This is an example of a relatively straightforward word problem that can be solved through a system of equations. That said, first look to logically eliminate any impossible answers.

Since we are seeking only the fixed annual expense, we can eliminate Choice E as $830,000 is the total expense in 2009. Additionally, since both years have total expenses in the hundreds of thousands of dollars, it is exceedingly unlikely that the fixed expense would not be in the hundreds of thousands, so eliminate Choice A and Choice B as well.

Now, to actually solve create two equations using the variable x to represent the fixed cost and y to represent the variable cost.

2009: x + 450y = 830,000
2010: x + 510y = 896,000

Then, subtract the 2009 equation from the 2009 equation to eliminate the variable x as follows:

x + 510y = 896,000
-(x + 450y = 830,000)
60y = 66,000

Divide that equation by 60 to determine that y = 1,100.

Finally plug 1,100 back into one of the original equations as the value for y and solve for x.

x + 450(1100) = 830,000 -> x + 495,000 = 830,000

Subtract 495,000 from 830,000 to determine that the fixed annual cost = 335,000, so select Choice C.