.

Your set up is correct. Looks good. Now, you will get a quadratic to solve.

\(\displaystyle 36=rt, \;\ 14=(r-4)(3-t)\)

Solve 36=rt for t and sub into the other equation and solve for r.

\(\displaystyle t=\frac{36}{r}\)

\(\displaystyle 14=(r-4)(3-\frac{36}{r})\)

\(\displaystyle 3r^{2}-62r+144=0\)

You will get two solutions, but only one will be of value.

Or, you could set up one equation from the outset. t=d/r

\(\displaystyle \frac{36}{r}+\frac{14}{r-4}=3\)

Solve for r. Same thing, only rearranged into one equation. If you make the sub you can see it's the same thing.

 

[serenity]

Thank you for your help. I had gotten the quadratic form but I thought it was a mistake. I didn't Know a quadratic form can come up in a distance formula problem yet with your illustrations I can see why. Thank you once again.