SOLUTION: Question: Mr. Hoerner spent $45.50 last month to see 11 movies. If video rentals cost $3, movie matinees cost $4.50, and evening movies cost $7, how many of each type of movie

Algebra ->  Systems-of-equations -> SOLUTION: Question: Mr. Hoerner spent $45.50 last month to see 11 movies. If video rentals cost $3, movie matinees cost $4.50, and evening movies cost $7, how many of each type of movie       Log On


   



Question 627773: Question: Mr. Hoerner spent $45.50 last month to see 11 movies.
If video rentals cost $3, movie matinees cost $4.50, and evening movies cost $7,
how many of each type of movie did he see?
I came up with 2 equations: r + m + e = 11
3r+4.5m+7e=45.50,
but using substitution or additive inverses didn't work for me.

Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You aren't going to be able to solve this one using ordinary means. You have three variables and only two equations. However, all is not lost. First realize that the solution must be three non-negative integers. You can't watch part of a movie (or at least you won't be able to pay only part of the cost of a movie), so there will be no fractional parts of any of the three types of movie. Also, you can't "unwatch" a movie, so no negative values in the solution either.

Now, notice that the cost of matinees is 4.50, so any odd number (1, 3, 5, 7, 9, or 11) matinees watched will have a cost component that ends in .50. The other types of movies cost whole dollars, and since the number of those watched is an integer, neither of the other types can contribute the 50 cents needed in the total. Hence, we have restricted the value of the variable to a positive odd number less than or equal to 11.

From here it is trial and error.

Let's assume that he watched exactly 1 matinee, and that would have been 4.50 of the total cost. Also, the sum of the other two types must then have been 10 movies and the total cost of the other two types had to be 45.50 minus 4.50 is 41.00.

Set up two equations:





-3 times the first equation:





So



and . Danger, Will Robinson!!! Not an integer. Keep looking.

Set up two new equations for the 3 matinee situation. 45.50 minus 3 times 4.50 = 32 and 11 minus 3 is 8, so:





Oooo...even numbers! looks promising.

-3 times the first equation:





So



and we have which means

Check. 6 times 3 is 18, 3 times 4.50 is 13.50, and 2 times 7 is 14. 18 plus 13.50 plus 14 is 45.50. Checks.

I'll leave it to you to check the other 4 possibilities to make sure there aren't multiple answers, but it wouldn't make sense for other answers to exist. We actually have two lines in 3 space that intersect in one point and we have already found it.

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You have the right equations, but you are
given 3 unknowns and only 2 equations can
be written, so you have to somehow make
up for the missing equation with logic.
-------------------------------
Notice he spent $45.50. The only way he can
end up with the 50 cents is if he saw an
odd number of the matinees, or
1,3,5,7,9, or 11 matinees.
---------------------
It couldn't have been 11 matinees, since
+11%2A4.5+=+49.5+ which is too much
It can't be 9 either since
+9%2A4.5+=+40.5+ and
+40.5+%2B+3+%2B+7+=+50.5+ already too much
-------------------------
I'll try 7
+7%2A4.5+=+31.5+
+31.5+%2B+3+%2B+7+=+41.5+ too low
Can't get an extra 4 to make it 45.5,
so it can't be 7
------------
I'll try 5
+5%2A4.5+=+22.5+
+22.5+%2B+3+%2B+7+=+32.5+ low
+22.5+%2B+3+%2B+14+=+39.5+ low
+22.5+%2B+3+%2B+21+=+46.5+ high
+22.5+%2B+6+%2B+14+=+42.5+ low
+22.5+%2B+18+%2B+14+=+54.5+ high
I'll try 3
+13.5+%2B+18+%2B+14+=+45.5+ OK
So +3%2A4.5+%2B+6%2A3+%2B+2%2A7+=+45.5+
He saw 3 matinees, 6 rentals, 2 evening movies
So, some logic and some guessing works