Question

1. $\frac{4x^{2}+x - 6}{x^{2}+3x + 2}-\frac{3x}{x + 1}+\frac{5}{x + 2}$

Explanation:

Step1: Factor the denominators

Factor $x^{2}+3x + 2=(x + 1)(x+2)$.

Step2: Find a common denominator

The common denominator of $\frac{4x^{2}+x - 6}{(x + 1)(x + 2)}$, $\frac{3x}{x + 1}$ and $\frac{5}{x + 2}$ is $(x + 1)(x + 2)$. Rewrite $\frac{3x}{x + 1}=\frac{3x(x + 2)}{(x + 1)(x + 2)}$ and $\frac{5}{x + 2}=\frac{5(x + 1)}{(x + 1)(x + 2)}$.

Step3: Combine the fractions

\[

$$\begin{align*} &\frac{4x^{2}+x - 6}{(x + 1)(x + 2)}-\frac{3x(x + 2)}{(x + 1)(x + 2)}+\frac{5(x + 1)}{(x + 1)(x + 2)}\\ =&\frac{4x^{2}+x - 6-3x^{2}-6x + 5x + 5}{(x + 1)(x + 2)}\\ =&\frac{4x^{2}-3x^{2}+x-6x + 5x-6 + 5}{(x + 1)(x + 2)}\\ =&\frac{x^{2}-1}{(x + 1)(x + 2)}\\ =&\frac{(x + 1)(x - 1)}{(x + 1)(x + 2)}\\ =&\frac{x - 1}{x + 2},x eq - 1 \end{align*}$$

\]

Answer:

$\frac{x - 1}{x + 2},x
eq - 1$