1 Answer

Majlinda · Answer 1 · 2019-07-19T01:48:09+0000

we are given

(y^2)/(y-3)*(y^2-y-6)/(y^2+1y)

we know that denominators can not be zero

so, restrictions would be values of y where denominator is not zero

$y-3\\eq 0$

$y\\eq 3$

$y^2+1y \\eq 0$

$y(y+1) \\eq 0$

$y \\eq 0$

$y \\eq -1$

so, restrictions are

$y \\eq 0$

$y \\eq -1$

$y\\eq 3$

(y^2)/(y-3)*(y^2-y-6)/(y^2+1y)

now, we can factor numerators and denominator

and then we can simplify it

(y^2)/(y-3)*(y^2-y-6)/(y^2+1y)=(y^2)/(y-3)*((y-3)(y+2))/(y(y+1))

now, we can cancel it

(y^2)/(y-3)*(y^2-y-6)/(y^2+1y)=(y(y+2))/(y+1)

(y^2)/(y-3)*(y^2-y-6)/(y^2+1y)=(y^2+2y)/(y+1)

so,

option-B..........Answer

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