MHB Solve Algebra Challenge: Find $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Algebra Challenge
AI Thread Summary
The discussion revolves around evaluating the expression $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$ under the condition that $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$. Participants are encouraged to engage with the problem, with hints provided to guide their thinking. The conversation emphasizes patience in allowing contributors to develop their solutions before revealing answers. The focus remains on collaborative problem-solving in algebra. Ultimately, the goal is to derive the value of the target expression based on the given condition.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.
 
Mathematics news on Phys.org
Hint:
Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...
 
the answer is:
$\dfrac {4031}{4030}$
correct ?
 
That is $\text{Correct}$.
 
anemone said:
If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}----(1)$.
let $x\geq 0, y\geq 0, z\geq 0$
and let :$a=\dfrac {x-y}{x+y}, b=\dfrac {y-z}{y+z}, c=\dfrac {z-x}{z+x}$
then $abc=\dfrac {2014}{2015}=-\dfrac {2014}{2015}\times 1\times(-1)$
$a,b,c$ must be 1 positive and 2 negatives (since 3 positives are impossible)
by observation if $z=0 $ then $c=-1<0, b=1>0($ independent of $x$ and $y$)
and $a=-\dfrac {2014}{ 2015}<0$
so we only have to find the values of $x$ and $y$
that is $x+y=2015---(2),x-y=-2014---(3)$
from (2)(3)$\therefore x=\dfrac {1}{2},y=\dfrac {4029}{2}$
and (1) becomes :
$\dfrac{\dfrac{1}{2}}{2015}+1+0=\dfrac {1}{4030}+1=\dfrac {4031}{4030}$
 
Last edited:
Albert said:
...
by observation if $z=0 $...

Hi again Albert,

I see that if you let $z=0$, then our target expression becomes undefined.:confused:
 
anemone said:
Hi again Albert,

I see that if you let $z=0$, then our target expression becomes undefined.:confused:
why ? Our target expression becomes undefined ?

take note my definiton of $a,b,c$
 
Last edited:
Albert said:
why ? our target expression becomes undefined ?

take note my definiton of $a,b,c$

Hi Albert,

Ops, I don't know where my head was today, I made a wrong assumption and replied hastily to your solution without thinking for more. I am sorry.
 
anemone said:
Hint:
Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...

Show your solution
 
  • #10
maxkor said:
Show your solution

It's only been 4 days since anemone posted a hint to the problem. She is trying to judiciously encourage participation in this problem. So, let's give her some more time before we command her to post the solution, okay?
 
  • #11
maxkor said:
Show your solution

I'm thinking perhaps the previous statement is a not-so-subtle hint, I will post for further hints before revealing the solution, okay?

anemone said:
Hint 1:
Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...

Hint 2:

Note that $a-c=x+y-z-x=y-z$, $b-a=y+z-x-y=z-x$ and $c-b=z+x-y-z=x-y$

Also $a-c+b=2y$, $b-a+c=2z$ and $c-b+a=2x$

Hint 3:

$(c-b)(a-c)(b-a)=a^2b-a^2c-ab^2+ac^2+b^2c-bc^2$
 
  • #12
My solution:

Let $a=x+y,\,b=y+z,\,c=z+x$, note that

$a-c=x+y-z-x=y-z$,

$b-a=y+z-x-y=z-x$ and

$c-b=z+x-y-z=x-y$.

Also,

$a-c+b=2y$,

$b-a+c=2z$, and

$c-b+a=2x$.

Therefore, the given equation becomes:$\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$\dfrac{(c-b)(a-c)(b-a)}{(a)(b)(c)}=\dfrac{2014}{2015}$And our target expression becomes:

$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

$=\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}$

$=\dfrac{c-b+a}{2a}+\dfrac{a-c+b}{2b}+\dfrac{b-a+c}{2c}$

$=\dfrac{a}{2a}+\dfrac{b}{2b}+\dfrac{c}{2c}+\dfrac{c-b}{2a}+\dfrac{a-c}{2b}+\dfrac{b-a}{2c}$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}\right)$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(-\dfrac{(c-b)(a-c)(b-a)}{abc}\right)$

$=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{2014}{2015}\right)$

$=\dfrac{4031}{4030}$

since $\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}=-\dfrac{(c-b)(a-c)(b-a)}{abc}$.
 

Similar threads

Replies
1
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
Back
Top