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If $x+y=2$ and $x^2+y^2=2$, what is the value of $xy$?

Interesting problem that I have no idea how to solve.

I know you're supposed to show working but considering it's only:

$x = 2-y$ and subbing that into the second equation, I doubt it's relevant...

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    If you do that, it works fine? Try follow the $x=2-y$ idea a bit further and see what you get. – 2017-01-14
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    It doesn't work like a normal simultaneous equations problem, that's the issue. – 2017-01-14
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    sure, but you get $(2-y)^2+y^2=2$, and you can definitely solve that equation for $y$, right? – 2017-01-14

5 Answers 5

6

Note that $(x+y)^2 = x^2+y^2+2xy$. This means that: $$xy = \frac{(x+y)^2-(x^2+y^2)}{2}$$ So, it follows that: $$xy = \frac{2^2-2}{2} = 1$$

5

Using the second equation

$x^2+y^2=x^2+y^2+2xy-2xy=(x+y)^2-2xy=2$

Using the first equation $$2^2-2xy=2$$

therefore $$xy=1$$

5

The laziest way is to notice that you're given $\color{red}{x^2+y^2=2},\ \color{blue}{x+y=2}$ and that $(x+y)^2=x^2+y^2+2xy$.

So squaring $({x+y})^2$, we get$$\begin{align*} & (\color{blue}{x+y})^2=\color{red}{x^2+y^2}+2xy\\ & \implies 4=2+2xy\\ & \implies xy=1\end{align*}\tag1$$

4

To continue your own method $$(2-y)^2+y^2=2$$ $$4-4y+2y^2=2$$ $$2-4y+2y^2=0$$ $$y^2-2y+1=0$$ $$(y-1)^2=0$$

Normally a quadratic would have two distinct solutions, but here $x$ and $y$ are forced to be equal. If there were two solutions the symmetry in the original equation would allow the values of $x$ and $y$ to be exchanged (here that changes nothing). But the symmetry does suggest that there will be ways of computing $xy$ without solving explicitly for $x$ and $y$ - in more complex situations (including problems with three or more variables) an explicit solution for the individual variables may be unwise, or even impossible.

1

Let $A=xy$. Let's try your substitution approach: \begin{aligned} A&=xy=x(2-x)=2x-x^2;\\ A&=xy=(2-y)y=2y-y^2. \end{aligned} Thus, $$ 2A=2(x+y)-(x^2+y^2)=2\times2-2=2. $$ So $A=\ldots$?