SOLUTION: I am having a hard time solving this problem. A movie theater had a certain number of tickets to give away. Five people got tickets. The first got one-third of the tickets, the

Algebra ->  Inequalities -> SOLUTION: I am having a hard time solving this problem. A movie theater had a certain number of tickets to give away. Five people got tickets. The first got one-third of the tickets, the       Log On


   

Question 176928: I am having a hard time solving this problem.
A movie theater had a certain number of tickets to give away. Five people got tickets. The first got one-third of the tickets, the second got one-fourth of the tickets, and the third got one-fifth of the tickets. The fourth person got eight tickets, and there were five left for the fifth person. Find the total number of tickets given away.
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the total number of tickets
:
(1/3)x+(1/4)x+(1/5)x+13---->represents all the tickets
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so if we subtract the total number of tickets represented by x then the result should be zero.
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here we go:
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(1/3)x+(1/4)x+(1/5)x+13)-x=0
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20x+15x+12x+780-60x=0.....multiplied all terms by 60 to get rid of fractions
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-13x=-780.....combined like terms subtracted 780 from both sides
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highlight%28x=60%29total number of tickets available distributed to 5 people