



 
Hi Don, How about $x^2 + y^2 = 0?$ $x = 0, y = 0$ is a solution, but is there another? Rewrite the equation as $y^2 = x^{2}. If $x \neq 0$ then $x^2$ is positive and hence $x^2$ is negative but $y^2$ can't be negative. Hence there is no solution with $x \neq 0.$ Similarly there is no solution with $y \neq 0.$ Hence $x = 0, y = 0$ is the only solution of $x^2 + y^2 = 0.$ Now look at $xy^2  3y^2  4x + 12 = 0.$ This can be rewritten as $y^2 (x  3) 4( x  3) = 0$ or $(x3)(y^2  4) = 0.$ Thus the product of the two numbers $x  3$ and $y^2  4$ is zero and hence at least one of them must be zero. Hence the solution of $y^2 (x  3) 4( x  3) = 0$ is $\left\{ (x,y): x=3, y=2 \mbox{ or } y = 2 \right\}. $ Geometrically that's all the points on the lines $x = 3, y = 2$ and $y = 2.$ Penny




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