2=|4+3x|  How do i solve problems like this? 

|5x+1|<5   How do you isolate x when it is not an equation
(<5 not =5)? Also, what needs to be manipulated to graph this
on a horizontal graph? 


3+|x-3|=7  How do I solve this algebraically?

Thankyou.

First here are the rules for getting rid of the absolute value symbols:

To remove absolute value symbols:

1.    |expression| = A  where A is a positive number write

      two separate equations:

      expression = A              expression = -A
      
      Solve each for the variable and there will usually be 
      two solutions.


2.    |expression| < A where A is a positive number, write
      this:

          -A  <  expression  <  A

      Then solve this for the variable in the middle.

      The answer will usually be like this C < x < D. The graph 
      will be like this

          -----------------(=======)-----------
                           C       D

      and in interval notation like this (C, D)

3.    |expression| < A where A is a positive number, write

      this:

          -A  <  expression  <  A

      Then you solve this for the variable in the middle.

      The answer will usually be like this C < x < D. The graph 
      will be like this

          -----------------[=======]-----------
                           C       D

      and in interval notation like this [C, D]


4.    |expression| > A  where A is a positive number becomes

      expression < -A   OR   expression > A

      Solve each for the variable with OR between them.
    
      The answer will usually be like this "x < C OR x > D". 
      The graph will be like this

         <=================)-------(=================>
                           C       D

      and in interval notation like this (-oo, C) U (D, oo)


5.    |expression| > A  where A is a positive number becomes

      expression < -A   OR   expression > A

      Solve each for the variable with OR between them.

      The answer will usually be like this "x < C OR x > D". 
      
      The graph will be like this

         <=================]-------[=================>
                           C       D

      and in interval notation like this (-oo, C] U [D, oo)


-----------------------------------------------

     2 = |4+3x|

This is 

     |4+3x| = 2 or case 1 above. So write the two equations:

      4+3x = 2         4+3x = -2
        3x = -2          3x = -6
         x = -2/3         x = -2

So the solution is these two values, -2/3 and -2

-------------------------------------------------

|5x+1| < 5

This is case 2:  So write this

         -5 < 5x+1 < 5

Add -1 to all three "sides":

         -5 < 5x+1 < 5
         -1     -1  -1
       ------------------
         -6 < 5x   < 4

Divide all three "sides" by 5

       -6/5 < 5x/5 < 4/5

         -6/5 < x < 4/5

The graph is

          -----------------(========)-----------:
                         -6/5     4/5



or in interval notation:   (-6/5, 4/5)

-----------------------------------------------

3 + |x-3| = 7

First get the absolute value term alone on the
left side by adding -3 to both sides, then you have

               |x-3| = 4

This is case 1 above. So write the two equations:

       x-3 = 4         x-3 = -4
         x = 7           x = -1
                 

So the solution is these two values, 7 and -1

Edwin