algabraic fraction?

Question

The $75 cost of a pizza party was to be shared equally by all those attending. Since five more people attended than was expected, the price per person dropped by $0.50. How many people attended the party?

I'm not really sure how to even set this up...

THANK YOU SOO MUCH MOONMAN

Answer 1

Start by letting p=number of people attending. You are given that:

75/p -.5 = 75/(p-5)

Multiply through by 2p(p-5) to get:
150(p-5) - p(p-5) = 150p

Simplify to get:
p^2 + 5p + 750 = 0

Factor (or use the quadratic formula) to get p=-30,25. Since p must be positive, p=25

Answer 2

1st
n = number of people at party
x = amount paid by each person for pizza

so

x = 75/n

but since 5 more people came the actual cost dropped by 0.5 dollars

now we have

x - .5 = 75/(n+5)

substitute x = 75/n from above for x

(75/n) - .5 = 75/(n+5)
(75 - .5n)/n = 75/(n+5) then cross multiply

75n = (75 - .5n)(n + 5)
75n = -.5n^2 + 375 - 2.5n + 75n the 75n cancel

so

0 = -.5n^2 - 2.5n + 375

use quadratic equation n = -30 or 25

so 25 people attended the party

Answer 3

Let

x = number of people actually attending party

Then

75/(x - 5) = 75/x + .50
75x = 75(x - 5) + (1/2)x(x - 5)
5*75 = (1/2)(x² - 5x) = x²/2 - 5x/2
375 = x²/2 - 5x/2
x²/2 - 5x/2 - 375 = 0
x² - 5x - 750 = 0
(x - 30)(x + 25) = 0
x = 30, the negative solution is rejected

30 people attended the party.

Answer 4

The original price per person p = 75/n where n= the number of people. With the five additional people teh equation is p-.5= 75/(n+5). Assuming you know hor to solce a system of equations, you can solve it from there.

Answer 5

z^2 - y^2.....z + y ------------ ÷ --------- (multiply via the reciprocal) a^2 - b^2....a - b z^2 - y^2....a - b ------------ × -------- (factor the numerator and denominator of the 1st fraction) a^2 - b^2....z + y (z + y)(z - y)....a - b ----------------- × ---------- (cancel out (z + y) diagonally and (a - b) diagonally) (a + b)(a - b)....z + y z - y -------- <===answer a + b