So let's first look a simple example. How do we figure out the value of x for which the absolute value of x+3 is greater than 6.
Making this in mathematical format, the question becomes:
#abs(x+3)>6#
If #abs(x+3)=6#, what would be the value of x?
It would be either #x+3=6# or #x+3=-6#...since if #abs(x)=6# it actually means the distance from x to zero is 6 units. On a axis x could be either -6 or 6, so x could attain a distance of 6.
Know let's go back to your question i.e #abs(-4y-3)<13#. As we seen before here instead of just a single variable like x or y we have an expression. So instead of the variable being the distance from zero here the expression's value is the distance from zero.
Know the expression #-4y-13# would have a distance of 13 from zero. So again either #-4y-13# or #-(4y-13)#, so the expression #-4y-13# could attain a distance of 13.
Expressing mathematically:
#-4y-13<13# or #-4y+13<13#
#-4y<26# or #-4y<0#
#y> -26/4# or #y>0# which would be the solution.
