Question 38838
{{{(2y+3)/(y^2-6y+5)}}} Anything divided by zero is undefined, so you would use the denominator.
{{{y^2-6y+5}}}
{{{(y^2-5y)+(-y+5)}}}
{{{y(y-5)-1(y-5)}}}
{{{(y-1)(y-5)}}}
For values 5 and 1, the expression is undefined.