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# MATH HELP? asap please!! which restriction is appropriate for the variable? 3y^2-3y/y+1 answer options below !!!?

answer options:

a. y =/ 1

b. y =/ 3

c. no restrictions are needed

d. y =/ -1

### 3 Answers

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- ted sLv 76 months agoFavorite Answer
obviously d) .................................{ but as typed it should be y ╪ 0 }...use (..) when needed !!!

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- Steve ALv 76 months ago
D.

y cannot = -1 because that causes division by zero

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- Jeff AaronLv 76 months ago
3y^2 - (3y/y) + 1 = 3y^2 - 3 + 1 = 3y^2 - 2, note that y can't be 0

Or did you mean:

(3y^2 - 3y) / (y + 1), note that y can't be -1, which I think answer (d) was meant to say.

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