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

 
Line 11: Line 11:
 
\textbf{(D) }3\qquad  
 
\textbf{(D) }3\qquad  
 
\textbf{(E) }4    </math>
 
\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.
 
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>
 
<math>2+2 = \boxed{\textbf{(E) 4}}</math>
 +
 +
==See Also==
 +
{{AHSME box|year=1978|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 12:01, 13 February 2021

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png