QUESTION: Let $x \neq y$ and the relation between x and y follows;
$\frac{x^2-y}{x^2y+xy}=\frac{x-y^2}{xy^2}=x-y$ $ \quad $ then
what is the exact result of
$\frac{x^2+y}{x^2(y^2-y+1)}=?$
a) $y+1 \quad $ b)$\frac{1}{y} \quad $ c) $y^2 \quad $ d) $y^2-1\quad $ e) $y$
I have done so far ;
$\frac{x^2-y}{x^2y+xy}=\frac{x-y^2}{xy^2}$
$ \implies x(y-1)=1-y^2 $
Case (if y=1):
We use second part of equality which is
$\frac{x-y^2}{xy^2}=x-y$
$ \implies \frac{x-1}{x}=x-1$
$\implies $ $x=1 $ which is contradiction because our assumption is $x \neq y$. So $y \neq 1$
So we simplify our expression ;
$ x(y-1)=1-y^2 \implies x=-(1+y)$ (this is the part that I stuck and I also don't understand by meaning of "exact result")
Thanks in advance for your guidances and help.
the first two equations are $$\frac{x^2-y}{xy(x+1)}=\frac{x-y^2}{xy^2}$$ multiplying by $xy\neq 0$ and $$x+1\neq 0$$ we obtain $$x^2y-y^2=(x-y^2)(x+1)$$ $$x^2y-y^2=x^2-xy^2+x-y^2$$ $$x^2y=x^2-xy^2+x$$ since $$x\neq 0$$ we have $$xy=x-y^2+1$$ $$x(y-1)=1-y^2$$ this gives $$x(y-1)=-(y-1)(y+1)$$ first case: $$y=1$$ then we get $$\frac{x-1}{x}=x-1$$ or $$(x-1)^2=0$$ thus $$x=1$$ and our term $$\frac{x^2+y}{x^2(y^2-y+1)}=2$$ the next case $$x+y+1=0$$ is for you! next case: $$x+y+1=0$$ and we get $$x=-y-1$$ and we get $$x-y^2=x^2y^2-xy^3$$ or $$x(1+y^3)=y^2(x^2+1)$$ with $$x=-y-1$$ we get5 $$(1+y^3)(-y-1)=y^2(y^2+2y+2)$$ $$-y-y^4-1-y^3=y^4-2y^3+2y^2$$ $$0=2y^4+3y^3+2y^2+y+1$$ and this equation has no real solutions, if i have made no mistake! i haven't made a mistake, therefore your problem has no real solutions!! (My program gives the same solution!)