Need help identifying which algebra rule was used

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    Algebra
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Discussion Overview

The discussion revolves around understanding a specific algebraic manipulation used in the context of derivatives, particularly how the expression involving -h is simplified to -1. Participants are exploring the rules of algebra that apply to this transformation, with a focus on the difference between working with equations and expressions.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how -h is changed to -1 in a previous post, indicating confusion about the algebraic steps taken.
  • Another participant clarifies that the transformation involves dividing both sides of an equation by h, rather than directly changing -h to -1.
  • A different participant points out that \frac{-h}{h} simplifies to -1, suggesting that this is a fundamental algebraic rule.
  • Some participants express that their prior education did not emphasize the use of equations in such simplifications, focusing instead on simpler factoring techniques.
  • One participant elaborates on the distinction between simplifying expressions and manipulating equations, noting the broader range of operations available when working with equations.

Areas of Agreement / Disagreement

Participants exhibit some disagreement regarding the clarity of the algebraic manipulation and the educational approaches to teaching these concepts. There is no consensus on the best method to understand or teach the simplification process.

Contextual Notes

Participants highlight a potential limitation in their understanding based on their educational backgrounds, which may not have covered the use of equations in the same way as presented in the current discussion.

LogarithmLuke
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https://www.physicsforums.com/threads/derivative-of-1-x.211489/

In the last post, three last steps. He changes -h into -1, how does this work? I know that if you for example have h/h it's equal to 1 regardless of what value h has, since they're both the same value. I don't see that being applied here though. Could anyone help me out?

By the way, i tried messaging the person who posted it, but he is no longer active on this forum. The thread is also not open for further replies.
 
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If
[tex]f(x+h)-f(x)=\frac{-h}{x(x+h)}[/tex]
then
[tex]\frac{f(x+h)-f(x)}{h}=\frac{-1}{x(x+h)}[/tex]

Is this what you're asking about?

To simplify it, it's basically equivalent to

If
[tex]A = \frac{-h}{B}[/tex]
then
[tex]\frac{A}{h}=\frac{-1}{B}[/tex]
because
[tex]\frac{A}{h}=\frac{\left(\frac{-h}{B}\right)}{h}=\frac{-h}{Bh}=\frac{-1}{B}[/tex]

But more simply, you can just notice that A has a factor of h in the numerator, so to divide A by h, you're then cancelling that factor.
 
LogarithmLuke said:
In the last post, three last steps. He changes -h into -1, how does this work?
That's not what happened. What he did do was divide both sides by h.
 
What are you puzzled by? [itex]\frac{-h}{h}=-1[/itex]
 
I know, i just didn't think about the equation that was there, because that's not how i learned to do it in school. It didn't cross my mind that he used the fact that there was an equation there to move further with the math problem. When I've done these we never used equations, just simple factoring.
 
LogarithmLuke said:
I know, i just didn't think about the equation that was there, because that's not how i learned to do it in school. It didn't cross my mind that he used the fact that there was an equation there to move further with the math problem. When I've done these we never used equations, just simple factoring.
There's a big difference between simplifying an expression (such as by factoring and combining terms and so on) and working with an equation. When you're working with an expression, there are only a few things you can do, but when you're working with an equation, there are lots of things you can do: add the same expression to both sides, multiply both sides by the same value, divide both sides by the same nonzero value, take logs of both sides, and many others.
 

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