Question

divide.
enter your answer as a single completely simplified rational expression with nonnegative exponents.
\\(\frac{x + 3}{4 - x} \div \frac{x^2 - 9}{x - 4}\\)

Explanation:

Step1: Rewrite division as multiplication

Recall that dividing by a fraction is the same as multiplying by its reciprocal. So, \(\frac{x + 3}{4 - x}\div\frac{x^{2}-9}{x - 4}=\frac{x + 3}{4 - x}\times\frac{x - 4}{x^{2}-9}\)

Step2: Factor expressions

Factor \(x^{2}-9\) using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(a = x\) and \(b = 3\), so \(x^{2}-9=(x + 3)(x - 3)\). Also, notice that \(4 - x=-(x - 4)\), so we can rewrite \(\frac{x + 3}{4 - x}\) as \(\frac{x + 3}{-(x - 4)}=-\frac{x + 3}{x - 4}\)
Now the expression becomes: \(-\frac{x + 3}{x - 4}\times\frac{x - 4}{(x + 3)(x - 3)}\)

Step3: Cancel common factors

Cancel out the common factors \((x + 3)\) and \((x - 4)\) from the numerator and the denominator.
After canceling, we are left with \(-\frac{1}{x - 3}\) or we can rewrite it as \(\frac{1}{3 - x}\) (by multiplying numerator and denominator by - 1)

Answer:

\(\frac{1}{3 - x}\) (or \(-\frac{1}{x - 3}\))