Do you know how to change a word problem to the equations?
Guys, its me again.
Once again professor leaving us stuff that inst on the curriculum/text book.
need a hand here got like 10 of these to do.
Heres one of them:
Jack and Jill spent two weeks (14 nights) touring Boston, New York City, And Philadelphia, and Washington DC. They Paid $120, $200, $80, and $100 Per night respectively. Their Total bill was $2020. The number of days spent in NYC was the same as the total number days spent in Boston and D.C. They spent 3 times as many days in NYC as they did in philly. How many days did they spend in each city?
Thanks in advance,
Gus
Hello, Gus!
Jack and Jill spent 14 days touring Boston, NYC, Phillly, and DC.
They paid $120, $200, $80, and $100 per day respectively. Their total bill was $2020.
The number of days spent in NYC was the same as the total number days spent in Boston and DC.
They spent 3 times as many days in NYC as they did in Philly.
How many days did they spend in each city?
We have: .
We have a system of four equations in four variables.
Solve it . . . you should get: .
I don't suppose you're taking linear algebra? Solving n linear equations in n variables would be covered near the beginning of an introductory course.
Usually in high school algebra we are limited to cases n = 2, maybe n = 3, and in fact the same methods apply to n > 3, but things can get cumbersome (hard to keep track of) if we don't do it systematically.
Basically, we multiply equations by constants and add equations together strategically to yield solutions. By using matrices we can work purely with the coefficients so we don't have to bother with the variable names (B, N, P, D). Gaussian elimination is simple enough to perform with paper and pencil, but it can take a little getting used to.
Not sure how much your professor expects you to know...
EDIT: Since the numbers are small, it is faster and easier in this case to just try out numbers. We know N must be a multiple of 3. There are only so many combinations. In particular, it can be seen quite quickly that N must equal 6. (N + N + N/3 = 14.) After that, we have one of the following cases:
(B, D) = (0, 6)
(B, D) = (1, 5)
(B, D) = (2, 4)
(B, D) = (3, 3)
(B, D) = (4, 2)
(B, D) = (5, 1)
(B, D) = (6, 0)
Im sorry i must be really stupid.
I just really dont see how to show the work for these.....
Guys im sorry.
As ive said before, math 45 and this professor is giving us stuff from math 140
and he batlantly says he is, many complain filed on him and he still doesnt care.
Everyone in the class is either failing or barely passing. He doesnt care though he still tests us on stuff that isnt even in our textbooks.
The three statements
B+P+N+D = 14
N = B+D
N = 3P
allow you to narrow a lot of things down, because you can make substitutions in the first equation based on the other two.
B+P+N+D = 14
N = B+D
====> N+N+P = 14
N+N+P = 14
N = 3P
====> N+N+(N/3) = 14
This last equation is in one variable, so you solve for N to get
N = 6
It immediately follows from
N = 3P
that
P = N/3 = 6/3 = 2
Now you need B and D. You must use the equation we haven't looked at yet,
120B + 200N + 80P + 100D = 2020
We know B and D have to add up to 6. I listed in my first post the possibilities for B and D (since they must be non-negative integers, there are only 7 possibilities). Just try out each combination until you get one that satisfies the last equation.
(Edited for typo: had written "nonzero" instead of "non-negative")