alexperi wrote:

A- original admission fee

P- original number of people

1) AP=(A-0.75)(P+100)

AP=AP+100A-0.75P -75

0=100A-0.75P-75

1 equation, 2 unknowns

2) AP=(A+1.5)(P-100)

AP=AP-100A+1.5P-150

0=-100A+1.5P-150

1 equation, 2 unknowns

1) + 2) 0=0+0.75P-225=> P=225*4/3=300

answer choice C

I completely agree with this, but it's a little too much. The moment you have the first equation, you should know you can't solve, and the same with the second one. Together, you've got two equations and two variables, so you should be able to solve.

The only risk here is that sometimes, with two equations, you can't solve. But if you learn the times when you can't, you can do a quick check for them and move on. They are:

1) When the equations are the same but look different

Example:

y = 3x + 6

6x = 2y - 12

These are actually the same equation.

2) When there are powers in one or both equation

Example:

y^2 + x = 9

x + 2y = 6

3) When one of the variables actually disappears, and you should have picked A or B

Example:

3x + 2y = 18

x + y = 4x + y + 12 <-- the y's cancel here so there's actually only 1 variable.

So look at the two equations and make sure quickly that you're not in any of those three situations (a very fast test when you get used to looking for them) and then the answer is C right away.