- #1

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THIS:

[tex]\frac{x-y}{2x} = \frac{dy}{dx}[/tex]

To THIS:

[tex]\frac{1}{2}(1-\frac{y}{x}) = \frac{dy}{dx}[/tex]

But I'm wondering... wouldn't It be this?

[tex]\frac{1}{2}(x-\frac{y}{x}) = \frac{dy}{dx}[/tex]

if not.. how come?

- Thread starter snowJT
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- #1

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THIS:

[tex]\frac{x-y}{2x} = \frac{dy}{dx}[/tex]

To THIS:

[tex]\frac{1}{2}(1-\frac{y}{x}) = \frac{dy}{dx}[/tex]

But I'm wondering... wouldn't It be this?

[tex]\frac{1}{2}(x-\frac{y}{x}) = \frac{dy}{dx}[/tex]

if not.. how come?

- #2

arildno

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Do you think that:

[tex]\frac{7-2}{7}=7-\frac{2}{7}[/tex]??

[tex]\frac{7-2}{7}=7-\frac{2}{7}[/tex]??

- #3

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- #4

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[tex]\frac{1}{2}(1-\frac{y}{x})= \frac{1}{2}-\frac{y}{2x}[/tex]

no?

- #5

arildno

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That is correct. So?

- #6

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Yes, but you can't multiply it back to get [tex]\frac{x}{x}[/tex] to get [tex]\frac{x-y}{2x}[/tex]

- #7

matt grime

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What......?

- #8

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well...

I don't think this makes sense... because how can you get x back when you go from this

[tex]\frac{1}{2}(1-\frac{y}{x})[/tex]

to this:

[tex]\frac{1}{2}\left(\frac{x}{x} - \frac{y}{x}\right)[/tex] <-- x over x appears from 1 over 1 which doesn't make sense

I don't think this makes sense... because how can you get x back when you go from this

[tex]\frac{1}{2}(1-\frac{y}{x})[/tex]

to this:

[tex]\frac{1}{2}\left(\frac{x}{x} - \frac{y}{x}\right)[/tex] <-- x over x appears from 1 over 1 which doesn't make sense

Last edited:

- #9

matt grime

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So you're saying that 2 divided by 2 is not 1?

- #10

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no, I'm saying you can't get x when you go backwards

- #11

cristo

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I don't know what your mean. This can be simplified as [tex]\frac{1}{2}(1-\frac{y}{x})=\frac{1}{2}-\frac{y}{2x}=\frac{2x-2y}{4x}=\frac{x-y}{2x}[/tex]no, I'm saying you can't get x when you go backwards

What's wrong with that?

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- #13

matt grime

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@snowJT: you were taught in elementary school or primary school, or some time before the age of 11 to put things over a common denominator, i.e. that a/b + c/d = (ad+bc)/bd, so use it, but sensibly so that you don't introduce unnecessary factors that are going to cancel out later. You are just adding together fractions. That is something you have been doing for years.

- #14

cristo

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How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.@cristo: what's wrong is that you needlessly introduce extra factors of 2 for no reason.

I understand your point, but sometimes, when someone doesn't see what is going on, it is *simpler* to follow rules that they will know, and then simplify later. Skipping steps, in this case, may have added confusion.

- #15

matt grime

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- #16

Gib Z

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it's not b

- #18

uart

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I found it interesting that the OP could understand the algebra when cristo included the redundant factor but could not understand it previously.How is that wrong? Clearly I know that the factor of two was included unnecessarily, but since the OP could not see the answer, I expanded out the fraction fully, then used the rule that he would be familiar with, namely that a/b+c/d=(ad+bc)/bd.

It seems to be a case of OP not being able to grasp that

- #19

uart

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Haha that's very funny, everyone knows that it's really

- #20

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That's actually true if you're talkling in ASCII.Haha that's very funny, everyone knows that it's reallya + 1that equalsb. :tongue:

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