1978 AHSME Problems/Problem 6: Difference between revisions

Revision as of 20:35, 21 January 2020

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.

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

@@ Line 1: / Line 1: @@
+== Problem 6 ==
+The number of distinct pairs <math>(x,y)</math> of real numbers satisfying both of the following equations:
+<cmath>x=x^2+y^2 \   \ y=2xy</cmath>
+is
+<math>\textbf{(A) }0\qquad
+\textbf{(B) }1\qquad
+\textbf{(C) }2\qquad
+\textbf{(D) }3\qquad
+\textbf{(E) }4     </math>
+If <math>x=x^2+y^2</math> and <math>y=2xy</math>, then we can break this into two cases.
+Case 1: <math>y = 0</math>
+If <math>y = 0</math>, then <math>x = x^2</math> and <math>0 = 0</math>
+Therefore, <math>x = 0</math> or <math>x = 1</math>
+This yields 2 solutions
+Case 2: <math>x = \frac{1}{2}</math>
+If <math>x = \frac{1}{2}</math>, this means that <math>y = y</math>, and <math>\frac{1}{2} = \frac{1}{4} + y^2</math>.
+Because y can be negative or positive, this yields <math>y = \frac{1}{2}</math> or <math>y = -\frac{1}{2}</math>
+This yields another 2 solutions.
+<math>2+2 = \boxed{\textbf{(E) 4}}</math>