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2016 AMC 12A Problems/Problem 7

Revision as of 21:56, 3 February 2016 by Ericzeng (talk | contribs)
The equation $x^2(x+y+1)=y^2(x+y+1)$ tells us $x^2=y^2$ or $x+y+1=0$.$x^2=y^2$ generates two lines $y=x$ and $y=-x$.$x+y+1=0$ is another straight line.The only intersection of $y=x$ and $y=-x$ is $(0,0)$,which is not on $x+y+1=0$.Therefore,the graph is three lines that do not have a common intersection,or $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$
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