Question

1. solve \\(\frac{x^2 - 81}{x + 9} \div \frac{x^2 + 11x + 18}{x + 2}\\). completely simplify your answer and state any restrictions on the variable.\

\\(\bigcirc\\) \\(1, x \
eq -9, x \
eq -2\\)\
\\(\bigcirc\\) \\(\frac{x - 9}{x + 9}, x \
eq -9, x \
eq -2\\)\
\\(\bigcirc\\) \\(\frac{x - 9}{x + 2}, x \
eq -9, x \
eq -2\\)\
\\(\bigcirc\\) \\(\frac{x + 9}{x - 9}, x \
eq -9, x \
eq -2\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{x^2 - 81}{x + 9} \times \frac{x + 2}{x^2 + 11x + 18}$

Step2: Factor numerators/denominators

$x^2 - 81 = (x-9)(x+9)$; $x^2 +11x+18=(x+2)(x+9)$
Expression becomes: $\frac{(x-9)(x+9)}{x+9} \times \frac{x+2}{(x+2)(x+9)}$

Step3: Cancel common factors

Cancel $(x+9)$ and $(x+2)$: $\frac{x-9}{x+9}$

Step4: Find restrictions

Denominators can’t be zero: $x+9
eq 0 \Rightarrow x
eq -9$; $x+2
eq 0 \Rightarrow x
eq -2$

Answer:

$\frac{x-9}{x+9}, x
eq -9, x
eq -2$ (Note: Assuming the second option has a typo and should be $\frac{x-9}{x+9}$ instead of $\frac{x-9}{z+9}$; if the original second option is $\frac{x-9}{x+9}$, then it is the correct choice)