Question

1. simplify the expression below.

\\(\frac{2n}{n + 1}+\frac{n - 3}{n^2 - 1}-\frac{7}{n - 1}\\)
a. \\(\frac{2(n - 5)}{n - 1}\\)
c. \\(\frac{2(n + 5)}{n + 1}\\)
b. \\(2(n - 5)(n - 1)\\)
d. \\(\frac{2}{(n - 5)(n + 1)}\\)

Explanation:

Step1: Factor the common denominator

Note that $n^2-1=(n+1)(n-1)$, so the least common denominator (LCD) is $(n+1)(n-1)$.

Step2: Rewrite each term with LCD

$$\frac{2n}{n+1} = \frac{2n(n-1)}{(n+1)(n-1)}, \quad \frac{n-3}{n^2-1} = \frac{n-3}{(n+1)(n-1)}, \quad \frac{7}{n-1} = \frac{7(n+1)}{(n+1)(n-1)}$$

Step3: Combine into single fraction

$$\frac{2n(n-1) + (n-3) - 7(n+1)}{(n+1)(n-1)}$$

Step4: Expand numerator terms

$$2n(n-1)=2n^2-2n, \quad 7(n+1)=7n+7$$
Substitute back:
$$\frac{2n^2-2n + n - 3 -7n -7}{(n+1)(n-1)}$$

Step5: Simplify numerator

Combine like terms:
$$2n^2 + (-2n+n-7n) + (-3-7) = 2n^2-8n-10$$
Factor numerator:
$$2n^2-8n-10=2(n^2-4n-5)=2(n-5)(n+1)$$

Step6: Cancel common factors

$$\frac{2(n-5)(n+1)}{(n+1)(n-1)} = \frac{2(n-5)}{n-1}$$

Answer:

A. $\frac{2(n-5)}{n-1}$