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$\frac{(x-y)^2}{(y-3)(3-x)} = 1$

That was my attempt, I can't think of anything else here. I'd prefer a hint

I I
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4 Answers4

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Reorder the equation as a second degree equation in $x$ and search for solutions $x$. You find that the discriminant $\Delta$ is always negative for all real value of $y \ne 3$ so there are no real solutions $x$ for real $y \ne 3$ and the solution $x=3$ for $y=3$.

1)Reorder: $$ x^2-x(3+y)+y^2+9-3y=0 $$ so: $a=1$, $b=-(3+y)$, $c=y^2+9-3y$.

2)Find $\Delta$:

$$ \Delta=b^2-4ac= (3+y)^2-4(y^2+9-3y)=-3y^2+18y-27=-3(y-3)^2 $$ 3) see: $\Delta <0 \quad \forall y \in \mathbb{R}-\{3\}$ because $(y-3)^2$ is always positive (for $y \ne 3$) being a sqare.

so there are no $x$ solutions (in $\mathbb{R}$) for all $y \ne 3$ and you can find the solution $x=3$ for $y=3$.

As an exercise do the same reordering for an equation in $y$ .... :)

If you done the exercise you see that all is the same. Why? Because the equation is symmetric for a change of $x$ and $y$, so every solution (if exists) is of the form $x=y=k$. Substituting in the given equation you have the simpler solution:

$2k^2+9=6k+k^2 \iff (k-3)^2=0 \iff k=3$

Emilio Novati
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  • more precise please – I I Feb 01 '15 at 17:58
  • Rearrange the equation to get a quadratic equation in $x$ in standard form. I think you will get $(1)x^2+(-y-3)x+(y^2-3y+9)=0$ (you better check). The discriminant for $ax^2+bx+c$ is $\Delta=b^2-4ac$. If $\Delta$ is negative there are no real solutions, if zero there is one, if positive there are two. Can you finish this yourself? – Rory Daulton Feb 01 '15 at 18:34
  • You didn't consider if $y=3$, which puts $\sqrt{\Delta}\in\mathbb{R}$. – 2'5 9'2 Feb 01 '15 at 19:16
  • yes!!! What a stupid ! I edit the ansver. Thank'you. – Emilio Novati Feb 01 '15 at 19:27
  • @RoryDaulton yes sir – I I Feb 01 '15 at 21:05
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Shift variables to eliminate the linear terms. $x=u+a$ and $y=v+b$, with $a,b$ yet-to-be-determined constants.

$$ \begin{align} x^2+y^2+9&=3(x+y)+xy\\ (u+a)^2+(v+b)^2+9&=3(u+a+v+b)+(u+a)(v+b)\\ u^2+2au+a^2+v^2+2bv+b^2+9&=3u+3a+3v+3b+uv+av+bu+ab \end{align} $$ We would like to be able to cancel the the terms that are linear in $u$ and $v$, so we want: $$ \left\{\begin{aligned} 2a&=3+b\\ 2b&=3+a \end{aligned}\right. $$ This system of linear equations in $a,b$ has solution $a=b=3$. So now we have $$ \begin{align} u^2+6u+9+v^2+6v+9+9&=3u+9+3v+9+uv+3v+3u+9\\ u^2+v^2&=uv\quad\text{(next, complete the square)}\\ 4u^2-4uv+4v^2&=0\quad\text{(multiplying by 4 avoids fractions)}\\ (2u-v)^2+3v^2=0 \end{align} $$ Since we have two squares adding to $0$, the only solution to this is $u=v=0$, which leads to $x=y=3$.

2'5 9'2
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  • what do you mean by cancel the terms that are linear in $u$ and $v$? – I I Feb 02 '15 at 11:38
  • For instance, at that point you have $(2a)u$ on the left and $3u+bu$ on the right. These are linear terms in $u$. And with the right choice for $a$ and $b$, you can make them go away. – 2'5 9'2 Feb 02 '15 at 15:36
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We have to check whether $$p(x,y):=x^2-xy-3x+y^2-3y+9$$ has real zeros $(x,y)$. This can be done by repeated completing the square: $$\eqalign{p(x,y) &=\left(x-{y\over2}-{3\over2}\right)^2-{1\over4}(y^2+6y+9)+y^2-3y+9\cr &=\left(x-{y\over2}-{3\over2}\right)^2+{3\over4}(y^2-6y+9)\cr &=\left(x-{y\over2}-{3\over2}\right)^2+{3\over4}(y-3)^2 \ .\cr}$$ Here the right hand side is $=0$ iff $y=3$ and $x=3$.

Christian Blatter
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0

Just did this yesterday: What's so special about the form $ax^2+2bxy+cy^2$?

Given some $g$ that is going to be set equal to $0,$ with $$ g = a x^2 + b x y + c y^2 + d x + e y + f $$ and taking the usual $$ \Delta = b^2 - 4 a c, $$ we get, assuming both $a, \Delta \neq 0$ for this amount of work to be necessary, $$ \color{red}{ -4a \Delta g} \; = \; \color{blue}{ (\Delta y + bd - 2ae)^2 - \Delta (2 a x + b y + d)^2} + \color{green}{ \left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)} $$ I put extra parentheses to emphasize that $\left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)$ is a single constant term, no $x,y$ involved. New variables to be called: $$ u = \Delta y + bd - 2ae, \; \; \; v = 2 a x + b y + d $$

For you, $$g= x^2 - xy + y^2 - 3 x - 3 y + 9 $$ is being set to $0,$ and $$ a=1, b=-1,c=1,d=-3,e=-3,f=9, \Delta = -3, $$ Go through it all, $$ -4ag = (-3y+9)^2 + 2 (2x-y-3)^2 $$ and this is being set to $0,$ the constant term came out to $0.$ So we have a single point, $y=3,$ then $2x-y-3 = 0$ says $x=3$ also.

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Will Jagy
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