## Algebra 2 (1st Edition)

$4x^{-1} y^3$
Here, we have $3 x^{3} y^{2}=\dfrac{12 x^{2} y^{5}}{a}$ Now, $a= \dfrac{12 x^{2} y^{5}}{3 x^{3} y^{2}}$ This gives: $a= 4 x^{2-3} \times y^{5-2}=4x^{-1} y^3$