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How does $-\frac{1}{x-2}+\frac{1}{x-3}$ become $\frac{1}{2-x}-\frac{1}{3-x}$
I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2}+\frac{1}{x-3}$ becomes $\frac{1}{2-x}-\frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$\begin{array}{rl}\frac{1}{{x}^{2}-5x+6}& =\frac{1}{\left(x-2\right)\left(x-3\right)}=\frac{1}{-3-\left(-2\right)}\left(\frac{1}{x-2}-\frac{1}{x-3}\right)=-\frac{1}{x-2}+\frac{1}{x-3}\\ & =\frac{1}{2-x}-\frac{1}{3-x}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{2}^{n+1}}{x}^{n}-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{3}^{n+1}}{x}^{n}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{{2}^{n+1}}-\frac{1}{{3}^{n+1}}\right){x}^{n}\end{array}$
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Aryanna Caldwell
Each of the terms was multiplied by $\frac{-1}{-1}$, which is really equal to 1, so it's a "legal" thing to do:
$-\frac{1}{x-2}+\frac{1}{x-3}$
$=-\frac{\left(-1\right)1}{\left(-1\right)\left(x-2\right)}+\frac{\left(-1\right)1}{\left(-1\right)\left(x-3\right)}$
$=-\frac{-1}{2-x}+\frac{-1}{3-x}$
$=\frac{1}{2-x}-\frac{1}{3-x}$
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Frederick Kramer
I am a grade 8 student, so I may not be able to explain really well.
First, I need to prove that $-\frac{1}{x-2}=\frac{1}{2-x}$
To prove, let's assume that "x" can be any number, for instance, I take x=8.
So by substituting,
$\begin{array}{rl}-\frac{1}{x-2}& =-\frac{1}{8-2}\\ & =-\frac{1}{6}\end{array}$
And same for this,
$\begin{array}{rl}\frac{1}{2-8}& =\frac{1}{-6}\\ & =-\frac{1}{6}\end{array}$
Therefore, we have proven that $-\frac{1}{x-2}=\frac{1}{2-x}$
I also need to prove that $\frac{1}{x-3}=-\frac{1}{3-x}$
So by substituting,
$\begin{array}{rl}\frac{1}{8-3}& =\frac{1}{5}\end{array}$
and the same for this,
$\begin{array}{rl}-\frac{1}{3-8}& =-\frac{1}{-5}\\ & =\frac{-1}{-5}\\ & =\frac{1}{5}\end{array}$
Therefore, we have proven that $\frac{1}{x-3}=-\frac{1}{3-x}$
By why it worked? The truth is, it is just having -1÷(-1)=1 (negative×negative=positive)(And anything times 1 is the same number)
So, from $-\frac{1}{x-2}$ to $\frac{1}{2-x}$ , they inserted both -1 for numerator and denominator as the following below.
$\begin{array}{rl}-\frac{1}{x-2}& =\frac{-1}{x-2}\\ & =\frac{-1\left(-1\right)}{-1\left(x-2\right)}\\ & =\frac{1}{-x+2}\\ & =\frac{1}{2-x}\end{array}$
same goes to $\frac{1}{x-3}=-\frac{1}{3-x}$