Let's go through the math problems in the image one by one.

Question 7

You are asked to find $x^2 + y^2$ given:

• $x + y = -\sqrt{3}$
• $xy = 1 - \sqrt{3}$

Using the identity: $x^2 + y^2 = (x + y)^2 - 2xy$

Substitute the given values: $x^2 + y^2 = (-\sqrt{3})^2 - 2(1 - \sqrt{3})$ $x^2 + y^2 = 3 - 2(1 - \sqrt{3}) = 3 - 2 + 2\sqrt{3} = 1 + 2\sqrt{3}$ So, $x^2 + y^2 = 1 + 2\sqrt{3}$.

Question 8

You are asked to find $x^2 + 2xy + y^2$ given:

• $x = \sqrt{2} + \sqrt{3}$
• $y = \sqrt{2} - \sqrt{3}$

Using the identity: $x^2 + 2xy + y^2 = (x + y)^2$

First, find $x + y$: $x + y = (\sqrt{2} + \sqrt{3}) + (\sqrt{2} - \sqrt{3}) = 2\sqrt{2}$

Now, square it: $(x + y)^2 = (2\sqrt{2})^2 = 8$

So, $x^2 + 2xy + y^2 = 8$.

Question 9 (1)

Solve for $x$ in the equation: $\sqrt{5}x - \sqrt{3}x = 2$

Factor out $x$: $x(\sqrt{5} - \sqrt{3}) = 2$

Solve for $x$: $x = \frac{2}{\sqrt{5} - \sqrt{3}}$

Question 9 (2)

Solve for $x$ in the equation: $\sqrt{3}x - 1 = x + 1$

Move $x$ to one side: $\sqrt{3}x - x = 2$

Factor out $x$: $x(\sqrt{3} - 1) = 2$

Solve for $x$: $x = \frac{2}{\sqrt{3} - 1}$

Would you like more details on any of these solutions or further questions? Let me know if you'd like to proceed with other problems.

Here are 5 additional questions related to these problems:

1. Can you explain why the identity $x^2 + y^2 = (x + y)^2 - 2xy$ works?
2. How can you simplify fractions with square roots like in question 9?
3. What are other identities involving squares and sums of variables?
4. How do we rationalize the denominator in a solution like $\frac{2}{\sqrt{5} - \sqrt{3}}$?
5. What happens if the signs of $x$ and $y$ are reversed in a sum of squares problem?

Tip: Always double-check for simplifications when dealing with square roots and quadratic identities, as it can make solving much easier.