The problem we are faced with is how to combine the first term with the other 2. We can do that with a creative use of the integer 1.
Let's start with a restatement of the original problem:
#(y+3)/(4y^2)+y-3#
So let's work with the #y-3# part of the equation. Let's first bracket up the terms so it's now one term (by doing this we aren't changing anything - order of operations isn't changed by making this a single quantity:
#(y+3)/(4y^2)+(y-3)#
And let's rewrite again, this time to remind ourselves that #y-3# is a fraction, albiet one that has a denominator of 1:
#(y+3)/(4y^2)+(y-3)/1#
Now let's multiply #y-3# by 1, or #1/1#:
#(y+3)/(4y^2)+(y-3)/1*1/1#
To have the #y-3#'s denominator match up with the other term, we need the denominator of the #1# to be #(4y^2)#. And to maintain the "one-ness" of 1, we need the numerator to also be #(4y^2)#:
#(y+3)/(4y^2)+(y-3)/1*(4y^2)/(4y^2)#
And now we can multiply:
#(y+3)/(4y^2)+((y-3)*(4y^2))/(4y^2)#
#(y+3)/(4y^2)+(4y^3-12y^2)/(4y^2)#
#(y+3+4y^3-12y^2)/(4y^2)=(4y^3-12y^2+y+3)/(4y^2)#
