Question

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

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

\frac{(x + 5)(x + 4)}{x^{3}-9x}

\frac{-2x + 11}{x^{3}-12x - 9}

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

Explanation:

Step1: Factor the denominators

$x^{2}-3x=x(x - 3)$, $x^{2}-9x=x(x - 9)$, $x^{2}-9=(x + 3)(x - 3)$

Step2: Find the least - common denominator (LCD)

The LCD of $x(x - 3)$, $x(x - 9)$ and $(x + 3)(x - 3)$ is $x(x - 3)(x - 3)(x + 3)=x(x^{2}-9)$

Step3: Rewrite each fraction with the LCD

$\frac{2x + 5}{x^{2}-3x}=\frac{(2x + 5)(x - 3)(x + 3)}{x(x - 3)(x - 3)(x + 3)}=\frac{(2x + 5)(x^{2}-9)}{x(x^{2}-9)}$
$\frac{3x + 5}{x^{2}-9x}=\frac{(3x + 5)(x - 3)(x + 3)}{x(x - 9)(x - 3)(x + 3)}=\frac{(3x + 5)(x^{2}-9)}{x(x^{2}-9)}$
$\frac{x + 1}{x^{2}-9}=\frac{(x + 1)\cdot x\cdot(x - 3)}{x(x - 3)(x + 3)(x - 3)}=\frac{(x + 1)x(x - 3)}{x(x^{2}-9)}$

Step4: Subtract the fractions

\[

$$\begin{align*} &\frac{(2x + 5)(x^{2}-9)-(3x + 5)(x^{2}-9)-(x + 1)x(x - 3)}{x(x^{2}-9)}\\ =&\frac{(2x + 5-3x - 5)(x^{2}-9)-(x^{2}+x)(x - 3)}{x(x^{2}-9)}\\ =&\frac{(-x)(x^{2}-9)-(x^{3}-3x^{2}+x^{2}-3x)}{x(x^{2}-9)}\\ =&\frac{-x^{3}+9x-(x^{3}-2x^{2}-3x)}{x(x^{2}-9)}\\ =&\frac{-x^{3}+9x - x^{3}+2x^{2}+3x}{x(x^{2}-9)}\\ =&\frac{-2x^{3}+2x^{2}+12x}{x(x^{2}-9)}\\ =&\frac{-2x^{2}+2x + 12}{x^{2}-9}\\ =&\frac{-2(x^{2}-x - 6)}{x^{2}-9}\\ =&\frac{-2(x - 3)(x+2)}{(x - 3)(x + 3)}\\ =&\frac{-2(x + 2)}{x + 3} \end{align*}$$

\]
After re - checking the steps, let's start over with a different approach:

Step1: Factor the denominators

$x^{2}-3x=x(x - 3)$, $x^{2}-9x=x(x - 9)$, $x^{2}-9=(x + 3)(x - 3)$
The LCD of $x^{2}-3x,x^{2}-9x,x^{2}-9$ is $x(x - 3)(x + 3)=x^{3}-9x$

Step2: Rewrite the fractions with the LCD

$\frac{2x + 5}{x^{2}-3x}=\frac{(2x + 5)(x + 3)}{x(x - 3)(x + 3)}=\frac{2x^{2}+6x+5x + 15}{x^{3}-9x}=\frac{2x^{2}+11x + 15}{x^{3}-9x}$
$\frac{3x + 5}{x^{2}-9x}=\frac{(3x + 5)(x - 3)}{x(x - 9)(x - 3)}=\frac{3x^{2}-9x+5x - 15}{x^{3}-9x}=\frac{3x^{2}-4x - 15}{x^{3}-9x}$
$\frac{x + 1}{x^{2}-9}=\frac{(x + 1)x}{x(x - 3)(x + 3)}=\frac{x^{2}+x}{x^{3}-9x}$

Step3: Subtract the fractions

\[

$$\begin{align*} &\frac{2x^{2}+11x + 15-(3x^{2}-4x - 15)-(x^{2}+x)}{x^{3}-9x}\\ =&\frac{2x^{2}+11x + 15-3x^{2}+4x + 15-x^{2}-x}{x^{3}-9x}\\ =&\frac{(2x^{2}-3x^{2}-x^{2})+(11x+4x - x)+(15 + 15)}{x^{3}-9x}\\ =&\frac{-2x^{2}+14x + 30}{x^{3}-9x}\\ =&\frac{-2(x^{2}-7x - 15)}{x^{3}-9x}\\ =&\frac{(x + 5)(x+2)}{x^{3}-9x} \end{align*}$$

\]

Answer:

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