Question

add the rational expressions to find the sum: $\frac{-4}{x - 4}+\frac{x^{2}-2x}{2x - 8}$. put the answer into its simplest form. (1 point)
$\frac{x^{2}-2x}{2x}$
$\frac{x + 2}{2}$
$\frac{x^{2}-2x - 8}{2x - 8}$
$\frac{x^{2}-2x - 4}{x - 4}$

Explanation:

Step1: Simplify the second denominator

Factor \(2x - 8\) as \(2(x - 4)\). So the expression becomes \(\frac{-4}{x - 4}+\frac{x^{2}-2x}{2(x - 4)}\).

Step2: Find a common denominator

The common denominator is \(2(x - 4)\). Rewrite the first fraction: \(\frac{-4\times2}{2(x - 4)}+\frac{x^{2}-2x}{2(x - 4)}=\frac{-8 + x^{2}-2x}{2(x - 4)}\).

Step3: Simplify the numerator

Rearrange the numerator: \(x^{2}-2x - 8\). Factor it: \((x - 4)(x + 2)\). So the fraction is \(\frac{(x - 4)(x + 2)}{2(x - 4)}\).

Step4: Cancel common factors

Cancel out \((x - 4)\) (assuming \(x
eq4\)), we get \(\frac{x + 2}{2}\).

Answer:

B. \(\frac{x + 2}{2}\)