1978 AHSME Problems/Problem 6

Problem 6

The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:

$x=x^2+y^2 \ \ y=2xy$ is

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }4$

If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.

Case 1: $y = 0$

If $y = 0$ , then $x = x^2$ and $0 = 0$

Therefore, $x = 0$ or $x = 1$

This yields 2 solutions

Case 2: $x = \frac{1}{2}$

If $x = \frac{1}{2}$ , this means that $y = y$ , and $\frac{1}{2} = \frac{1}{4} + y^2$ .

Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$

This yields another 2 solutions.

$2+2 = \boxed{\textbf{(E) 4}}$

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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