Question 67483
Let x =  number of tickets sold before the game and y = the number of tickets sold the day of the game.  Now from the information provided, we know that tickets purchased before the day of the game cost $1.00 and those purchased the day of the game cost $1.50.  Also $700 was sold from 600 in sales of tickets.  From this information, we have:

x + y = 600, and
$1.00(x) + $1.50(y) = $700.

From the first of these equations we can solve for x
x + y = 600 ==> x = 600 - y (subtracting y from both sides)

Now, since we know x, we can plug this into the second equation above
$1.00(x) + $1.50(y) = $700 ==> $1.00(600-y) + $1.50(y) = $700

Now multiply the $1.00 into the 600 - y as follows:

$1.00(600-y) + $1.50(y) = $700 ==> 600 - y + 1.50y = $700
==> 600 + 0.5y = 700.  So now we can solve for y which is the number of tickets sold at the gate!

600 + 0.5y = 700 (subtract 600 from both sides)
0.5y = 100 (now divde both sides by 0.5)
y = 200

So, there were 200 tickets sold at the gate.  You could also figure out how many were sold before the game day by plugging this value of y into one of the original equations and solving for x, but this is not required in the question.  Only need to solve for y (the number of tickets sold at the gate)

Hope this helps!!