## Intermediate Algebra (6th Edition)

$\dfrac{5}{4y}$
RECALL: $x^{-m}=\dfrac{1}{x^m},m \gt 0$ Use the rule above to have: $\\=\dfrac{1}{y^1}+\dfrac{1}{(4y)^1} \\=\dfrac{1}{y}+\dfrac{1}{4y}$ Make the expressions similar using the LCD which is $4y$ to have: $\\=\dfrac{1(4)}{y(4)} + \dfrac{1}{4y} \\=\dfrac{4}{4y} + \dfrac{1}{4y} \\=\dfrac{5}{4y}$