1. I need help in solving the following compound
inequality.
I also need to write solution set in interval notation 
and graph. 

 -1 < 3 - 2x < 11

The idea is to isolate x in the middle.
Get rid of the 3 by adding -3 to all three sides:

 -1 < 3 - 2x < 11
 -3  -3        -3
-------------------
 -4 <    -2x < 8

So we have
 
    -4 < -2x < 8

Now we divide all three siade by the coefficient of x,
which is -2.  But since we are dividing by a negative
number that reverses the inequalities:

    -4     -2x      8
   ———— > ————— > ————
    -2     -2      -2

or

        2 > x > -4

Which is the same as

       -4 < x < 2

On a number line this is graphed as

--------o=================·-------
 -6 -5 -4 -3 -2 -1  0  1  2  3  4        

It is an open circle at -4, because
it is < and not <, and closed (dark
circle) at 2, because it is < and not <.

The interval notation is a shorthand
for this graph:

       (-4, 2]

The "(" on the left represents the
open circle, the -4 is the left endpoint
of the shaded part, the 2 is the right endpoint
of the shaded part, and the "]" represents
the closed circle.


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2. I need to graph the following absolute value 
inequality:

|x+2y|(greater than or equal to) 6

First we draw the boundary graph, which is found by
replacing the > sign by =

|x + 2y| = 6 

x + 2y = 6 or x + 2y = -6

So we graph both these lines, which are parallel:


 

This gives us three regions, the region above the red 
line, the region between the two lines, and the region 
below the green line.  We select an arbitrary test point 
in each of the three regions to see which we shade as 
the solution set, and which we don't.

I arbitrarily select the point (0,5) as a test point 
for the region above the red line and substitute it into 
the original inequality:

   |x + 2y| > 6
 |0 + 2(5)| > 6
         10 > 6

This is true so we shade the part above the red line.

I arbitrarily select the point (0,0) as a test point for 
the region between the two lines and substitute it into 
the original inequality:

   |x + 2y| > 6
 |0 + 2(0)| > 6
          0 > 6

This is false so we do not shade the region between the 
lines:

I arbitrarily select the point (0,-5) as a test point 
for the region below the green line and substitute it 
into the original inequality:

    |x + 2y| > 6
 |0 + 2(-5)| > 6
          10 > 6

This is true so we shade the part below the green line.

Now I can't shade on here but you can on your paper. 
Shade the big area above the red line, and the big area 
below the green line. But do not shade the region 
between the two lines.  Also draw both lines solid 
because the inequality symbol is > and not >.
If it had been > instead you would draw the line
dotted.

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3. I need to graph the following compound inequality:
 3-xy+5 

Sorry, you mistyped that.  There's no >, <, > or <

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4. I need to determine whether -4 satisfies the 
following compound equality: 

   -3x > 0 AND 3x - 4 < 11

Plug it in and see if it's true or false:

   -3x > 0 AND    3x - 4 < 11

-3(-4) > 0 AND 3(-4) - 4 < 11

     4 > 0 AND   -12 - 4 < 11

     4 > 0 AND       -16 < 11

They are BOTH true so yes, it satisfies the
compound inequality.  If either had been false
then it would not have satisfied it. If it had
been OR instead of AND only one side would have
needed to be true for it to have been a solution.

Edwin