# 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?

• I just realized, assuming that x and y aren't equal, then x must equal +/- (y-4) and y must equal +/- (x-4), otherwise there's no way x(x-4)=y (y-4). – Hockeyfan19 Nov 27 '17 at 2:11
• Be weary of the line of reasoning; it’s fallacious. Just because $ab = cd$ and $a \neq b$, doesn’t mean $a = c$. That trick only works when we have something like $ab = 0$, whence we can conclude $a$ or $b$ is zero. See the most recent answer to this question for the right application of this idea. It’s subtle. – Bob Krueger Nov 28 '17 at 1:49

## 4 Answers

\begin{align} x^2 - 4x &= y^2 - 4y \\ x^2 - y^2 &= 4x - 4y \\ (x-y)(x+y) &= 4(x-y) \\ x+y &= 4\end{align} where dividing by $x-y$ is allowed since $x \neq y$.

• It seems so obvious now, thanks for showing the way! I wasn't seeing it at all. – Hockeyfan19 Nov 26 '17 at 21:17

As an alternative, the solution that struck me first was completing the square in both $x$ and $y$. This is common when dealing with quadratics, especially once there are no $xy$ cross-terms. $$x^2 - 4x = y^2 - 4y$$ $$x^2 - 4x + 4 = y^2 - 4y + 4$$ $$(x-2)^2 = (y-2)^2$$ This means that either $x-2 = y-2$ or $x-2 = -(y-2)$, which means either $x = y$ or $x+y = 4$, as desired.

• I actually saw that I could complete the square when I was working on it. I missed that it would progress towards the solution though. Thanks for the alternative approach! – Hockeyfan19 Nov 27 '17 at 1:38
• You're welcome. Look into quadratic forms and conic sections for more general problems of this type. – Bob Krueger Nov 27 '17 at 1:44
• This is the way I first went, so +1 ;) – Lamar Latrell Nov 27 '17 at 4:37

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 $-xy$.)

• Very interesting approach, but where did the -4t come from in your polynomial? – Hockeyfan19 Nov 27 '17 at 2:04
• I'm not sure what you mean. Since $x^2-4x=c$, $x^2-4x-c=0$, and similarly for $y$. So $x$ and $y$ are roots of $t^2-4t-c$. – Eric Wofsey Nov 27 '17 at 2:32
• Okay so it comes from the original equation, I see now – Hockeyfan19 Nov 27 '17 at 2:43
• And you're using the t for clarities sake I think? – Hockeyfan19 Nov 27 '17 at 2:45

Write $x+y=d$. Then we have $y=d-x$ and so: $$(d-x)^2-4(d-x) = x^2-4x$$ so $$d^2-2dx-4d =-8x$$thus $$d(d-2x)-4(d-2x)=0$$ so $$(d-2x)(d-4)=0$$ If $d=2x$ we get $x=y$ which is impossible. So $d=4$ and we have $x+y=4$.