How do I solve $3xy=4xz=5yz$ ${x+y \over x-z}=?$

Question

$3xy=4xz=5yz$

$ {x+y \over x-z}=?$

The answer is ${9 \over 2}$

But how to solve?

I am preparing for exam.

This question comes from metropol mathematics 1 testbook

Answer 1

$x=\frac{5y}{4}$ and $z=\frac{3y}{4}$

Does this help?

Answer 2

Suppose

$$3xy=4xz=5yz$$

then

$$3xy=4xz \implies y=\frac{4z}{3}=\frac{4x}{5}$$ $$4xz=5yz \implies x=\frac{5y}{4}=\frac{5z}{3}$$ $$3xy=5yz \implies z=\frac{3x}{5}=\frac{3y}{4}$$

and

$$x+y=x+\frac{4x}{5}=\frac{9x}{5}$$ $$x-z=x-\frac{3x}{5}=\frac{2x}{5}$$

so provided $x\neq 0$

$$\frac{x+y}{x-z}={\frac{45x}{10x}}=\frac{9}{2}$$

Answer 3

There is no single answer without additional restrictions on $x,y,z$. First we notice that if a constant in each of the three expressions is considered a missing variable, we quickly get a solution $z=3,y=4,x=5$, then the answer would be $(x+y)/(x-z)=9/2.$ But notice that $x=1,y=0,z=0$ is also a solution, but then $(x+y)/(x-z)=1$.