Absolute Value Simplifications

Simplify, leaving as little as possible inside absolute value signs:

I 9x I

I 2x^2 I

I -2x^2 I

I -6y I

I x^2 over -y I

fraction^

Solve:

I q I = 0

I x I = 3

10 - I 2x - 1 I = 4

I 3x + 4 = I x - 7 I

If anyone could show me how to do these examples so I can do the real work I'd be appreciative!
the more detail the better ^^

|9x| = |9|.|x| = 9x ... (recall that |xy| = |x|.|y|)

|2x^2| = |2.x.x| = |2|.|x|.|x| = 2|x|^2

|-2x^2| = |-2.x.x| = |-2|.|x|.|x| = 2|x|^2

|-6y| = |-6|.|y| = 6|y|

|x^2 / y| = |x.x / y| = |x|.|x| / |y| = |x|^2 / |y| ... (recall that |x/y| = |x|/|y|)

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To solve |x| = a for x, there are a couple of cases. If a is negative, then there is no solution, since |x| is always positive. Otherwise, x = a or -a. For example |x| = 3 means that x = 3 or -3.

|q| = 0
q = 0 or -0
q = 0

10 - |2x - 1| = 4
|2x - 1| = 6
2x - 1 = 6 or -6
2x = 7 or -5
x = 7/2 or -5/2

To do the last one, remember that |x| = |y| means that either x = y or x = -y. In this case:

3x + 4 = x - 7, OR 3x + 4 = -(x - 7)
2x = -11 OR 4x = 3
x = -11/2 OR 4/3