x^2+y^2=2, xy=1, how to find x and y

Question

I have problems when doing these equations when I don't know any variable's value. Can someone please explain how to do this and possibly give some tips when it comes to solving these problems?

Well, I know x = 1 and y = 1, but what about this one: a-b=3, a:b=3:2, find a^2-b^2 (I am not quite is a:b=3:2 = a/b:3/2 or it was meant to be that way)

Answer 1

Observe that $(x+y)^2 = x^2+y^2+2xy$ and $(x-y)^2 = x^2 + y^2 - 2 xy.$ So $(x+y)^2 = 4$ and $(x-y)^2 = 0.$ Therefore $x+y= \pm 2$ and $x - y = 0.$ So either $x=y=1$ or $x=y=-1.$

Answer 2

Since $x^2+y^2=2$ and $xy=1$, then$$(x-y)^2=x^2+y^2-2xy=2-2=0.$$Therefore, $x=y$. But now the second equation becomes $x^2=1$. So, the only solutions are $x=y=1$ and $x=y=-1$.

Answer 3

Since $xy=1$ it is $x^2+y^2=2xy$, which gives us $x^2-2xy+y^2=0\Leftrightarrow (x-y)^2=0$

This means, it has to be $x=y$. From the condition $xy=1$ we deduce $x^2=1\Leftrightarrow x=y=\pm 1$, which gives the only two solution $(1,1), (-1,-1)$.

Answer 4

First you can see that $x \ne 0 $ (otherwise $xy = 0$). So we get to $y = \frac{1}{x}$ and then substitute: $x^2 + \frac{1}{x^2} = 1$ and then $x^4 + 1 = 2x^2$ which lead to $$x^4 - 2x^2 + 1 = 0$$ whcih means $(x^2-1)^2 = 0$

so $x^2 = 1$ and we get $x = y = 1 \text{ or } x= y=-1$