A doubt in using Modulus

Question

|x-12| = 12 - 3x

Solving This gives 
 x-12 = 3x-12; 
2x = 0 => x = 0;

and

|x-12| = 12 -3x;
 
4x = 24;
x = 6; 
However, if I sub X as in 6 in the above equation

|6-12| = 12 - 18;
|-6| = -6 ; ( LHS will be always positive, because of the mod)

I don't know how to proceed after this, Please help 
Also, can someone please share a few questions which apply this concept?

aniket16c · Answer

|x-12|=12-3x

The procedure to solve this question is to consider two cases:
x-12>0 and x-12<0
Case 1: x-12>0 = x>12
x-12=12-3x --> x=0.
This solution is not valid as x>12 is the pre- condition for this solution

Case 2: x-12<0 = x<12
-(x-12)=12-3x
12-x=12-3x --> x=0
This solution is valid as the pre-condition of x<12 is satisfied

saukrit · Answer

Well if you consider the basics of modulus

|x| = x if x>0

|x| = -x if x<0

Now when considering |x-12| for the case when x-12> 0, You must have x>12 to have a valid solution.  (1)

Now when you solve
|x-12| = 12 -3x;
4x = 24;
x = 6
You know from  (1)  that x has to be greater that 12. Hence the solution x=6 is no longer valid. Hope this explains your doubt.

A doubt in using Modulus

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