Answer
$\dfrac{8y + xy + 4}{2xy^2}$
Restriction: $x \ne 0; y \ne 0$
Work Step by Step
Before adding the two expressions, we need to find the least common denominator. Let's factor the denominators completely first:
$\dfrac{5y + 2}{xy^2} + \dfrac{2x - 4}{4xy}$
The denominators are already completely factored.
The least common multiple for the two expressions is the product of the prime factors of those polynomials expressed to the greatest power present, which is $4xy^2$. Now, we have to multiply the numerator of each polynomial with whatever factor is lacking between its denominator and the least common multiple:
$\dfrac{(5y + 2)(4)}{4xy^2} + \dfrac{(2x - 4)(y)}{4xy^2}$
Multiply out the terms to simplify:
$\dfrac{20y + 8}{4xy^2} + \dfrac{2xy - 4y}{4xy^2}$
Now, we add the numerators and keep the denominator as-is:
$\dfrac{20y + 8 + (2xy - 4y)}{4xy^2}$
Distribute to get rid of the parentheses, paying attention to the signs:
$\dfrac{20y + 8 + 2xy - 4y}{4xy^2}$
Combine like terms:
$\dfrac{16y + 2xy + 8}{4xy^2}$
Factor out a $2$ from the numerator:
$\dfrac{2(8y + xy + 4)}{4xy^2}$
Divide both numerator and denominator by $2$ to simplify:
$\dfrac{8y + xy + 4}{2xy^2}$
Now, we need to find the restrictions by seeing what values of the variable will make the denominator equal to zero because if the denominator of any rational expression is zero, the expression is undefined.
If $4xy^2$ cannot be zero, then $x$ cannot be zero and $y$ cannot be zero:
Restriction: $x \ne 0; y \ne 0$