I decided to create a similar question using different numbers. Let me know if my set up is correct and if it makes sense.
Question:
On the first part of a trip to Coney Island driving on the freeway, Rita averaged 65 mph. And the rest of the trip, which was 25 miles longer than the first part, she averaged 60 mph. Find the total distance to Coney Island if the second part of the trip took 20 minutes more than the first part.
Let x = first part of trip
Let y = time of trip
x = 65y....Equation 1
x + 25 = 60(1/3 + y)...Equation 2
Yes or no?
Thanks.
Yes, it works, but the problem makes no sense.
[MATH]x = 65y \text { and } x + 25 = 60(1/3 + y) \implies 65y + 25 = 20 + 60y \implies[/MATH]
[MATH]5y = -\ 5 \implies y = -\ 1.[/MATH]
What does a negative number of hours mean?
In any case, this is
NOT the way that I recommend beginning students approach such problems. I suggest this way: start by identifying and naming
EACH potentially relevant but unknown number
[MATH]\text d = \text{first distance.}[/MATH]
[MATH]\text e = \text{second distance.}[/MATH]
[MATH]\text t = \text{first time.}[/MATH]
[MATH]\text u = \text{second time.}[/MATH]
Four unknowns so I need four equations. Write those down in mathematical language.
[MATH]d = 65t.[/MATH]
[MATH]e = 60u.[/MATH]
[MATH]d + 25 = e[/MATH]
[MATH]t + \dfrac{20}{60} = u.[/MATH]
In an earlier post, you asked how to find two equations in the verbiage. My approach is not to worry about equations until after you have explicitly identified all unknowns. Then you know how many equations you must find and can start looking for them. Now combine and simplify.
[MATH]e = d + 25 \text { and } e = 60u \implies 60u = d + 25.[/MATH]
[MATH]\text {BUT } u = t + \dfrac{20}{60} \ implies 60 * \left ( t + \dfrac{20}{60} \right ) = d + 25 \implies[/MATH]
[MATH]60t + 20 = d + 25 \implies 60t - d = 5.[/MATH]
[MATH]d = 65t \implies 60t - 65t = 5 \implies -\ 5t = 5 \implies t = -\ 1.[/MATH]
It is a bit slower, but it is absolutely sure. Very organized and logical.
