SOLUTION: Desparate need of help. Thank you in advance! Some students planned for a get-together. The budget for food was $500. Five of the students failed to come because of the distance

Question 69989: Desparate need of help. Thank you in advance!
Some students planned for a get-together. The budget for food was $500. Five of the students failed to come because of the distance and therefore the cost of food for each member increased by $5. How many students attended the get-together?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
"Some students planned for a get-together. The budget for food was $500. Five of
the students failed to come because of the distance and therefore the cost of
food for each member increased by $5. How many students attended the
get-together?"
:
Let x = Number of actual students that attended
(x+5) = number that planned to attend
:
Original cost per student: 500/(x+5)
Actual cost per student: 500/x
:
Actual cost = anticipated cost + $5
500/x = (500/(x+5)) + 5
:
Mult equation by x(x+5) to get rid of the denominators, resulting in:
500(x+5) = 500x + 5(x(x+5))
:
500x + 2500 = 500x + 5x^2 + 25x
:
0 = 5x^2 + 25x + 500x - 500x - 2500
:
A quadratic equation:
5x^2 + 25x - 2500 = 0
:
Simplify, divide equation by 5
x^2 + 5x - 500 = 0
:
Factor this to:
(x+25)(x-20) = 0
:
x = 20 actually made it (positive solution is what we want here)
:
:
Check our solution using the cost:
Actual cost: 500/20 = $25 each
:
Anticipated cost: 500/25 = $20 each

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