Difference between revisions of "2016 AMC 12A Problems/Problem 7"
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Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ? | Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ? | ||
− | <math>\textbf{(A)}\ \text{two parallel lines}\ \qquad\textbf{(B)}\ \text{two intersecting lines}\ \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\ \qquad\textbf{(D)}\ \text{three lines that do not all pass through a | + | <math>\textbf{(A)}\ \text{two parallel lines}\ \qquad\textbf{(B)}\ \text{two intersecting lines}\ \qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\ \qquad\textbf{(D)}\ \text{three lines that do not all pass through a common point}\ \qquad\textbf{(E)}\ \text{a line and a parabola}</math> |
==Solution 1== | ==Solution 1== |
Revision as of 22:40, 29 January 2017
Problem
Which of these describes the graph of ?
Solution 1
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
Solution 2
If , then dividing both sides of the equation by gives us . Rearranging and factoring, we get . If , then the equation is satisfied. Thus either , , or . These equations can be rearranged into the lines , , and , respectively. Since these three lines are distinct, the answer is .
Solution 3
Subtract on both sides of the equation to get . Factoring gives us , so either , , or . Continue on with the second half of solution 2.
Diagram:
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 |
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