Question 349557
you want to raise 4000.


let x = original number of members.


let y = amount required from each member.


you get:


y = 4000 / x


with an increase of 20 members, the amount each member needed to raise went down by 10.


you get:


y-10 = 4000 / (x + 20)


you have 2 equations that need to be solved simultaneously.


since y = 4000 / x, you can substitute for y in the second equation to get:


(4000 / x) - 10 = 4000 / (x + 20)


multiply both sides of this equation by (x + 20) * x to get:


4000 * (x + 20) - (10 * (x + 20) * x) = 4000 * x


Simplify by removing parentheses to get:


4000*x + 80,000 - 10*x^2 - 200*x = 4000 * x


subtract 4000 * x from both sides of this equation to get:


80,000 - 10*x^2 - 200*x = 0


re=order the terms to get:


-10*x^2 - 200*x + 80,000 = 0


multiply both sides of this equation by -1 to get:


10x^2 + 200*x - 80,000 = 0


you don't have to do that, but it's easier to factor the equation if the coefficient of the x^2 term is positive.


divide both sides of the equation by 10 to get:


x^2 + 20*x - 8000 = 0


factor the equation to get:


(x + 100) * (x - 80) = 0


your values of x can be either -100 or +80.


since x has to be positive, then your answer is x = 80.


substitute in your original equation to confirm this is a good solution.


y = 4000 / x becomes y = 4000 / 80 becomes y = 50


with the original number of members, each member is required to provide 50.


y - 10 would then equal 40.


your second equation is:


y-10 = 4000 / (x + 20)


y - 10 = 40
x + 20 = 100


you get:


40 = 4000 / 100 which becomes:


40 = 40 which is true.


when you add 20 members, each member now has to pay 40.


solution is confirmed as good.


there were 80 original members in the club.