Is it possible to express "$x$" in this equation using only "$a$"?

Question

I have following equations:

$$\frac{(x-1)}{(a-1)}=\frac{(y-1)}{(b-1)}$$ $$\frac{(y-1)}{(b-1)}=\frac{(z-1)}{(c-1)}$$

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = k$$

$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$$

I need to solve by $x$ from this equation where $(x,y,z,a,b,c,k \in \mathbb{R})$, and the best I can get is this:

$$\frac{1}{x} + \frac{1}{\frac{(x - 1) (b - 1)}{a - 1} + 1} + \frac{1}{\frac{(x - 1) (c - 1)}{a - 1} + 1} = k$$

which can be transformed to this (by replacing $a$):

$$\frac{1}{x} + \frac{1}{\frac{(x - 1) (b - 1)}{\frac{b*c}{b*c - c - b} - 1} + 1} + \frac{1}{\frac{(x - 1) (c - 1)}{\frac{b*c}{b*c - c - b} - 1} + 1} = k$$

Is it possible to somehow remove those $b$ and $c$ and leave only $a$ in equation?

Answer 1

You can't. If this were possible, then the value of $x$ would depend only by $a$, independently by the values of $b,c$.

But now, fix any value of $k$, and consider the two cases

1. $a=2, b=c=4$

your equation becomes $$\frac{1}{x}+\frac{1}{\frac{3(x-1)}{1}+1}+\frac{1}{\frac{3(x-1)}{1}+1}=k $$ which gives you some solution $x_1$.

2. $a=2, b=3, c=6$

your equation becomes $$\frac{1}{x}+\frac{1}{\frac{2(x-1)}{1}+1}+\frac{1}{\frac{5(x-1)}{1}+1}=k $$ which gives you a different solution $x_2 \neq x_1$.

Thus knowing $a=2$ in both cases cannot give you explicitely the value of $x$.