Answer
$\dfrac{6\sqrt{5}-4\sqrt{3}}{33} $
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{2}{3\sqrt{5}+2\sqrt{3}}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2}{3\sqrt{5}+2\sqrt{3}} \cdot\dfrac{3\sqrt{5}-2\sqrt{3}}{3\sqrt{5}-2\sqrt{3}}
\\\\=
\dfrac{2(3\sqrt{5}-2\sqrt{3})}{(3\sqrt{5}+2\sqrt{3})(3\sqrt{5}-2\sqrt{3})}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{2(3\sqrt{5}-2\sqrt{3})}{(3\sqrt{5})^2-(2\sqrt{3})^2}
\\\\=
\dfrac{2(3\sqrt{5}-2\sqrt{3})}{9(5)-4(3)}
\\\\=
\dfrac{2(3\sqrt{5}-2\sqrt{3})}{45-12}
\\\\=
\dfrac{2(3\sqrt{5}-2\sqrt{3})}{33}
.\end{array}
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2(3\sqrt{5})+2(-2\sqrt{3})}{33}
\\\\=
\dfrac{6\sqrt{5}-4\sqrt{3}}{33}
.\end{array}