Answer
Answer:
$\frac{2y^{2}-5y+3}{(2y+3)^{2}}$
= $\frac{2y^{2}-5y+3}{4y^{2}+12y+ 9}$
(Both answers are the same, only the second one has its denominator expanded).
Work Step by Step
Given:
$\frac{y-1}{2y+3}$ + $\frac{3}{(2y+3)^{2}}$
Evaluating the sum and simplifying:
$\frac{y-1}{2y+3}$ + $\frac{3}{(2y+3)^{2}}$
= $\frac{(y-1)(2y-3) + 3}{(2y+3)^{2}}$
= $\frac{y(2y)-3(y)-1(2y)-1(-3)}{(2y+3)^{2}}$
= $\frac{2y^{2}-3y-2y+3}{(2y+3)^{2}}$
= $\frac{2y^{2}-5y+3}{(2y+3)^{2}}$
= $\frac{2y^{2}-5y+3}{(2y)^{2}+2(2y)(3)+ 3^{2}}$
= $\frac{2y^{2}-5y+3}{4y^{2}+12y+ 9}$
