SOLUTION: 150 people attended a conference. $4,200 was collected in registration fees. If members were charged $25 and non-members were charged $35, how many members attended?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: 150 people attended a conference. $4,200 was collected in registration fees. If members were charged $25 and non-members were charged $35, how many members attended?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 262771: 150 people attended a conference. $4,200 was collected in registration fees. If members were charged $25 and non-members were charged $35, how many members attended?
Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
We can set up 2 equations like this: Let x represent the members and y be the non-members.
x+y=150
25x+35y=4200
Now, let's multiply the first equation by -25 to give us:
-25x-25y=-3750 Add this to the second equation:
25x+35y=4200 And we get:
10y=450 Divide each side by 10 to get:
y=45
There we go!
We now know there were 45 non-members. Plug this into either original equation to find the number of members.
I'll let you figure this part.
Make sense?
RJ
Check out a new book I wrote at:
www.math-unlock.com