Hi there! I need one more help please? Thanks!

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Answer 1 · 2017-09-20T02:47:50

#( y^2 + 2y)/(y+1)#

# y^2-y-6) = (y -3)(y+2)#

# y^2+y = y ( y + 1)# so

# { y^2(y-3)(y+2)}/{(y-3)y(y+1)}#

# (y-3)/(y-3) = 1#

# y^2/y = y/1#

Dividing out common terms gives

# {y(y+2)/(y+1)# multiplying across the parenthesis gives

# (y^2 + 2y)/ (y+1) #

y cannot be -1 because the denominator would become zero and the term would become undefined or infinite.

# ( -1+1) = 0#