Question

Answer 1

a: \(=\dfrac{3\left(x+3\right)+6x+8\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9+6x+8x-24}{\left(x-3\right)\left(x+3\right)}=\dfrac{17x-15}{\left(x-3\right)\left(x+3\right)}\)

b: \(=\dfrac{-20x}{3y^2}\cdot\dfrac{6y}{5x}=\dfrac{-8}{y}\)