Chikis Posted April 15, 2016 Share Posted April 15, 2016 (edited) 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 April 15, 2016 by Chikis Link to comment Share on other sites More sharing options...

timo Posted April 15, 2016 Share Posted April 15, 2016 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. Link to comment Share on other sites More sharing options...

Chikis Posted April 15, 2016 Author Share Posted April 15, 2016 (edited) Do I take it that [math]\frac{-bp}{cp}=\frac{bp}{-cp}[/math]? Edited April 15, 2016 by Chikis Link to comment Share on other sites More sharing options...

timo Posted April 15, 2016 Share Posted April 15, 2016 Yes, that is correct. Link to comment Share on other sites More sharing options...

Sensei Posted April 15, 2016 Share Posted April 15, 2016 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] Link to comment Share on other sites More sharing options...

Chikis Posted April 15, 2016 Author Share Posted April 15, 2016 Alright, thank you for the asistance. Link to comment Share on other sites More sharing options...

TransientResponse Posted May 15, 2016 Share Posted May 15, 2016 (edited) 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 May 15, 2016 by TransientResponse Link to comment Share on other sites More sharing options...

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