Question 270126: I have been trying to figure out this problem for an hour now I cannot get the answer the book gives me in the back. I found the problem here on your site already but the answers are not the same as the back of my book either.
This is the link to the problem on your site, http://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.164399.html
The problem is the same as in my book but the answers in the back say it should be : 2 hr at 50 mph and 3 hr at 40 mph.
I am so lost!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! There are three answers on algebra.com to your problem.
The top answer is wrong.
The bottom two answers are right.
My method for solving this problem would be as follows:
Let T1 equal the time traveled at 50 miles per hour.
Let T2 equal the time traveled at 40 miles per hour.
T1 + T2 must be equal to 5 since the total time traveled is 5 hours.
Since Rate * Time equals distance, and since the total distance equals 220 miles, we get:
50*T1 + 40*T2 = 220
You have two equations that need to be solved simultaneously.
They are:
50*T1 + 40*T2 = 220
T1 + T2 = 5
From the second equation, we can get T1 in terms of T2.
We get T1 = 5 - T2.
We substitute 5 - T2 for T1 in the first equation to get:
50*(5-T2) + 40*T2 = 220
We simplify by removing parentheses to get:
50*5 - 50*T2 + 40*T2 = 220
We simplify further and combine like terms to get:
250 - 10*T2 = 220
We subtract 250 from both sides of the equation to get:
-10*T2 = 220 = 250 = -30
We multiply both sides of this equation by -1 to get:
10*T2 = 30
We divide both sides of this equation by 10 to get:
T2 = 3
Since T2 = 3, then T1 = 2, and our original equations of:
50*T1 + 40*T2 = 220 becomes 50*2 + 40*3 = 220 becomes 100 + 120 = 220 which is true.
T1 + T2 = 5 becomes 2 + 3 = 5 which is also true.
The answer of 2 hours at 50 miles per hour and 3 hours at 40 miles per hour is good.
If you check the solutions again, you will see that 3 people answered and 2 out of the 3 were correct.
I looked at the first answer again (the wrong one) and I'm not sure he was answering the question that was asked.
Hopefully this will end your confusion.
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