I'm quite young and haven't used this site before, so I apologise if I'm not laying things out correctly or any other problem really (tips would be appreciated). Regardless, here's something I'm quite confused about.
Problem:
$x^2 + xy = 20$
$y^2 + xy = 30$
Solve: $xy$
The solution to this problem is xy=12. However, I found a way where I got the answer of ±12. It is very easy to see why -12 is impossible, but the algebra that I wrote down still makes sense. If possible, can someone please explain where I went wrong.
Algebraic "proof" of ±12:
Adding the two equations gives:
$x^2 + 2xy + y^2 = 50$
$(x+y)^2 = 50$
$x+y = ±5√2$
Since:
$x^2 +xy = 20 => x(x+y) = 20 => x(±5√2)$
then
$x = 20/(±5√2) = ±2√2$
following the same logic,
$y = ±3√2 $
So surely, $xy = ±6 * 2 = ±12 $?
It obviously doesn't, and proving that it doesn't is easy - but where did I go wrong with the algebra above? Again, sorry if I didn't set things out correctly I don't know LaTeX or anything like that.
When $x =2\sqrt{2}$ this mandates $y=3\sqrt{2}$ in the way you have solved it, because $x+y$ must equal $+5\sqrt{2}$ or $-5\sqrt{2}$.
Therefore the product xy is either $2\sqrt{2} \times 3\sqrt{2}$ OR $-2\sqrt{2} \times -3\sqrt2$ which are both 12.
Apart from adding 20 and 10 to get 50, your algebra was absolutely fine. It's just that small nuance that you were forgetting ($x+y=\pm5\sqrt{2}$ is a constraint on your solutions).