Good day everyone.

I have three questions relating to simplifying algebraic questions, and what intimidates me the most are the denominators. Here are the questions:

1. $\displaystyle \dfrac{4}{(x -2)(x -4)}$ $\displaystyle - \dfrac{2}{(x -2)(x -3)}$

2. $\displaystyle \dfrac{3x}{x -3}$ $\displaystyle - \dfrac{x}{x^2 -9}$

3. $\displaystyle \dfrac{x -4}{(x +1)(x -5)}$ $\displaystyle -\dfrac{x +5}{(x +1)(x +3)}

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Here are my workings respectively:

1. $\displaystyle \dfrac{4}{(x -2)(x -4)}$ $\displaystyle - \dfrac{2}{(x -2)(x -3)}$

$\displaystyle =\dfrac{4(x -3) -2(x-4)}{(x -2)(x -3)(x -4)}$

$\displaystyle =\dfrac{4x -12 -2x +8}{(x-2)(x-3)(x-4)}$

$\displaystyle =\dfrac{4x-2x-12+8}{(x-2)(x-3)(x-4)}$

$\displaystyle =\dfrac{2x -4}{(x-2)(x-3)(x-4)}$ ?

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2. $\displaystyle \dfrac{3x}{x -3}$ $\displaystyle - \dfrac{x}{x^2 -9}$

$\displaystyle =\dfrac{3x}{x-3}$ $\displaystyle -\dfrac{x}{(x-3)(x+3)}$

$\displaystyle =\dfrac{3x(x-3)(x+3) -x(x-3)}{(x-3)^2(x+3)}$

$\displaystyle =\dfrac{3x^3+9x^2-9x^2-27x-x^2+3x}{(x-3)^2(x+3)}$

$\displaystyle =\dfrac{3x^3-24x-x^2}{(x-3)^2(x+3)}$ ?

I am so sorry that I couldn't do it further. I felt mortified when my answers are wrong. Please find my faults and explain how I should have done. Thank you.

PythagorasNeophyte