Simplify equation with negative exponents

AI Thread Summary
The discussion revolves around simplifying the expression (x^-2 - y^-2) / (x^-1 + y^-1). Participants explore different methods, with one person factoring the numerator and another opting to eliminate negative exponents first. Both approaches yield equivalent results, though they appear different. There is also clarification that expressions with negative exponents are not classified as polynomials but rather as rational functions. Ultimately, the conversation highlights the flexibility in mathematical representation and simplification methods.
Boba-Feet
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Homework Statement


Simplify (x-2 - y-2) / (x-1 + y-1)


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The Attempt at a Solution


So I just factorised the numerator into x-1 - y-1 and x-1 + y-1. And was left with x-1 - y-1 as an answer. The textbook gives (y - x) / (xy) as the answer (no working shown). So after some substitution I've realized my answer and the textbook's are the same. But I would like to know how they got theirs and is it a simpler form of my answer?

And I read on the internet that polynomials with negative exponents are not called polynomials, is that true and what should they be called? Thanks a lot in advance :)
 
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To simplifying this equation is like simplifying any other fraction with a plus or minus in between it.

evaluate this:
\frac{1}{10} - \frac{1}{5} = ?

Then, imagine that x = 10 and y = 5
 
Starting with your result, you can write ##x^{-1}=\frac{1}{x}## and do the same with y, and combine the sum to a single fraction afterwards.

And I read on the internet that polynomials with negative exponents are not called polynomials, is that true and what should they be called? Thanks a lot in advance :)
Well, they are polynomials in a different variable (##x^{-1}## instead of x)...
As functions, they are called rational functions.
 
MY first thought was to get rid of those negative exponents by multiplying both numerator and denominator by x^2y^2. That gives
\frac{y^2- x^2}{xy^2- x^2y}= \frac{(y- x)(y+x)}{xy(x- y)}= \frac{y+ x}{xy}
 
OK I am able to see how the textbook arrived at it's answer.
I looked at it and initially thought to factorise because I saw the difference of two squares (x^-2 - y^-2).
But the textbook writer thought to remove the negative exponents first. Strange that both answers look so different but mean the same thing.

Thanks for the quick and helpful responses.
 
Last edited:
Boba-Feet said:
OK I am able to see how the textbook arrived at it's answer.
I looked at it and initially thought to factorise because I saw the difference of two squares (x^-2 - y^-2).
But the textbook writer thought to remove the negative exponents first. Strange that both answers look so different but mean the same thing.

Thanks for the quick and helpful responses.

It is really not strange at all; do you prefer to write (1/3) + (1/5) or 8/15? Which way is "better" depends on what you want to do with the answer.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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