The given equation is:

$\frac{-1}{\sqrt{2} + 1 - \sqrt{3}} = \frac{\sqrt{p} + \sqrt{q} + r}{t}$

To simplify and solve, we will start by simplifying the left-hand side (LHS):

$\text{LHS} = \frac{-1}{\sqrt{2} + 1 - \sqrt{3}}$

We'll rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2} + 1 + \sqrt{3}$. Let's proceed with that.

I'll perform this simplification first.The left-hand side (LHS) simplifies to:

$\frac{-\sqrt{2}}{4}$

Thus, the equation becomes:

$\frac{-\sqrt{2}}{4} = \frac{\sqrt{p} + \sqrt{q} + r}{t}$

Now, to match both sides, we need to express the right-hand side such that it equals $\frac{-\sqrt{2}}{4}$. This implies:

• $\sqrt{p} = -\sqrt{2}$
• $\sqrt{q} + r = 0$
• $t = 4$

From $\sqrt{p} = -\sqrt{2}$, we find $p = 2$. Since $\sqrt{q} + r = 0$, we can set $q = r = 0$.

Thus, the values are:

• $p = 2$
• $q = 0$
• $r = 0$
• $t = 4$

Let me know if you'd like further details or have more questions!

Here are some related questions for you:

1. How can you rationalize a denominator in a more complex expression?
2. Why do we use conjugates in rationalization?
3. How can this method apply to solve other radical equations?
4. What are alternative methods to rationalizing fractions with square roots?
5. How can we solve equations involving cube roots or higher-order radicals?

Tip: Always double-check signs when dealing with square roots and conjugates to avoid common mistakes.