How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, but I don't know how, I thought it should be 9, but my answer was wrong.
Q2. If: $\frac{x}{x + y} = 5$ then what is $\frac{y}{x + y}$ equal to?
The answer for this is -5, I also don't know how.
Q3. If: $x^4 + y^4 = 6   x^2  y^2 \land x\neq y$ then what is $\frac{x^2 + y^2}{x^2 - y^2}$ equal to?
I guess the answer for this was $\sqrt{2}$ but I'm not sure.
Any body can explain how to solve these questions, and questions of the same pattern?

For the first one :
$$\left(x-\frac{1}{x}\right)^2 = x^2  - 2 + \frac{1}{x^2} \Leftrightarrow x^2 + \frac{1}{x^2} = \left(x-\frac{1}{x}\right)^2 + 2 \implies x^2+ \frac{1}{x^2} = 11$$
For the second one, observe that :
$$\frac{x}{x+y} + \frac{y}{x+y} = 1 \Rightarrow 5 + \frac{y}{x+y} = 1 \Leftrightarrow \frac{y}{x+y} = -4 $$
For the third one, a small hint :
$$(x^2-y^2)^2 = x^4 - 2x^2y^2 + y^4$$
Alternativelly, observe that it also is :
$$x^4 + y^4 = 6 x^2  y^2 \Leftrightarrow \frac{x^2}{y^2} + \frac{y^2}{x^2} = 6$$
