Question

What is the correct answer: Add the rational expressions to find the sum: -4/x-4 +x2-2x/2x-8 put the answer into its simplest form.

Answer 1

To add the rational expressions \(-\frac{4}{x-4} + \frac{x^2 - 2x}{2x - 8}\), we need to get a common denominator and simplify the expression.

First, observe that:
\[
2x - 8 = 2(x - 4).
\]

This allows us to rewrite the second fraction:
\[
\frac{x^2 - 2x}{2x - 8} = \frac{x^2 - 2x}{2(x - 4)}.
\]

Now we combine the two fractions:
\[
-\frac{4}{x-4} + \frac{x^2 - 2x}{2(x - 4)}.
\]

The common denominator is \(2(x - 4)\). We will convert the first fraction:
\[
-\frac{4}{x - 4} = -\frac{4 \cdot 2}{2(x - 4)} = -\frac{8}{2(x - 4)}.
\]

Now we can add the two fractions:
\[
-\frac{8}{2(x - 4)} + \frac{x^2 - 2x}{2(x - 4)} = \frac{-8 + (x^2 - 2x)}{2(x - 4)} = \frac{x^2 - 2x - 8}{2(x - 4)}.
\]

Next, we simplify the numerator \(x^2 - 2x - 8\). We factor it as follows:
\[
x^2 - 2x - 8 = (x - 4)(x + 2).
\]

So, we substitute that back into our fraction:
\[
\frac{(x - 4)(x + 2)}{2(x - 4)}.
\]

We can cancel \((x - 4)\) from the numerator and denominator (as long as \(x \neq 4\)):
\[
\frac{x + 2}{2}, \text{ for } x \neq 4.
\]

Thus, the final answer in its simplest form is:
\[
\frac{x + 2}{2}, \quad (x \neq 4).
\]