Basic Algebra Test $\mathfrak{P}$ $\mathfrak{a}$ $\mathfrak{r}$ $\mathfrak{t}$ $\mathfrak{3}$

$\dfrac{ax+by+cz}{ax+by+cz}=1$

$\dfrac{ax}{ax+by+cz} +\dfrac{by}{by+cz+ax} +\dfrac{cz}{cz+ax+by}=1$

$\dfrac{ax}{ax+a^2}+\dfrac{by}{by+b^2}+\dfrac{cz}{cz+c^2}=1$

$\dfrac{x}{a+x}+\dfrac{y}{b+y}+\dfrac{z}{c+z}=1$

$\dfrac{x}{a+x}+\dfrac{y}{b+y}+\dfrac{z}{c+z}-3=\boxed{-2}$

Let a = b = c =1. Then we may suppose x = y = z = 1/2. ( satisfies all the equations and conditions) Plug in values. You get the answer :)

$\begin{aligned} a^2&=by+cz\\ b^2&=cz+ax\\ c^2&=ax+by \end{aligned}$ $x=\frac{-a^{2}+b^{2}+c^{2}}{2a}$ $y=\frac{a^{2}-b^{2}+c^{2}}{2b}$ $z=\frac{a^{2}+b^{2}-c^{2}}{2c}$ $\therefore$ $\displaystyle \dfrac{x}{a+x} +\dfrac{y}{b+y} + \dfrac{z}{c+z} = 1$ $1-3 = \boxed{\boxed{\boxed{-2}}}$

If you add the constraint that $a=b=c$ and $x=y=z$ (which we can do since the answer is constant, so long that a solution still exists) you'll notice that the three equations given boil down to one equation (which still has a solution):

$a^2=2ax \\ a=2x$

and the expression we are solving for looks like: $\frac{x}{2x+x} + \frac{x}{2x+x} + \frac{x}{2x+x} - 3 \\ = 1 - 3 = \boxed{-2}$

I got $a=2x$