Help please!Algebra 2 is killing me I need help!

8)  Absolute value is the distance from zero on a number line (and has no reference to which direction left/right from zero), so this means the value is always positive:

 abs(2) = 2
abs(-2) = 2

9) Solve the equation for x

(3x + y)/z = 2

*multiply both sides by z

(3x + y) = 2z

*subtract y from both sides

3x = 2z - y

*divide both sides by 3

x = (2z - y)/3

10) Which point is a solution to the equation 6x - 5y = 4? Justify your choice
A.  (1, 2)
B.  (1, -2)
C.  (-1, -2)
D.  (-1, 2)

*plug (x, y) coordinates into equation and see if the result is a valid equation:

*start with A.  (1, 2):

6(1) - 5(2) = 4
6 - 10 = 4
-4 = 4            [NO GOOD]

*now try B.  (1, -2):

6(1) - 5(-2) = 4
6 - (-10) = 4
6 + 10 = 4
16 = 4            [NO GOOD]

*now try C.  (-1, -2):

6(-1) - 5(-2) = 4
-6 - (-10) = 4
-6 + 10 = 4
 4 = 4            [OK]

*just for fun let's also verify D.  (-1, 2) is not the solution, since we found that C. was:

6(-1) - 5(2) = 4
-6 - 10 = 4
-16 = 4           [NO GOOD]

The answer is C.  (-1, -2) (and the justification is that we solved for it to be true)

11)  Domain is all values 'x' (i.e. input)
     Range is all values 'y' (i.e. output)

a.) y = 2x + 1 is a line with a slope of 2:1 (vert:horiz) and a y-intercept of y = 1, but because it is a line, it extends from -infinty to +infinity for both 'x' and 'y', so . .

Domain = (-infinity ≤ x ≤ +infinity)
Range = (-infinity ≤ y ≤ +infinity)

b.)  This table on shows discrete values of input/output, so the domain/range is also discrete . .

Domain = (3, 7, 11)
Range = (-1, -3, -5)

c.) Just from visual confirmation of the plot's extents . .

Domain = (-5 ≤ x ≤ 5)
Range = (-1 ≤ y ≤ 1)

d.) Again using visual confirmation of the plot's extents . .

Domain = (-2 ≤ x ≤ 2) *note extents are limited by vertical asymptote
Range = (-infinity ≤ y ≤ +infinity)

12)  There are 2 lines of symmetry (they are the vertical line drawn at x = 0, and the horizontal line drawn at y = 0 that bisect the ellipse)