Answer
$\dfrac{y}{y^{2}+2y-3}-\dfrac{1}{y^{2}+4y+3}=\dfrac{y^{2}+1}{(y+3)(y-1)(y+1)}$
Work Step by Step
$\dfrac{y}{y^{2}+2y-3}-\dfrac{1}{y^{2}+4y+3}$
Factor the denominators of both fractions:
$\dfrac{y}{y^{2}+2y-3}-\dfrac{1}{y^{2}+4y+3}=...$
$...=\dfrac{y}{(y+3)(y-1)}-\dfrac{1}{(y+3)(y+1)}=...$
Evaluate the subtraction of these two fractions using the LCD, which is $(y+3)(y-1)(y+1)$ in this case:
$...=\dfrac{y(y+1)-(y-1)}{(y+3)(y-1)(y+1)}=...$
Simplify:
$...=\dfrac{y^{2}+y-y+1}{(y+3)(y-1)(y+1)}=\dfrac{y^{2}+1}{(y+3)(y-1)(y+1)}$
