The number of distinct pairs (x,y) of real numbers satisfying both of the following equations: x=x2+y2y=2xy is
Pick one
Solution
If x=x2+y2 and y=2xy, then we can break this into two cases. Case 1: y=0 If y=0, then x=x2 and 0=0 Therefore, x=0 or x=1 This yields 2 solutions Case 2: x=21 If x=21, this means that y=y, and 21=41+y2. Because y can be negative or positive, this yields y=21 or y=−21 This yields another 2 solutions. 2+2=(E) 4
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