Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total of people took the trip. She was able to purchase coach tickets for $260 and first class tickets for $950. She used her total budget for airfare for the trip, which was $3720. How many first class tickets did she buy? How many coach tickets did she buy?
Let's assume that Sarah bought x first class tickets and y coach tickets.
The total budget for airfare is $3720, so we can write the equation:
950x + 260y = 3720.
Including Sarah, there are a total of n people, so the number of tickets bought is x + y = n - 1 (excluding Sarah).
Simplifying the equation, we have:
260y = 3720 - 950x,
which can be rearranged to:
y = (3720 - 950x)/260.
Substituting this expression into the equation x + y = n - 1, we get:
x + (3720 - 950x)/260 = n - 1.
Multiplying through by 260 to get rid of the fraction, we have:
260x + 3720 - 950x = 260(n - 1).
Simplifying further, we have:
-690x = 260n - 3980.
Dividing by -690, we get:
x = (260n - 3980)/-690.
Since x represents the number of first class tickets Sarah bought, it must be a positive integer. Thus, 260n - 3980 must be a multiple of 690.
For x to be an integer, 260n - 3980 must be divisible by 690. Checking for values of n (1, 2, 3...), we find that it is divisible by 690 for n = 5.
Substituting n = 5 into the equation for x, we have:
x = (260 * 5 - 3980)/-690 = -30/69 = -10/23.
Since x must be a positive integer, we discard this solution.
Therefore, there is no possible solution for the given scenario.