How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$

Question

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 $$ \small\begin{align} \frac1{x^2-5x+6} &=\frac1{(x-2)(x-3)} =\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right) =\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\ &=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}} =\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n =\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n} \end{align} $$ Original image

Answer 1

Each of the terms was multiplied by $\frac{-1}{-1}$, which is really equal to $1$, so it's a "legal" thing to do:

$-\dfrac{1}{x - 2} + \dfrac{1}{x - 3}$

$ = -\dfrac{(-1)1}{(-1)(x - 2)} + \dfrac{(-1)1}{(-1)(x - 3)}$

$ = -\dfrac{-1}{2 - x} + \dfrac{-1}{3 - x}$

$ = \dfrac{1}{2 - x} - \dfrac{1}{3 - x} $

Answer 2

$$ \frac{1}{x-a} = \frac{1}{-(a - x)} = - \frac{1}{a - x} $$

Answer 3

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{align} -\frac {1} {x-2} & = -\frac {1} {8-2}\\ & = -\frac {1} {6} \end{align}

And same for this,

\begin{align} \frac {1} {2-8} & =\frac {1} {-6}\\ & = -\frac {1} {6} \end{align}

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{align} \frac {1} {8-3} & =\frac {1} {5}\\ \end{align}

and the same for this,

\begin{align} -\frac {1} {3-8} & =-\frac {1} {-5}\\ & = \frac {-1} {-5}\\ & = \frac {1} {5}\\ \end{align}

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$\times$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{align} -\frac {1} {x-2} & = \frac {-1} {x-2}\\ & = \frac {-1(-1)} {-1(x-2)}\\ & = \frac {1} {-x+2}\\ & = \frac {1} {2-x}\\ \end{align}

same goes to $\frac {1} {x-3}=-\frac {1} {3-x}$

Answer 4

This problem all boils down to the following relationship $$-1 = \frac{-1}{1}=\frac{1}{-1}$$

Part one is easy if you just express the division as a multiplication $$x=\frac{-1}{1}\implies -1=1\cdot x\implies -1=x$$ For part two, $$x=\frac{1}{-1}\implies1=-1\cdot x$$ $$1+(-1)+x=-1\cdot x+(-1)+x$$ $$x+0=-1\cdot x + 1\cdot x + (-1)$$ $$x=x((-1)+1)+(-1)$$ $$x=x((-1)+1)+(-1)$$ $$x=0x+(-1)$$ $$x=0+(-1)$$ $$x=-1$$ This assumes that $0x=0$ $$0x=(0+0)x$$ $$0x=0x+0x$$ $$0x+(-0x)=0x+0x+(-0x)$$ $$0=0x+0$$ $$0=0x$$

Answer 5

We have $- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{-1} \times \frac{1}{x-2} + \frac{-1}{-1} \times \frac{1}{x-3} = \frac{1}{(-1)\times (x-2)} + \frac{(-1)\times 1}{(-1)\times (x-3)} = \frac{1}{-x+2} + \frac{-1}{-x+3}= \frac{1}{-x+2} - \frac{1}{-x+3} $. And we have the result by just nature of your own question.