Sarah took the advertising department from her company on a round-trip to meet with a potential client. Including Sarah, a total of 12 people took this trip. She was able to purchase coach tickets for $300 and first class tickets for $940. She used her total budget for airfare for the trip, which was $6800. How many first class tickets did she buy? How many coach tickets did she buy?
I am very confused and need some help please.
Let F = # first class passengers and
let C = # coach class passengers.
F + C = 12
940F + 300C = 6800
Solve these two equations simultaneously for F and C.
Post your work if you get stuck.
So I just find a number for both F and C and make sure that it doesn’t go over 6800 and there is enough tickets for 12 people?
No, you don't guess; you solve the two equations simultaneously.
F + C = 12
940F + 300C = 6800
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Solve the first equation for C. That will be
C = 12-F and substitute this for C in equation 2 like this
940F + 300(12-F) = 6800. Now you have an equation that has just one unknown (that's F), solve for that like this.
940F + 3600 - 300F = 6800
940F-300F = 6800-3600
640F = 3200
F = 3200/640 = 5 first class tickets. Now plug that number for F back into equation 1 and solve for C. That gives you #F and #C tickets.
Then check it to see that 5*940 + C*300 = 6800. If that checks you know you solved the two equations correctly.
Ohhhhh thank you so much!!!
To solve this problem, let's assign some variables to the unknown quantities.
Let's say Sarah bought x first class tickets and y coach tickets.
The cost of a first class ticket is $940, so the total cost of first class tickets is 940x.
The cost of a coach ticket is $300, so the total cost of coach tickets is 300y.
According to the problem, the total cost of all the tickets is $6800. So, we can write the equation:
940x + 300y = 6800
Since we know that a total of 12 people took the trip, and Sarah was one of them, the number of tickets purchased must add up to 12. So we have the second equation:
x + y + 1 = 12
Now we have a system of equations:
940x + 300y = 6800 --------(1)
x + y + 1 = 12 --------(2)
We can solve this system of equations using substitution or elimination methods.
Let's solve the second equation for x:
x = 12 - y - 1
x = 11 - y --------(3)
Substituting equation (3) into equation (1), we get:
940(11 - y) + 300y = 6800
Expanding and simplifying:
10340 - 940y + 300y = 6800
-640y = -3540
y = (-3540)/(-640)
y = 5.53125
Since the number of coach tickets cannot be a fraction, we will round down to the nearest whole number because it doesn't make sense to buy a fraction of a coach ticket. So, y = 5.
Substituting y = 5 back into equation (3), we find:
x = 11 - y
x = 11 - 5
x = 6
Therefore, Sarah bought 6 first class tickets and 5 coach tickets.
To summarize:
Sarah bought 6 first class tickets and 5 coach tickets.