<s>[INST] How would I solve $x^2-4x=y^2-4y$ without knowing the answer beforehand? The equation is $x^2-4x=y^2-4y$ in the case where $x\ne y$. The answer is $x+y=4$. 
I can start from $x+y=4$ and create the equation very easily, and I can substitute $x+4=y$ into the equation and show both sides are equal easily. I just don't get how I would find the answer if I didn't know it before hand, and all I had was the equation? Any advice? 
 [/INST] Let $c$ be the common value of $x^2-4x$ and $y^2-4y$.  Then $x$ and $y$ are both roots of the polynomial $t^2-4t-c$.  Since we are assuming $x$ and $y$ are distinct, they are all of the roots, so $t^2-4t-c$ factors as $(t-x)(t-y)$.  Since $(t-x)(t-y)$ expands to $t^2-(x+y)t+xy$, comparing the coefficients of $t$ gives $x+y=4$.
(Conversely, if $x+y=4$, then since $x$ and $y$ are both roots of $(t-x)(t-y)=t^2-(x+y)t+xy=t^2-4t+xy$, $x^2-4x$ and $y^2-4y$ are both equal to $-xy$.)
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