Question

what is the simplest form of \\(\frac{x^2 + 5x - 6}{x^2 + 9x + 18}\\)?\
\\(\circ\\) \\(\frac{x + 2}{x + 6}\\)\
\\(\circ\\) \\(\frac{1}{3}\\)\
\\(\circ\\) \\(\frac{x - 1}{x + 3}\\)\
\\(\circ\\) \\(-\frac{1}{3}\\)

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 + 5x - 6\): Find two numbers that multiply to \(-6\) and add to \(5\), which are \(6\) and \(-1\). So \(x^2 + 5x - 6=(x + 6)(x - 1)\).
Factor \(x^2 + 9x + 18\): Find two numbers that multiply to \(18\) and add to \(9\), which are \(6\) and \(3\). So \(x^2 + 9x + 18=(x + 6)(x + 3)\).

Step2: Simplify the fraction

Substitute the factored forms into the original fraction:
\(\frac{(x + 6)(x - 1)}{(x + 6)(x + 3)}\)
Cancel out the common factor \((x + 6)\) (assuming \(x
eq - 6\)):
\(\frac{x - 1}{x + 3}\)

Answer:

\(\frac{x - 1}{x + 3}\) (corresponding to the option with this expression)