If $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ find the ratio of $x$, $y$ and $ z$ I have this question from higher algebra by Hall and Knight:
if $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} $ then find the ratio of x,y and z?
There are two answers given for this question, the first is $\frac x4 =\frac y2 =\frac z3$ and the second is $\frac x1 =\frac y{-1} =\frac z0 $. Now I solve this question in the following manner:
Adding numerator and denominator gives $ \frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y}= 2\frac{x+y}{x+y} =2$  (when x+y is not zero).This gives the first answer.
When (x+y ) is zero => $ y= -x $ that is  $\frac xy = -1$ and now there are two things
(i)If I put these values in the original expression I get $ \frac{y}{x-z} =\frac 0z= -1$ this implies 0= 1 where am I making the mistake?
(ii)Also from the original expression I have$  \frac{x+y}{z} = \frac{x}{y}$ multiplying by z I get $ x+y =z\frac xy$. This implies that z=0 and x:y:z =x:-x:0 = 1:-1:0 and this gives the second answer.But the problem is how z can be zero when it appears in the denominator in the expression
I think these things are very basic but still I am stuck.Could anyone please help me in knowing where the mistake is?
 A: I too am concerned about an answer of $\frac{z}{0}$ being listed. That is a poor way of writing that $z$ is not related to $x$ and $y$ (if that is actually what it means).
What the second answer should say is "If both $x$ and $y$ are zero then $z$ can be any value." Asking this as a ratio question makes it very awkward.
A: It's not clear what the OP means by "adding" the numerator and denominator but there is a way using componendo and dividendo. If $\frac{x+y}{z} = \frac{x}{y}$ then $\frac{(x + y) - x}{z - y} = \frac{x}{y} \implies \frac{y}{z - y} = \frac{x}{y}$.
This implies that $\frac{y}{z - y} = \frac{y}{x - z} \implies z - y = x - z \implies 2z = x + y$. Thus $\frac{2z}{z} = \frac{x}{y}$ and we have two cases: either $z = 0$ which implies $\frac{y}{x} = \frac{x}{y} \implies (y + x)(y - x) = 0$. This results in $y = -x$, or $y = x$ in which case $z = 2x$ and $x = y = z = 0$ which does not make any sense.
The other case is when $z = 2y$ and hence $\frac{3y}{z} = 2$ so $x = 2y = \frac{2}{3} \cdot 2z$, which is equivalent to the form above.
A: y/(x-z)=(y+x)/z=x/y = be k 
y= kx - kz ..... 1 
y+x = kz ...... 2 
x=yk ................3 
Add 1 & 2 , 2y+ x = kx , substitute 3 , 2y+ yk = k *yk 
2 +k = k² , k²- k - 2 = 0 , k= 2 or k = -1 
for k = 2 , x = 2y , substitute in 2 , 3y = 2z 
x:y:z = 2y:y:3y/2 = 4:2: 3 
