Homework Help: math

Posted by Janice - Please Help on Thursday, March 24, 2011 at 9:00pm.

In 2010 a city raised $5,999,400 in tax revenue by charging each taxpayer the same amount. For 2011 there has been a net loss of 12 taxpayers, necessitating an increase fo $200 per taxpayer in order to maintain the tax revenue of $5,999,400. How many taxpayers were there in 2010?

math - MathMate, Saturday, March 26, 2011 at 10:05pm
Let n = number of taxpayers in 2010.
So
tax for each in 2010 = 5999400/n

For 2011, there are 12 taxpayers less, and each one pays $200 more, so
(n-12)*(5999400/n+200)=5999400
Multiply each side by n:
(n-12)(5999400+200n)=5999400
Solve for n to get:
-594 or 606. The negative root is discarded to leave 606 as the number of taxpayers in 2010.

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