Let us take an example:

Find value of x: {|x+3| + x}/(x+2) > 1

We know that 

      |x|  = x if x > 0

and

      |x|  = -x if x < 0

Case 1. if |x+3| is positive then

     {|x+3| + x}/(x+2) > 1

=> (x+3 + x)/(x+2) > 1

=> (2x + 3) > x+2

=> 2x - x > 2-3

=> x > -1

Case 2. if |x+3| is negative then

     {|x+3| + x}/(x+2) > 1

=> {-(x+3) + x}/(x+2) > 1

=> (-x - 3 + x) > x+2

=> -3 > x+ 2

=> x + 2 < -3

=>  x < -3-2

=> x < -5

So value of x lies between -5 and -1

In this way, we solve the modulus questions.

Hope you got the point.