Question #N503
If \(\frac{x^2 - 9}{x + 3} = 8\), and \(x \neq -3\), what is the value of \(x\)?
A. -11
B. -5
C. 5
D. 11
Correct Answer is: D
We can factor the numerator of the fraction, giving us \(\frac{(x + 3)(x - 3)}{x + 3} = 8\). Since \(x \neq -3\), we can cancel the \(x + 3\) terms, leaving us with \(x - 3 = 8\). Adding 3 to both sides gives us \(x = 11\).