# 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 \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

• $\frac{-1}{x-2}=\frac{(-1)(-1)}{(-1)(x-2)}=\frac1{2-x}$
– robjohn
May 10 '15 at 3:41
• So one of the two (-1)'s in the numerator and the (-1) in the denominator don't cancel one of the (-1)'s in the numerator out? It looks like that would leave a negative one back in the numerator. May 10 '15 at 3:45

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

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

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

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

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

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 $\frac {1} {x-3}=-\frac {1} {3-x}$

• @MartinSleziak isn't your edit...a bit the same? May 10 '15 at 5:19
• I have just added space after commas, full stops, etc. (And, accidentally, introduced one typo. Sorry for that.) May 10 '15 at 5:24
• @MartinSleziak Ahh,I see.Doesn't matter :) May 10 '15 at 5:33

It's very easy. It comes by using factorization and simplification rules in general. In the case of your question, we have

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

and in the case of $\frac{1}{x-3}$, by multiplying both denominator and numerator with $(-1)$, we have that

$\frac{1}{x-3} = \frac{(-1)}{(-1)} \times \frac{1}{x-3} = \frac{(-1)}{(-x+3)}= -\frac{1}{(3 - x)}$. So we will have

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

And we are done.

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.

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

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 $0x=0$ $$0x=(0+0)x$$ $$0x=0x+0x$$ $$0x+(-0x)=0x+0x+(-0x)$$ $$0=0x+0$$ $$0=0x$$

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

Thus $$-\frac{1}{x-2} + \frac{1}{x - 3} = \frac{1}{2-x} - \frac{1}{3-x}$$