Hello

chetan2u,

I Read your post about the absolute Modulus and accordingly I tried to solve this problem by critical value method. Hereunder my solution and please correct me if I did something wrong.

Statement1:-

the critical value is 2 so we are looking for values : x>=2 or x<2.

x>=2

If x>=2 ,let x=3 for example , the modulus would be negative. so by assigning -ive sign to the modulus we will get -(6-3x)=x-2 , so x=2, which is in the region x>=2 so we will accept this value.

x<2

If x<2, let x=-3 for example , the modulus would be positive. so by assigning +ve sign to the modulus we will get (6-3x)=x-2 , so x=2, which is not in the region x<2 , so this value is rejected.

The same as for Statement 2 , the critical value is -3/5 . so x>=-3/5 or x<-3/5 after solving it we will get 2 values , one accepted "x=2" and one rejected x=-12/17