Lösung 4.1:7d

Aus Online Mathematik Brückenkurs 1

We rewrite the equation in standard form by completing the square for the x- and y-terms,

x2−2xy2+2y=(x−1)2−12=(y+1)2−12.

Now, the equation is

(x−1)2−1+(y+1)2−1(x−1)2+(y+1)2=−2=0.

The only point which satisfies this equation is (xy)=(1−1) because, for all other values of x and y, the left-hand side is strictly positive and therefore not zero.