Question 8495
 |2x-1|-|x+5|=3

 The root of 2x-1 =0 is x= 1/2
 and the root of x+5 =0 is x= -5.
 As
  -oo        -5              1/2            +oo
   -----------*---------------*---------------

 Note, -5 < 1/2 and these two roots divide the real line into 3 parts
 (-oo,-5) , [-5,1/2] and (1/2,+oo).

 Case (i) if x is in (-oo,-5) , since x+5 < 0 and 2x-1 <0 
      then  |2x-1| = -(2x-1) = -2x+1, and |x+5| = -x-5.
      So,|2x-1|-|x+5|=3 is equivalent to -2x+1 -x-5 =3
      and we get -3x = 7, x = -7/3 .
      But -7/3> -5 , so it is an invalid answer.

 Case (ii) if x is in [-5,1/2] , since x+5 >= 0 and 2x-1 <= 0 
      then  |2x-1| = -(2x-1) = -2x+1, and |x+5| = x+5.
      So,|2x-1|-|x+5|=3 is equivalent to -2x+1 +x+5 =3
      and we get -x = -3,  x = 3.
      But -3 is not in [-5,1/2] , so it is an invalid answer.
 
 Case (iii) if x is in (1/2,+oo) , since x+5 < 0 and 2x-1 > 0 
      then  |2x-1| = 2x-1, and |x+5| = -x-5.
      So,|2x-1|-|x+5|=3 is equivalent to 2x-1 -x-5 =3
      and we get x = 9.
      But 9 is in (1/2,+oo) , so it is a valid answer.
 
 We claim that there is only one solution x = 9.

 If you have  trouble understanding ,try to review the definition about
 the absolute value, think carfully and draw  
 a diagram on the real line. Sorry ,I won't give you further explanations.

 Kenny