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Algebra Difficulty 2.9 Junior Find the answer

The number of distinct pairs (x,y)(x,y) of real numbers satisfying both of the following equations:
x=x2+y2  y=2xyx=x^2+y^2 \ \ y=2xy
is

Pick one

Solution

If x=x2+y2x=x^2+y^2 and y=2xyy=2xy, then we can break this into two cases.
Case 1: y=0y = 0
If y=0y = 0, then x=x2x = x^2 and 0=00 = 0
Therefore, x=0x = 0 or x=1x = 1
This yields 2 solutions
Case 2: x=12x = \frac{1}{2}
If x=12x = \frac{1}{2}, this means that y=yy = y, and 12=14+y2\frac{1}{2} = \frac{1}{4} + y^2.
Because y can be negative or positive, this yields y=12y = \frac{1}{2} or y=12y = -\frac{1}{2}
This yields another 2 solutions.
2+2=(E) 42+2 = \boxed{\textbf{(E) 4}}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.