If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .

Question

If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .

What I Tried :

I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .

All all $3$ to get :-

$$2(x + y + z) = 5xy + 6yz + 7zx$$ Or, $$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$ That is, $$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$

Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$

So $$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$

$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$

I tried till this , then I have no idea . Can anyone help?

Answer 1

Assuming $x,y,z \neq 0$, we have that,

$$x + y = 5xy \iff \frac1x+\frac1y =5$$

$$y + z = 6yz \iff \frac1y+\frac1z =6$$

$$z + x = 7zx \iff \frac1z+\frac1x =7$$

then solve for $1/x$, $1/y$, $1/z$.

The case $x=0 \lor y=0 \lor z=0$ is trivial.

Answer 2

Multiply both sides of $x+y=5xy$, $y+z=6yz$ and $z+x=7zx$ by $z,x$ and $y$ respectively and add them together to obtain:

$xz+zy+xy+xz+zy+xy=2(xz+zy+xy)=18xyz$ so $xz+zy+xy=xy+z(x+y)=xy+5xyz=9xyz$.

So $xy=4xyz$ and hence $z=\frac{1}{4}$, etc.

Answer 3

Taking @user's hint a little further (and also assuming $x,y,z,\neq 0$), writing \begin{align} u &=\frac{1}{x}\\ v&=\frac{1}{y}\\ w&=\frac{1}{z} \end{align} Yields \begin{align} u+v &= 5\\ v+w &= 6\\ z+u&=7 \end{align} thus $$\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 1 & 0 & 1\\ \end{pmatrix} \begin{pmatrix} u\\ v\\ w\\ \end{pmatrix}= \begin{pmatrix} 5\\ 6\\ 7 \end{pmatrix} $$ Thus $$ \begin{pmatrix} u\\ v\\ w\\ \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & -1 & 1\\ 1 & 1 & -1\\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 5\\ 6\\ 7 \end{pmatrix}$$ Whence $$\begin{pmatrix} u\\ v\\ w\\ \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 6\\ 4\\ 8\\ \end{pmatrix}$$

And returning to the representation of these $u, v, w$ in terms of $x, y, z$ gives the desired result.