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The given expression is:

\[
\frac{2\sqrt{2} - 3}{\sqrt{5} + \sqrt{3}}
\]

To simplify this, we need to rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator, \(\sqrt{5} - \sqrt{3}\):

\[
\frac{2\sqrt{2} - 3}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}}
\]

This multiplication will eliminate the square roots in the denominator. Let's proceed with the steps.

1. **Expand the Numerator**:
   \[
   (2\sqrt{2} - 3)(\sqrt{5} - \sqrt{3}) = 2\sqrt{2} \cdot \sqrt{5} - 2\sqrt{2} \cdot \sqrt{3} - 3\cdot\sqrt{5} + 3\cdot\sqrt{3}
   \]
   Simplifying each term:
   \[
   = 2\sqrt{10} - 2\sqrt{6} - 3\sqrt{5} + 3\sqrt{3}
   \]

2. **Expand the Denominator**:
   \[
   (\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2
   \]

So, the expression becomes:

\[
\frac{2\sqrt{10} - 2\sqrt{6} - 3\sqrt{5} + 3\sqrt{3}}{2}
\]

Now, divide each term in the numerator by 2:

\[
= \sqrt{10} - \sqrt{6} - \frac{3}{2}\sqrt{5} + \frac{3}{2}\sqrt{3}
\]

Thus, the simplified form of the expression is:

\[
\sqrt{10} - \sqrt{6} - \frac{3}{2}\sqrt{5} + \frac{3}{2}\sqrt{3}
\]

Would you like a more detailed explanation on any of the steps? 

Here are five related questions you might find interesting:

1. How do we rationalize a denominator with a binomial?
2. Can you explain why multiplying by the conjugate removes square roots in the denominator?
3. What are some other methods to simplify expressions with square roots?
4. How would this process differ if the denominator had more than two terms?
5. Are there cases where rationalizing the denominator isn’t necessary?

**Tip:** When dealing with expressions involving square roots, always look for opportunities to simplify by using conjugates to eliminate roots from the denominator.

Simplify \( \frac{2\sqrt{2} - 3}{\sqrt{5} + \sqrt{3}} \)

Learn how to simplify the expression \( \frac{2\sqrt{2} - 3}{\sqrt{5} + \sqrt{3}} \) by rationalizing the denominator using conjugates. This solution covers algebraic concepts involving radicals and rationalization. Suitable for students in Grades 9-11.

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