Given the sum and the product of two variables; what's the sum of their reciprocals? I'm sorry if it's a simple problem, but I need an explanation..
If
$$ x + y = 2 $$
$$ xy = 3$$
Then:
$${1\over x} + {1\over y} = z  $$
$${z\in  \{ \Bbb R}\}$$
My attempt was as simple like that:
$$x=2-y$$
$$(2-y)y=3$$
$$y^2 -2y + 3 = 0$$
$$(y^2 -2y +1) +2 = 0$$
$$(y-1)^2 = -2$$
$$(y-1) = \pm \sqrt{-2} = \pm i\sqrt{2}$$
$$y= 1\pm i\sqrt{2}$$
$$x=2-(1\pm i\sqrt{2})$$
$$x= 1\pm i\sqrt{2}$$
Since $$x = y$$ Therefore:
$${1\over x} + {1\over y} = {2\over 1\pm i\sqrt{2}}$$
My question is about how to proceed after? or maybe I did a mistake before? How can I reach the final real solution?
Thank you.
 A: Hint for easier solution:
$$
\frac1x + \frac1y = \frac{x+y}{xy}
$$
As for your solution, we have $x = 2-y$ and $y = 1\pm i\sqrt2$. This means $x = 1\mp i\sqrt2$ ($x$ must here necessarily be the complex conjugate of $y$; the sign from $\pm$ and the sign from $\mp$ must be chosen oppositely). Regardless of which is which, we get
$$
\frac1x + \frac1y = \frac1{1+i\sqrt2} + \frac1{1-\sqrt2}\\
= \frac{1-i\sqrt2}{(1+i\sqrt2)(1-i\sqrt2)} + \frac{1+i\sqrt2}{(1+i\sqrt2)(1-i\sqrt2)}\\
= \frac2{1+2}
$$
A: Hint:
Multiplying both sides with $xy$ we find:$$y+x=zxy$$
A: @Arthur 's answer is the easiest. Here is where you went astray:
Think for a minute. If, as you say, 
$$
x = y,
$$
then since they sum to $2$, you have $x = y = 1$ and the product isn't $3$.
You have to deal more sensibly with the $\pm$ sign in the solution to your quadratic: use $+$ for $x$ and $-$ for $y$.
A: If $x+y=S$ and $xy=P$, then $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}= S/P$
A: Your approach is correct up to where you solved for the values of $y$ and $x$. The sum $\frac{1}{x} + \frac{1}{y}$ will always be equal to $\frac{1}{1 - i \sqrt 2} + \frac{1}{1 + i \sqrt 2}$. Cross multiplying gives you $\frac{1 + i \sqrt 2 + 1 - i \sqrt 2}{(1 - i \sqrt 2)(1 + i \sqrt 2)} = \frac{2}{1 + 2} = \frac{2}{3}$.
