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>. | ||
− | == | + | == 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
Contents
[hide]Problem
Define . Which of the following describes the set of points for which ?
Solution
and . Therefore, we have the equation Factoring out a gives Factoring both sides further, . It follows that if , , or , both sides of the equation equal 0. By this, there are 3 lines (, , or ) so the answer is .
Solution 2
Following from the previous solution, . Then, . Factoring, . Now, the solutions are obviously , , or , which each correspond to a line. Thus, the answer is .
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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