Question

State any restrictions on the variables.


3y^2-3y
-------
y+1


A.
y ≠ 1

B.
y ≠ –1

C.
y ≠ 3

D.
no restrictions are needed

Answer 1

@Perl @Mr.Expensive

Answer 2

Though I cannot understand this problem very much, probably I will be like that.
\[\frac{ 3^y{2}-3y }{ y+1 }=\frac{ 3(y^{2}-y) }{ y+1 }=\frac{ 3(y+1)(y-1) }{ y+1 }=3(y-1)\]
\[\frac{ 3y^{2} -3y}{ y+1 }=0, y=1\]

So the answer is "A".

Answer 3

Thank you @Tsuyo

Answer 4

A

Answer 5

Oh, don't worry about it.

Answer 6

Actually, the answer was B...