Answer
The rationalized value of the denominator of \[\frac{3-\sqrt{y}}{2-\sqrt{y}}\] is
\[\frac{6-y+\sqrt{y}}{4-y}\]
Work Step by Step
Consider the radical expression.
\[\frac{3-\sqrt{y}}{2-\sqrt{y}}\]
The conjugate of the denominator is \[2+\sqrt{y}\].
Multiply the numerator and the denominator by \[2+\sqrt{y}\].
\[\begin{align}
& \frac{3-\sqrt{y}}{2-\sqrt{y}}=\frac{3-\sqrt{y}}{2-\sqrt{y}}\cdot \frac{2+\sqrt{y}}{2+\sqrt{y}} \\
& =\frac{6+3\sqrt{y}-2\sqrt{y}-y}{{{2}^{2}}-{{\left( \sqrt{y} \right)}^{2}}} \\
& =\frac{6-y+\sqrt{y}}{4-y}
\end{align}\]
