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How do I understand this?


Chikis

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How do I show, prove, see or be convinced that

[math]-\frac{bp}{cp-a}=\frac{bp}{a-cp}[/math]

I am concerned with what is going on in the minus sign there. I do remember that the following holds:

[math](-)(-)=+[/math]

Or that

[math](-1)(-1)=+1[/math] or just 1.

I also remember that [math]\frac{(-1)}{(-1)}=1[/math] or just 1 or that [math]\frac{(-)}{(-)}=+[/math]. What can I make of these identities in being convinced or cleared of what is obscured to me.

Edited by Chikis
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If you are used to re-formulating terms the identity is obvious. So it is a bit hard for me to guess what could be the blocker to your understanding. Perhaps this very simple statement helps:

 

The minus sign in front of your fraction "belongs" to the nominator, i.e.

[math] - \frac{bp}{cp - a} = \frac{-bp}{cp - a} [/math]

 

If that did not help yet (I recommend trying to go on from this first step by yourself before reading the 2nd hint):

 

 

Multiplying a term by 1 does not change it. Neither does multiplying with [math]\frac{-1}{-1}[/math], since that equals 1.

 

 

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How do I show, prove, see or be convinced that

If we have equation in form:

[math]\frac{a}{b}=\frac{c}{d}[/math]

 

We can multiply both sides by b and receive:

[math]a=\frac{b*c}{d}[/math]

 

Then multiply both sides by d and receive:

[math]a*d=b*c[/math]

 

 

In your case it's:

[math]-\frac{bp}{cp-a}=\frac{bp}{a-cp}[/math]

 

so after rearranging it's:

[math]-(bp*(a-cp))=bp*(cp-a)[/math]

[math]-(bp*a-bp*cp)=bp*cp-bp*a[/math]

[math]-bp*a+bp*cp=bp*cp-bp*a[/math]

[math]bp*cp-bp*a=bp*cp-bp*a[/math]

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  • 5 weeks later...

I am fond of interpreting the situation like "distribution of the minus(literally -1)", rather than multiplying the nominator and denominator by -1.

If you "move" the - to the denominator space, where (cp-a) becomes (a-cp) that do the trick.

Note: You can multiply right side denominator by -1 which is the same. However, I suggest getting rid of minuses, than creating new minus signs on the leading left side numbers..

Edited by TransientResponse
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