Difference between revisions of "2016 AMC 12A Problems/Problem 7"
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The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math> . <math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x</math> . <math>x+y+1=0</math> is another straight line. The only intersection of <math>y=x</math> and <math>y=-x</math> is <math>(0,0)</math> , which is not on <math>x+y+1=0</math> . Therefore, the graph is three lines that do not have a common intersection,or <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math> | The equation <math>x^2(x+y+1)=y^2(x+y+1)</math> tells us <math>x^2=y^2</math> or <math>x+y+1=0</math> . <math>x^2=y^2</math> generates two lines <math>y=x</math> and <math>y=-x</math> . <math>x+y+1=0</math> is another straight line. The only intersection of <math>y=x</math> and <math>y=-x</math> is <math>(0,0)</math> , which is not on <math>x+y+1=0</math> . Therefore, the graph is three lines that do not have a common intersection,or <math>\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}</math> | ||
Revision as of 22:20, 3 February 2016
Problem
Which of these describes the graph of 
?
Solution
The equation tells us 
or 
. generates two lines 
and 
. is another straight line. The only intersection of and is 
, which is not on . Therefore, the graph is three lines that do not have a common intersection,or 
See Also
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
