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Difference between revisions of "2013 AMC 10B Problems/Problem 14"

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<math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>.
 
<math>x\clubsuit y = x^2y-xy^2 </math> and <math>y\clubsuit x = y^2x-yx^2</math>. Therefore, we have the equation <math> x^2y-xy^2 = y^2x-yx^2</math> Factoring out a <math>-1</math> gives <math> x^2y-xy^2 = -(x^2y-xy^2)</math> Factoring both sides further, <math>xy(x-y) = -xy(x-y)</math>. It follows that if <math>x=0</math>, <math>y=0</math>, or <math>(x-y)=0</math>, both sides of the equation equal 0. By this, there are 3 lines (<math>x=0</math>, <math>y=0</math>, or <math>x=y</math>) so the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>.
  
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== Alternate Solution ==
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== Solution 2==
 
Following from the previous solution, <math> x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>.
 
Following from the previous solution, <math> x^2y-xy^2 = y^2x-yx^2</math>. Then, <math>2x^2y-2xy^2=0</math>. Factoring, <math>2xy(x-y)=0</math>. Now, the solutions are obviously <math>x=0</math>, <math>y=0</math>, or <math>x=y</math>, which each correspond to a line. Thus, the answer is <math>\boxed{\textbf{(E)}\text{ three lines}}</math>.
  

Revision as of 18:15, 30 December 2017

Problem

Define $a\clubsuit b=a^2b-ab^2$. Which of the following describes the set of points $(x, y)$ for which $x\clubsuit y=y\clubsuit x$?

$\textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel lines}\\ \qquad\textbf{(D)}\ \text{two intersecting lines}\\ \qquad\textbf{(E)}\ \text{three lines}$

Solution

$x\clubsuit y = x^2y-xy^2$ and $y\clubsuit x = y^2x-yx^2$. Therefore, we have the equation $x^2y-xy^2 = y^2x-yx^2$ Factoring out a $-1$ gives $x^2y-xy^2 = -(x^2y-xy^2)$ Factoring both sides further, $xy(x-y) = -xy(x-y)$. It follows that if $x=0$, $y=0$, or $(x-y)=0$, both sides of the equation equal 0. By this, there are 3 lines ($x=0$, $y=0$, or $x=y$) so the answer is $\boxed{\textbf{(E)}\text{ three lines}}$.

Solution 2

Following from the previous solution, $x^2y-xy^2 = y^2x-yx^2$. Then, $2x^2y-2xy^2=0$. Factoring, $2xy(x-y)=0$. Now, the solutions are obviously $x=0$, $y=0$, or $x=y$, which each correspond to a line. Thus, the answer is $\boxed{\textbf{(E)}\text{ three lines}}$.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AMC 10 Problems and Solutions

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