Solving Three equations for 3 Unknowns Today I have a question and I am really curious to know about this.
Question:
$$ 16y+39z+50zy=0$$
$$ 85x-78z+95zx=0$$
$$ 85x+32y+70xy=0$$
$$\text{Are The Equations like these can be solve for unknowns? If not Why?If so How?}$$
I derived those from these:
$$ 1/x+1/y-1/z=-1$$
$$ 4/x+3/y+2/z=16$$
$$ 2/x-2/y-3/z=5$$
 A: If you derived the system of the equations correctly, you then get a normal 3x3 system equation with first-degree set of variables. This is a common linear algebra problem and can be solved in a ton of ways. 
First of all, express $\frac{1}{x} = i , \frac{1}{y} = j , \frac{1}{z} = k$
Then you will have the following : 
$ i + j + k = 1, 4i + 3j + 2k = 16, 2i - 2j - 3k = 5 $
In order to show that these can be solved or not, you can apply the Gauss Method on the matrix of their coefficients. Be careful after that, on probable solutions that may not apply for the starting fractions ($\frac{1}{x}$etc )
A: If you have to solve $$ 16y+39z+50zy=0\tag 1$$
$$ 85x-78z+95zx=0\tag 2$$
$$ 85x+32y+70xy=0\tag 3$$ use $(2)$ and solve it for $z$ to get $$z=-\frac{85 x}{95 x-78}\tag 4$$ (which assumes $95x-78\neq 0 $ )
Do the same using $(3)$ for $y$ to get $$y=-\frac{85 x}{70 x+32}\tag 5$$ (which assumes $70x+32\neq 0 $ )
Use $(4)$ and $(5)$ in $(1)$ which will lead, after simplification to the beautiful $$0=0$$ So, your system has solutions which depend on an arbitrary value of $x$. You can notice that the trivial solution $x=y=z=0$ is a result of the above.
Edit
If the first equation was $$a y+b z+c zy=0$$ reducing to same denominator would lead to $$x^2 (-8075 a-5950 b+7225 c)+x (6630 a-2720 b)=0$$ which shows that $x=0$ is always a solution, the other one being $$x=\frac{2 (39 a-16 b)}{5 (19 a+14 b-17 c)}$$ which is again $0$ if, as in your case, $39a=16b$ and very problematic if $19a+14b-17c=0$.
