Answer
$\frac{y^2-2y-3}{y^2+y-1}$
Work Step by Step
Step 1. Perform operations to the numerator:
$\frac{y+3}{y}-\frac{4}{y-1}=\frac{y^2-y+3y-3-4y}{y(y-1)}=\frac{y^2-2y-3}{y(y-1)}=\frac{(y-3)(y+1)}{y(y-1)}$
Step 2. Perform operations to the denominator:
$\frac{y}{y-1}+\frac{1}{y}=\frac{y^2+y-1}{y(y-1)}$
Step 3. Use the above results in the original expression, we have:
$\frac{\frac{y+3}{y}-\frac{4}{y-1}}{\frac{y}{y-1}+\frac{1}{y}}=\frac{\frac{(y-3)(y+1)}{y(y-1)}}{\frac{y^2+y-1}{y(y-1)}}=\frac{(y-3)(y+1)}{y(y-1)}\cdot\frac{y(y-1)}{y^2+y-1}=\frac{(y-3)(y+1)}{y^2+y-1}=\frac{y^2-2y-3}{y^2+y-1}$
