Find the largest value of \(c\) such that \(\frac{c^2 + 6c -27}{c-3} +2c = 23\).
Note that c^2 + 6c - 27 factors as (c + 9) (c - 3)
So we have
(c + 9) (c - 3)
__________ + 2c = 23
c - 3
c + 9 + 2c = 23
3c + 9 = 23 subtract 9 from both sides
3c = 14 divide both sides by 3
c = 14 / 3
