can i say that $x=y$ if both have the same terms? If I got that,
$$x=\dfrac{y^3+1}{2}\textrm{ and }y=\dfrac{x^3+1}{2}$$
then can I say that $x=y$ ? it looks correct but I'm not sure why.
Thanks.
 A: Yes, you can certainly say so (contrary to what others have answered), provided you're interested only in real solutions. Not in general, but in this particular case, I mean.
If this system had a solution $(a,b)$ with $a\ne b$, we would have
$$
2a=b^3+1,\quad 2b=a^3+1
$$
from which we get $2(a-b)=b^3-a^3=(b-a)(b^2+ab+a^2)$ and, since $a\ne b$,
$$
a^2+ab+b^2=-2
$$
that has no real solution, a contradiction.
So the intersections of these curves can also be computed as the solutions of
$$
\begin{cases}
2y=x^3+1\\
y=x
\end{cases}
$$
and the resolvent equation is $x^3-2x+1=0$, which factors as $(x-1)(x^2+x-1)=0$ giving
$$
x=1
\qquad\text{or}\qquad
x=\frac{-1-\sqrt{5}}{2}
\qquad\text{or}\qquad
x=\frac{-1+\sqrt{5}}{2}
$$

More generally, if we have equations $y=f(x)$ and $x=f(y)$, where $f$ is a polynomial, we can certainly write
$$
y-x=f(x)-f(y)
$$
and, since $f(x)-f(y)$, as a polynomial in two variables, is divisible by $x-y$, we have
$$
x=y \qquad\text{or}\qquad \frac{f(x)-f(y)}{x-y}=-1
$$
where the fraction denotes the quotient polynomial.
This can be done unless the equation $y-x=f(x)-f(y)$ is trivial, that is, when $f(x)-f(y)=y-x$ as polynomials. This is exactly the case when $f(z)=k-z$, which is the one mentioned by Hagen von Eltzen.
Of course things are more complicated if non polynomial functions are involved.
A: ![you can solve a system of equation and find the solution(s) . or ,you can put x from 1st relation into 2nd relation ,to find roots then you have x,y .the picture may help you ][2]
or ,you can put x from 1st relation into 2nd relation ,to find roots then you have x,y .the picture may help you .try this 
A: No you can't. You have $x=f(y)$ ans $y=f(x)$ for some function $f$. If you had (simpler) $x=10-y$ and $y=10-x$ you could not say $x0y$ either. After all it might be that $x=3$ and $y=7$.
(At least $x$ (and $y$) are fixed points of the iterated function $x\mapsto f(f(x))$.)
A: Rewriting the expression gives:
$$
x=\frac{Y^3+1}{2} \Rightarrow \\
Y = \left\{
\begin{array}{rc} 
\sqrt[3]{2x-1} & \mbox{for } x \ge 1/2 \\
-\sqrt[3]{1-2x} & \mbox{for } x < 1/2
\end{array}
\right.
$$ 
Compare this with
$$
y=\frac{x^3+1}{2}
$$
which should be different in general.

The plot indicates three common points, which have to lie on the identity line.
A: No, you cannot. Solve for $x$ and $y$ for both functions. When doing so, you'll get $y=\sqrt [3]{2x-1}$ which does not equal ${x^3 + 1}\over 2$. And $x=\sqrt [3] {2y-1}$ which does not equal ${y^3 + 1}\over 2$.
A: Yes, you can say this, and this applies more generally to any increasing function:
If $f$ is an increasing function and $f(x)=y$ and $f(y)=x$, then $x=y$.
$\underline{Pf}$ Suppose instead that $x\ne y$, so WLOG we can assume that $x<y$.
$\;\;\;\;$Then $f(x)<f(y)\implies y<x$, which gives a contradiction; so $x=y$.
A: Yes the graphs cut along $ x=y $ with real roots/ intersections.
