Solve x/y+y/z+z/x for Real Numbers with 24,10

[anemone]

Let $x,\,y,\,z$ be real numbers satisfying

$\dfrac{(x+y)(y+z)(z+x)}{xyz}=24$

$\dfrac{(x-2y)(y-2z)(z-2x)}{xyz}=10$

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

 

[Sudharaka]

Hi anemone,

Thanks for the challenge. :)

Spoiler

Simplifying the first equation we get,

\[\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=22\]

Simplifying the second equation we get,

\[\frac{2x-z}{y}+\frac{2y-x}{z}+\frac{2z-y}{x}=\frac{17}{2}\]

Adding these two equations we get,

\[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{17}{6}\]

 

[anemone]

Hi anemone,

Thanks for the challenge. :)

Spoiler

Simplifying the first equation we get,

\[\frac{$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$}{z}+\frac{x+z}{y}+\frac{y+z}{x}=22\]

Simplifying the second equation we get,

\[\frac{2x-z}{y}+\frac{2y-x}{z}+\frac{2z-y}{x}=\frac{17}{2}\]

Adding these two equations we get,

\[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{17}{6}\]

Hi Sudharaka, :)

Thanks for participating. You've used a systematic approach to solve for this challenge. Unfortunately you've made a minor addition error, but it is a good solution nevertheless. Well done, Sud!

Spoiler

$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$

 

[Sudharaka]

Hi Sudharaka, :)

Thanks for participating. You've used a systematic approach to solve for this challenge. Unfortunately you've made a minor addition error, but it is a good solution nevertheless. Well done, Sud!

Spoiler

$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$

Ah... silly algebra... :p

 

[CellCoree]

I would first analyze the given information and understand the equations provided. The first equation states that the product of the three expressions, (x+y), (y+z), and (z+x), when divided by the product of x, y, and z, is equal to 24. Similarly, the second equation states that the product of the three expressions, (x-2y), (y-2z), and (z-2x), when divided by the product of x, y, and z, is equal to 10.

To solve for the required expression, $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$, I would first manipulate the given equations to get rid of the fractions. For the first equation, I would multiply both sides by xyz to get:

$(x+y)(y+z)(z+x)=24xyz$

Similarly, for the second equation, I would multiply both sides by xyz to get:

$(x-2y)(y-2z)(z-2x)=10xyz$

Next, I would expand the expressions on both sides of the equations and simplify to get:

$(x^2y+xy^2+2xyz+y^2z+yz^2+xz^2+2xyz)=24xyz$

$(x^2y-2xy^2-4xyz+2y^2z-4yz^2+8xyz-2xz^2)=10xyz$

By comparing the coefficients of the like terms, I would get the following system of equations:

$x^2y+xy^2+y^2z+xz^2=22xyz$

$x^2y-2xy^2+2y^2z-2xz^2=14xyz$

I would then solve this system of equations to get the values of x, y, and z. Once I have these values, I can substitute them in the expression $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$ to get the final answer.

In conclusion, to find the value of $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$ for the given equations, I would first manipulate the equations to get rid of the fractions and then solve the resulting system of equations to find the values of x, y, and z