Difference between revisions of "1978 AHSME Problems/Problem 6"

Latest revision as of 11:01, 13 February 2021

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

@@ Line 11: / Line 11: @@
 \textbf{(D) }3\qquad
 \textbf{(E) }4     </math>
+== Solution ==
 If <math>x=x^2+y^2</math> and <math>y=2xy</math>, then we can break this into two cases.
@@ Line 31: / Line 33: @@
 <math>2+2 = \boxed{\textbf{(E) 4}}</math>
+==See Also==
+{{AHSME box|year=1978|num-b=5|num-a=7}}
+{{MAA Notice}}