An algebra problem by Abhijith Asokan

By multiplying the three equations:

(xyz + y + x + 1/z + z + 1/x + 1/y + 1/xyz) = 28/3

xyz + 1/xyz = 2

then

xyz = 1

You have two approaches to deal with such problems :

1-The easy non-frustrating approach

2- The long hard to solve approach

let's begin with the easy non-frustrating one first :

By multiplying the equations : $\\ (x+\frac { 1 }{ y } )(y+\frac { 1 }{ z } )(z+\frac { 1 }{ x } )=(\frac { 7 }{ 3 } )(4)(1)$

$xyz+y+x+\frac { 1 }{ z } +z+\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ xyz } =\frac { 28 }{ 3 }$

Looks pretty scattered , doesn't it ? let's give it some order

$xyz+\frac { 1 }{ xyz }+(x+\frac { 1 }{ y } )+(y+\frac { 1 }{ z } )+(z+\frac { 1 }{ x } )=\frac { 28 }{ 3 }$

$xyz+\frac { 1 }{ xyz } +\frac { 7 }{ 3 } +4+1=\frac { 28 }{ 3 }$

$xyz+\frac { 1 }{ xyz } =\frac { 6 }{ 3 } =2$

Multiplying by $xyz$

${ (xyz) }^{ 2 }-2(xyz)+1=0$

$\boxed { xyz=1 }$

As For the other approach , It basically counts on finding the value of $x,y,z$ , but I'll solve it by finding only one of them

We Begin it by multiplying the first equation by $y$ :

$xy+1=\frac { 7y }{ 3 }$

$xy=\frac { 7y }{ 3 } -1\rightarrow \quad (1)$

We'll need this one , but not now

Back to the first equation :

$x=\frac { 1 }{ y } -\frac { 7y }{ 3 } =\frac { 7y-3 }{ 3y }$

Use this to substitute x in the third equation :

$z+\frac { 1 }{ \frac { 7y-3 }{ 3y } } =1$

$z+\frac { 3y }{ 7y-3 } =1$

And in the second equation :

$\frac { 1 }{ z } =4-y\quad \Longleftrightarrow \quad z=\frac { 1 }{ 4-y }$

$\therefore \frac { 1 }{ 4-y } +\frac { 3y }{ 7y-3 } =1$

After simplifying , it should be like this :

$4{ y }^{ 2 }-9y+12=0\\ { (2y-3) }^{ 2 }=0\\ y=\frac { 3 }{ 2 }$

We have every thing in need . now , Let's get to what we want :

Back to the third equation and multiply it with $xy$ :

$xyz+y=xy$

From (1) :

$xyz+y=\frac { 7y }{ 3 } -1$

$xyz=\frac { 4y }{ 3 } -1$

$xyz=\frac { 4*3 }{ 3*2 } -1$

$\boxed { xyz=1 }$

Well i din't find the value of x, y & z but found the ans the other way..

Multiplying equation 1, 2 & 3..

Later simplifying we get an equation

xyz+x+y+z+1/x+1/y+1/z+1/xyz=28/3

Again simplifying & considering a=xyz

We get, a^2-2a+1=0

Rest is simple..

$\begin{cases}x+\frac 1 y=\frac 7 3\\y+\frac 1 z=4\\z + \frac 1 x=1\end{cases}\;(1)$ multiplying 1st and 2nd members of the equations of system $1\,$ :

$(x+\frac 1 y)(y+\frac 1 z)(z + \frac 1 x)=\frac {28} 3, (2)$

$xyz+y +x +\frac 1 z+z+\frac 1 x+\frac 1 y+\frac 1 {xyz}=\frac {28} 3, (3)$

now introducing $(1)$ in $(3)$ :

$xyz + \frac 1 {xyz}=\frac{28} 3-\frac 7 3-4-1=2, (4)$ denoting $xyz=t,\,t \ne 0, (4)$ become:

$t+\frac1 t=2, (5)$ $t^2-2t+1=0, (6)$ $(t-1)^2=0, (7)$ equation. $(7)\,$ has solution: $t=xyz=\boxed 1$

x + 1/y = 7/3 x = 7/3 - 1/y = (7y-3)/3y

z + 1/x = 1 z = 1 - 1/x = 1 - 3y/(7y-3) = (4y - 3)/(7y - 3)

y + 1/z = 4 y = 4 - 1/z = 4 - (7y - 3)/(4y - 3) y = (9y - 9)/(4y - 3)

Cross-Multiplying and simplifying we get: 4y^2 - 12y + 9 = 0 Therefore, y = 3/2

Using the value of y, we get x = 5/3; z = 2/5

x y z = 5/3 * 3/2 * 2/5 = 1

$z=1-1/x=(x-1)/x,$ $y=4-1/z=(4z-1)/z,$ $(4z-1)/z-x/(x-1),$ $1/x=1-z,$ $x=1/(1-z),$ $1/(1-z)+z/(4z-1)=7/3$ $<=>(4z-1)+z(1-z)=(7/3)(1-z)(4z-1)$ $<=>-z^2+5z-1=(7/3)(-4z^2+5z-1)$ $<=>3(-z^2+5z-1)=7(-4z^2+5z-1)$ $25z^2-20z+4=0$ $(5z-2)^2=0$ $z=2/5$ $1/x=1-2/5=3/5$ $x=5/3$ $y=(8/5-1)/(2/5)=(3/5)/(2/5)=3/2$ $xyz=2/5*5/3*3/2=\boxed{1}$

all's well, ends well :D

first find out the value of z from 3rd equation and value of y from 1st equation and substitute the values of y and z in 2nd equation you will get the equation 9X^2 - 30X + 25= 0 so we get X= 5/3 now put the value of x in third equation we get z = 2/5 and from first equation we get y = 3/2 and Hence XYZ = (5/3)(3/2)(2/5)= 1