Answer
$22y^5+9y^4+7y^3-13y^2+3y-11$
Work Step by Step
Consider the given polynomial
$(13y^5+9y^4-5y^2+3y+6)-(-9y^5-7y^3+8y^2+11)$
Need to drop the parentheses and will have to change the sign of each term of the second polynomial because of the presence of the negative outside.
Thus,
$(13y^5+9y^4-5y^2+3y+6)-(-9y^5-7y^3+8y^2+11)=13y^5+9y^4-5y^2+3y+6+9y^5+7y^3-8y^2-11$
or, $=13y^5+9y^5+9y^4+7y^3-5y^2+3y+6-8y^2-11$
or, $=22y^5+9y^4+7y^3-13y^2+3y-11$
