Proportions and Equations: Can You Prove a Ratio with this Equation?

[jewel]

if

x/(b+c-a)=y/(c+a-b)=z/(a+b-c)

prove that..

x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz)

 

[MarkFL]

Re: hi i am stuck up with this numerical of fraction

Can you show us what you have tried?

Showing your work allows our helpers to see where you are stuck and what mistake(s) you may be making, so that they can offer suggestions to get you going again.

 

[jewel]

Re: hi i am stuck up with this numerical of fraction

Can you show us what you have tried?

Showing your work allows our helpers to see where you are stuck and what mistake(s) you may be making, so that they can offer suggestions to get you going again.

i am stuck in the initial stage itself not getting how will i proceed ...:confused

 

[Opalg]

if

x/(b+c-a)=y/(c+a-b)=z/(a+b-c)

prove that..

x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz)

Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.

 

[jewel]

Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.

thanks for advice... i will try that way and see

 

[jewel]

thanks for advice... i will try that way and see

i tried...its not working out.i am unable to solve .please help (Worried)

 

[Opalg]

Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.

There are probably several ways to attack this problem. The method I used was to start with \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) Then $$x = \lambda(b+c-a),$$ $$y = \lambda(c+a-b),$$ $$z = \lambda(a+b-c).$$ Add those equations to get $x+y+z = \lambda(a+b+c)$. Now subtract the first of those three displayed equations, giving $y+z = 2\lambda a$. Next, multiply that by $x$: $$xy+xz = 2\lambda ax.$$ In a similar way, $$yx + yz = 2\lambda by,$$ $$zx+zy = 2\lambda cz.$$ At this stage, take a look at what you are trying to prove, and see if you can get there.

 

[jewel]

There are probably several ways to attack this problem. The method I used was to start with \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) Then $$x = \lambda(b+c-a),$$ $$y = \lambda(c+a-b),$$ $$z = \lambda(a+b-c).$$ Add those equations to get $x+y+z = \lambda(a+b+c)$. Now subtract the first of those three displayed equations, giving $y+z = 2\lambda a$. Next, multiply that by $x$: $$xy+xz = 2\lambda ax.$$ In a similar way, $$yx + yz = 2\lambda by,$$ $$zx+zy = 2\lambda cz.$$ At this stage, take a look at what you are trying to prove, and see if you can get there.

thanks a lot ... i got it ...:)

 

[MarkFL]

thanks a lot ... i got it ...:)

I removed the question you tagged on, as it was posted word for word (including the attempt) by another user, and the new topic can be found here:

http://mathhelpboards.com/pre-algebra-algebra-2/continued-proportion-problem-7531.html