Express as a single fraction: \(\frac{x}{x-2}-\frac{x+2}{x+3}\)
- \(\frac{2x^2 - 3x - 4}{(x-2)(x+3)}\)
- \(\frac{2x^2 + 3x - 4}{(x-2)(x+3)}\)
- \(\frac{2}{(x-2)(x+3)}\)
-
\(\frac{ 3x + 4}{(x-2)(x+3)}\)
- \(\frac{3x - 4}{(x-2)(x+3)}\)
The correct answer is: D
Explanation
\(\frac{x}{x-2}-\frac{x+2}{x+3}\)
\(\frac{[x][x+3] - ([x+2][x-2])}{[x-2][x+3]}\)
= \(\frac{x^2 + 3x - (x^2 - 4)}{[x-2][x+3]}\)
= \(\frac{3x + 4}{[x-2][x+3]}\)