Question

find the difference
(1/x+3)-(2/x-1)-(2x/x^2+2x-3)

Answer 1

To find the difference

\[
\left( \frac{1}{x+3} \right) - \left( \frac{2}{x-1} \right) - \left( \frac{2x}{x^2 + 2x - 3} \right)
\]

we will first simplify each term and then combine them.

1. **Factor the third term's denominator:**
The expression \( x^2 + 2x - 3 \) can be factored:
\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\]
So we rewrite the third term:
\[
\frac{2x}{(x + 3)(x - 1)}
\]

2. **Rewrite the expression:**
Now our expression looks like this:
\[
\frac{1}{x+3} - \frac{2}{x-1} - \frac{2x}{(x + 3)(x - 1)}
\]

3. **Find a common denominator:**
The common denominator for \( \frac{1}{x+3} \) and \( \frac{2}{x-1} \) is \( (x+3)(x-1) \). Therefore, we will rewrite the fractions:
- For \( \frac{1}{x+3} \):
\[
\frac{1}{x+3} = \frac{x-1}{(x+3)(x-1)}
\]
- For \( \frac{2}{x-1} \):
\[
\frac{2}{x-1} = \frac{2(x+3)}{(x-1)(x+3)} = \frac{2x + 6}{(x-1)(x+3)}
\]

4. **Combine the fractions:**
Now substitute back into the expression:
\[
\frac{x-1}{(x+3)(x-1)} - \frac{2x + 6}{(x+3)(x-1)} - \frac{2x}{(x + 3)(x - 1)}
\]
Combine them under a single denominator:
\[
\frac{x-1 - (2x + 6) - 2x}{(x+3)(x-1)}
\]
Simplifying the numerator:
\[
= \frac{x - 1 - 2x - 6 - 2x}{(x+3)(x-1)} = \frac{-3x - 7}{(x+3)(x-1)}
\]

5. **Final answer:**
So, the difference is:
\[
\frac{-3x - 7}{(x+3)(x-1)}
\]

In simplified form, this would be the final result for the given expression.