# Math Help - Simultaneous equations

1. ## Simultaneous equations

Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

Question: If $x$ and $y$ are distinct real numbers such that $x^3=1-y$ and $y^3=1-x,$ find all possible values of $xy$ .

Solution: $x^3=1-y$
$y^3=1-x,$

$x^3-y^3=x-y$
(Dividing by $(x-y)$ ): $x^2+xy+y^2=1$
$(x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy
$

$(x+y+1)(x+y-1)=xy
$

Case 1: $x+y+1=x$, $x+y-1=y$,
$y=-1,x=1$
Case 2: $x+y+1=y$, $x+y-1=x$,
$x=-1,y=1$

Only possible value of $xy$ is $-1$

First post with latex.

2. Originally Posted by I-Think
Buenas noches. I have a question here that I have answered, so I hope that the forum could rate my solution, detect errors and provide alternatives. Thanks.

Question: If $x$ and $y$ are distinct real numbers such that $x^3=1-y$ and $y^3=1-x,$ find all possible values of $xy$ .

Solution: $x^3=1-y$
$y^3=1-x,$

$x^3-y^3=x-y$
(Dividing by $(x-y)$ ): $x^2+xy+y^2=1$
$(x+y)^2-xy=1 \Rightarrow(x+y)^2-1^2=xy
$

$(x+y+1)(x+y-1)=xy
$

Isomorphism: Whatever follows is wrong.
Case 1: $x+y+1=x$, $x+y-1=y$,
$y=-1,x=1$
Case 2: $x+y+1=y$, $x+y-1=x$,
$x=-1,y=1$

Only possible value of $xy$ is $-1$

First post with latex.
One possible line of attack:

$x^2 + xy + y^2 = 1$

See it as a quadratic in x, then $x = \dfrac{-y \pm \sqrt{4 - 3y^2}}{2}$

Now for a real solution to exist, $4 \ge 4 - 3y^2 \ge 0$
Thus $y^2 \le \frac{4}{3}$

And then $xy = \dfrac{-y^2 \pm \sqrt{4y^2 - 3y^4}}{2}$.....

3. ## Question

I was wondering if that method was correct...
May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)

Also, does this mean there is an infinite number if values of $xy$ when both $x$and $y$are real numbers?

4. Originally Posted by I-Think
I was wondering if that method was correct...
May I ask what makes this method wrong in this case, I've seen it used in other questions. (Is it that two variables are being related to each other and not to a real number?)
Here's a counterexample: $x = 0, y = 1$,then $(x+y+1)(x+y-1)=xy$. However neither of your cases are true!!

Also, does this mean there is an infinite number if values of $xy$ when both $x$and $y$are real numbers?
Yes, I think they asked for a range of values...