1978 AHSME Problems/Problem 6

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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

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

Solution

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.

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

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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