Question 1235
 Case (i) If 3x - 4 >= 0,
  then |3x-4| > 11 converts to 
  3x - 4 > 11 
  or 3x > 11 + 4 = 15,
  So, x > 5...(*)
  But,by 3x - 4 >= 0, or 3x >= 4 or x >= 4/3...(1)
  Hence, the numbers x > 5 satisfying the given condition  (1)

 Case (ii) If 3x - 4 < 0,
  then |3x-4| > 11 converts to 
  -(3x - 4) > 11 , or -3x + 4 > 11,
  or -3x > 11 - 4 = 7,
  Dividing both sides by -3(have to change the direction of the inequality)
  We have x < -7/3 ...(**)
  
  But,by 3x - 4 < 0, or 3x < 4 or x < 4/3...(2)
  We see that the numbers x < -7/3 satisfying the given condition  (2)

 Answer: x > 5 or x < -7/3
        The solution set is (-oo, -7/3) U (5,+oo)


 Another way: USe |ax -b| > c (note a > 0,c >=0) is equivalent to
 x > (b+c)/a or x < (c -b)/a

 Now a = 3, b = 4, c = 11 and we have
  x > 5 or x < -7/3

 Moreover, if c < 0, |ax -b| > c is true for all x and the solution
 set is R(all real numbers)