If $\frac{y}{x}=\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$ with $x,y,z,w>0$, $6x\ne y+z$, $7y\ne x-z$, then what is the value of $\frac{x}{y}$?

Question

If $\frac{y}{x}=\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$ with $x,y,z,w>0$, $6x\ne y+z$, $7y\ne x-z$, then what is the value of $\frac{x}{y}$?

I tried to solve it as follows:

From these equations we have the following equations:

$\frac{y}{x}=\frac{3x+y-w}{6x-y-z}$, $\frac{y}{x}=\frac{y+w}{7y-x+z}$, $\frac{3x+y-w}{6x-y-z}=\frac{y+w}{7y-x+z}$

Hence we have the following:

$y(6x-y-z)=x(3x+y-w)$, $y(7y-x+z)=x(y+w)$, $(3x+y-w)(7y-x+z)=(y+w)(6x-y-z)$

$6xy-y^2-yz=3x^2+xy-xw$, $7y^2-yx+zy=xy+xw$, $21xy-3x^2+3xz+7y^2-yx+yz-7yw+wx-wz=6xy-y^2-yz+6xw-wy-wz$

And I continued along these lines, but it didn't lead me anywhere. Could you please explain to me how to solve this question?

Answer 1

Hint :

Using $$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$

we get

$$\frac{y}{x}=\frac{3x+2y}{\ldots}$$

a quadratic in $x,y$ which can be solved for positive $x,y$.