Question 719571
Generally you cross multiply when you have an equal sign between single terms
Let's do it this way
:
{{{(y-3)/(y^2+y)}}}-{{{(y+1)/(2y)}}}
Factor out y in the 1st denominator
{{{(y-3)/(y(y+1))}}}-{{{(y+1)/(2y)}}}
the common denominator is 2y(y+1), so we have:
{{{(2(y-3) - (y+1)(y+1))/(2y(y+1))}}}
FOIL
{{{((2y-6) - (y^2+2y+1))/(2y(y+1))}}}
removing brackets changes the signs
{{{((2y - 6 - y^2 - 2y - 1))/(2y(y+1))}}}
combine like terms
{{{((-y^2 - 7))/(2y(y+1))}}}