## Intermediate Algebra (6th Edition)

$\dfrac{6}{y\sqrt{2}}$
Multiplying both the numerator and the denominator by a factor that will make the numerator a perfect power of the radical, the rationalized-numerator form of the given expression, $\dfrac{\sqrt[]{18}}{y} ,$ is \begin{array}{l}\require{cancel} \dfrac{\sqrt[]{18}}{y}\cdot\dfrac{\sqrt[]{2}}{\sqrt[]{2}} \\\\= \dfrac{\sqrt[]{36}}{y\sqrt{2}} \\\\= \dfrac{\sqrt[]{(6)^2}}{y\sqrt{2}} \\\\= \dfrac{6}{y\sqrt{2}} .\end{array}