[tex]\dfrac{2-\frac{1}{y}}{3+\frac{1}{y}}=\left(2-\dfrac{1}{y}\right):\left(3+\dfrac{1}{y}\right)=\left(\dfrac{2y}{y}-\dfrac{1}{y}\right):\left(\dfrac{3y}{y}-\dfrac{1}{y}\right)\\\\=\dfrac{2y-1}{y}:\dfrac{3y+1}{y}=\dfrac{2y-1}{y}\cdot\dfrac{y}{3y+1}=\dfrac{2y-1}{3y+1} [/tex]
