what is the difference?\n$\frac{2x + 5}{x^{2}-3x}-\frac{3x + 5}{x^{3}-9x}-\frac{x + 1}{x^{2}-9}$\n$\frac{(x…

what is the difference?\n$\frac{2x + 5}{x^{2}-3x}-\frac{3x + 5}{x^{3}-9x}-\frac{x + 1}{x^{2}-9}$\n$\frac{(x + 5)(x + 2)}{x^{3}-9x}$\n$\frac{(x + 5)(x + 4)}{x^{3}-9x}$\n$\frac{-2x + 11}{x^{3}-12x - 9}$\n$\frac{3(x + 2)}{x^{2}-3x}$

what is the difference?\n$\frac{2x + 5}{x^{2}-3x}-\frac{3x + 5}{x^{3}-9x}-\frac{x + 1}{x^{2}-9}$\n$\frac{(x + 5)(x + 2)}{x^{3}-9x}$\n$\frac{(x + 5)(x + 4)}{x^{3}-9x}$\n$\frac{-2x + 11}{x^{3}-12x - 9}$\n$\frac{3(x + 2)}{x^{2}-3x}$

Answer

Explanation:

Step1: Factor the denominators

$x^{2}-3x=x(x - 3)$, $x^{3}-9x=x(x^{2}-9)=x(x + 3)(x - 3)$, $x^{2}-9=(x + 3)(x - 3)$ The least - common denominator (LCD) of the three fractions is $x(x + 3)(x - 3)=x^{3}-9x$.

Step2: Rewrite each fraction with the LCD

$\frac{2x + 5}{x^{2}-3x}=\frac{(2x + 5)(x + 3)}{x(x - 3)(x + 3)}=\frac{2x^{2}+6x+5x + 15}{x^{3}-9x}=\frac{2x^{2}+11x + 15}{x^{3}-9x}$ $\frac{3x + 5}{x^{3}-9x}$ remains the same. $\frac{x + 1}{x^{2}-9}=\frac{(x + 1)x}{(x + 3)(x - 3)x}=\frac{x^{2}+x}{x^{3}-9x}$

Step3: Subtract the fractions

$\frac{2x^{2}+11x + 15-(3x + 5)-(x^{2}+x)}{x^{3}-9x}=\frac{2x^{2}+11x + 15-3x - 5-x^{2}-x}{x^{3}-9x}$ $=\frac{(2x^{2}-x^{2})+(11x-3x - x)+(15 - 5)}{x^{3}-9x}=\frac{x^{2}+7x + 10}{x^{3}-9x}$ Factor the numerator: $x^{2}+7x + 10=(x + 5)(x+2)$ So the result is $\frac{(x + 5)(x + 2)}{x^{3}-9x}$

Answer:

$\frac{(x + 5)(x + 2)}{x^{3}-9x}$