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Chemistry 8 Online
OpenStudy (anonymous):

restrictions for y^2 + y + 6/ y^2 -4 ? I know when you break it down it y+3/ Y+2

OpenStudy (preetha):

Should post this in Math

OpenStudy (anonymous):

By restrictions, I am guessing that you want to know when the denominator is zero. So, set up that equation and solve it: y^2 - 4 = 0 (y +2)(y - 2) = 0 y = +/- 2

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