x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1  TOPIC_SOLVED

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby japiga on Mon Oct 18, 2010 8:00 pm

Help to solve this:
If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0
japiga
 
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

Sponsor

Sponsor
 

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby Martingale on Mon Oct 18, 2010 8:36 pm

japiga wrote:Help to solve this:
If x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1. Prove: xy+yz+zx=0


hint 8-) :



therefore



hence

User avatar
Martingale
 
Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby japiga on Tue Oct 19, 2010 7:47 am

:confused: And then? How to include this in the first equation x/a=y/b=z/c ?
japiga
 
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby Martingale on Tue Oct 19, 2010 12:45 pm

japiga wrote::confused: And then? How to include this in the first equation x/a=y/b=z/c ?





solve for x in terms of z and solve for y in terms of z


then plug those into
User avatar
Martingale
 
Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby japiga on Tue Oct 19, 2010 1:49 pm

But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?
japiga
 
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby Martingale on Tue Oct 19, 2010 2:19 pm

japiga wrote:But what is a purpose of plugging into the equation of xy + yz + zx? Actually, how to reach this equation?



:confused:


Don't you want to show that xy+yz+xz=0? ....Yes!!

so start with xy+yz+xz and show that is has to be zero using the hints I provided.
User avatar
Martingale
 
Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby japiga on Tue Oct 19, 2010 2:47 pm

You mean that this equation x/a=y/b=z/c should be split in two equations such as x/a=y/b and y/b=z/c. So x= (az)/c and y=(bz)/c. Does it what you mean?
japiga
 
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby Martingale on Tue Oct 19, 2010 4:13 pm

japiga wrote:... So x= (az)/c and y=(bz)/c. Does it what you mean?


yes
User avatar
Martingale
 
Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1

Postby japiga on Tue Oct 19, 2010 4:27 pm

Sorry, but it is not so helpful! :confused:
Please give me step by step procedure.
japiga
 
Posts: 32
Joined: Mon Sep 20, 2010 3:28 pm

Re: x/a=y/b=z/c, a+b+c=1 and a^2+b^2+ c^2=1  TOPIC_SOLVED

Postby Martingale on Tue Oct 19, 2010 4:34 pm

japiga wrote:...So x= (az)/c and y=(bz)/c. Does it what you mean?



japiga wrote:Sorry, but it is not so helpful! :confused:
Please give me step by step procedure.






...
User avatar
Martingale
 
Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

Next

Return to Advanced Algebra ("pre-calculus")

cron