5
$\begingroup$

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

$\endgroup$
2
  • 5
    $\begingroup$ $\frac{-1}{x-2}=\frac{(-1)(-1)}{(-1)(x-2)}=\frac1{2-x}$ $\endgroup$
    – robjohn
    May 10 '15 at 3:41
  • $\begingroup$ 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. $\endgroup$
    – zavtra
    May 10 '15 at 3:45
6
$\begingroup$

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} $

$\endgroup$
0
5
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ @MartinSleziak isn't your edit...a bit the same? $\endgroup$ May 10 '15 at 5:19
  • $\begingroup$ I have just added space after commas, full stops, etc. (And, accidentally, introduced one typo. Sorry for that.) $\endgroup$ May 10 '15 at 5:24
  • $\begingroup$ @MartinSleziak Ahh,I see.Doesn't matter :) $\endgroup$ May 10 '15 at 5:33
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.