Does $x= y$ hold in this mathematical relation? If 
$$x(x-100)=y(y-100)$$
then can we say $x=y$?
I assume them to be equal because I can't seem to find any value which disproves my hypothesis.
 A: You can systematically work through this to find the solutions.
If $x(x-100)=y(y-100)$ then
$$x^2-100x=y^2-100y$$ If you did quadratic equations in high school, this is crying out for 'completing the square':
$$x^2-100x+2500=y^2-100y+2500\\(x-50)^2=(y-50)^2$$
This boils down to 2 cases:


*

*Case 1: $(x-50)=(y-50)$. This leads to $x=y$.

*Case 2:  $(x-50)=-(y-50)$. This leads to $x+y=100$.


Edit Please see Harish Chandra Rajpoot's answer - it contains a cleaner derivation of the two cases by producing the factorisation $(x-y)(x+y-100)=0$. Note that the two cases overlap when $x=y=50$.
A: Notice, one can easily factorize as follows $$x(x-100)=y(y-100)$$
$$x^2-y^2-100(x-y)=0$$
$$(x-y)(x+y)-100(x-y)=0$$
$$(x-y)(x+y-100)=0$$
$$\color{red}{x=y}$$
or $$\color{blue}{x+y-100=0}$$
A: If $x(x-100)=y(y-100)$, then not necessarily $x=y$. What this means exactly is that the quadratic function $f(x)=x(x-100)$ is not injective in the domain $\mathbb R$.
To see why, look at the graph of $f(x)$. You should be a aware it is a parabola and how it looks, but here's the graph on WolframAlpha.
You can use the Horizontal Line Test. Draw some horizontal lines (i.e. parallel to the X-axis) on the graph. The function $f(x)$ will be injective if and only if every horizontal line intersects $f(x)$ at exactly either $0$ or $1$ point.
Here's a generalization: $$f(x)=a_{2k}x^{2k}+a_{2k-1}x^{2k-1}+\cdots+a_1x+a_0$$
The polynomial function $f(x)$ for any $a_i\in\mathbb R$, $a_{2k}\neq 0$, $k\in\mathbb Z^+$ is not injective in the domain $\mathbb R$, i.e. if $f(x)=f(y)$, then not necessarily $x=y$.
To see why, notice that if $a_{2k}>0$, then $\lim_{x\to +\infty}f(x)=+\infty$ and $\lim_{x\to -\infty}f(x)=+\infty$, and if $a_{2k}<0$, then $\lim_{x\to +\infty}f(x)=-\infty$ and $\lim_{x\to -\infty}f(x)=-\infty$.
