#### Answer

$3y^2 +5y+\frac{29}{3}+\frac{124}{3(3y-5)}$

#### Work Step by Step

$(4y+9y^3-7) \div (-5+3y)$
$=(9y^3+4y-7) \div (3y-5)$
$3y^2+5y$
_________________________
$3y-5)9y^3+4y-7$
$-9y^3+15y^2$
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$15y^2+4y$
$-15y^2+25y$
_________________________
$29y-7$
$-29y+\frac{145}{3}$
_________________________
$\frac{124}{3}$
The answer is $3y^2 +5y+\frac{29}{3}+\frac{124}{3(3y-5)}$.