Question 1171682
<pre>
I like his method better.  But here is an alternate way to approach the
problem.

a group of people planned a trip that had a fixed cost of $30,000. later 20
more people wanted to go causing the average cost to decrease by $50 per
person. How many people were planning to go originally?
<pre>
At first, there were x people each paying $y and that made up the $30,000. To
say that mathematically, we put down what we do in equation form, multiply
people by dollars to get total dollars: 

(x)($y) = $30000

or just

     xy = 30000
 
But later, instead of only x people going on the trip, there were now x+20
people going.  Then the price per person went down from $y to $(y-50).  Again,
to say that mathematically, we put down what we do in equation form, multiply
people by dollars to get total dollars: 

     (x+20)(y-$50) = $30,000

or just

     (x+20)(y-50) = $30,000

So we have the system of equations

{{{system(xy=30000,(x+20)(y-50)=30000)}}}

Can you solve that system?  You'll get two pairs of answers for x and y, one
positive and one negative, and you'll discard the negative answer.

If you have trouble solving it, tell me about it down below.

Edwsin</pre>