Question

simplify the complex fraction.
\\(\frac{\frac{9}{x + 4}-\frac{6x}{x + 5}}{\frac{3}{x - 4}-\frac{2x}{x - 3}}\\)

Explanation:

Step1: Find common - denominators for numerator and denominator

For the numerator $\frac{9}{x + 4}-\frac{6x}{x + 5}=\frac{9(x + 5)-6x(x + 4)}{(x + 4)(x + 5)}=\frac{9x+45-6x^{2}-24x}{(x + 4)(x + 5)}=\frac{-6x^{2}-15x + 45}{(x + 4)(x + 5)}$.
For the denominator $\frac{3}{x - 4}-\frac{2x}{x - 3}=\frac{3(x - 3)-2x(x - 4)}{(x - 4)(x - 3)}=\frac{3x-9-2x^{2}+8x}{(x - 4)(x - 3)}=\frac{-2x^{2}+11x - 9}{(x - 4)(x - 3)}$.

Step2: Rewrite the complex - fraction

The complex fraction $\frac{\frac{9}{x + 4}-\frac{6x}{x + 5}}{\frac{3}{x - 4}-\frac{2x}{x - 3}}$ becomes $\frac{\frac{-6x^{2}-15x + 45}{(x + 4)(x + 5)}}{\frac{-2x^{2}+11x - 9}{(x - 4)(x - 3)}}=\frac{-6x^{2}-15x + 45}{(x + 4)(x + 5)}\times\frac{(x - 4)(x - 3)}{-2x^{2}+11x - 9}$.
Factor out a $- 3$ from the numerator of the first fraction: $\frac{-3(2x^{2}+5x - 15)}{(x + 4)(x + 5)}\times\frac{(x - 4)(x - 3)}{-2x^{2}+11x - 9}$.

Step3: Simplify if possible

After factoring and simplifying (assuming no further common factors can be cancelled out), the simplified form is $\frac{(-6x^{2}-15x + 45)(x - 4)(x - 3)}{(-2x^{2}+11x - 9)(x + 4)(x + 5)}$.

Answer:

$\frac{(-6x^{2}-15x + 45)(x - 4)(x - 3)}{(-2x^{2}+11x - 9)(x + 4)(x + 5)}$