Question

1. $\frac{-3x - 9}{x^{2}+x - 6}-\frac{x + 3}{2 - x}$

Explanation:

Step1: Factor the denominator

Factor \(x^{2}+x - 6=(x + 3)(x-2)\).

Step2: Rewrite the second - fraction

\(\frac{x + 3}{2 - x}=-\frac{x + 3}{x - 2}\).

Step3: Find a common denominator

The common denominator of \(\frac{-3x - 9}{(x + 3)(x - 2)}\) and \(-\frac{x + 3}{x - 2}\) is \((x + 3)(x - 2)\). Rewrite \(-\frac{x + 3}{x - 2}\) as \(-\frac{(x + 3)(x + 3)}{(x + 3)(x - 2)}\).

Step4: Combine the fractions

\(\frac{-3x-9+(x + 3)(x + 3)}{(x + 3)(x - 2)}=\frac{-3x-9+x^{2}+6x + 9}{(x + 3)(x - 2)}\).

Step5: Simplify the numerator

Combine like - terms in the numerator: \(x^{2}+3x=x(x + 3)\).

Step6: Simplify the fraction

\(\frac{x(x + 3)}{(x + 3)(x - 2)}=\frac{x}{x - 2},x
eq - 3\).

Answer:

\(\frac{x}{x - 2},x
eq - 3\)