**Quote:**

BN1989 wrote:

I have a general question.

If I have an inequality with an absolute value expression, why can't I simplify the absolute value expression.

First I can devide by -2, which gives me |3-2x|>-7

Now why can't I check the two cases for the absolute value expression that I have to check when absolute value expression are in equalities?

Please read my responses above: YOU DO NOT NEED TO DO THAT, since LHS is an absolute value then it's ALWAYS more than negative number -7, so it'll be more for ALL values of x.

I understand the reason why we dont need to do the lengthy process since LHS is an absolute value. Since this topic is difficult for me I try to solve with the longer way just to get more grasp of the concept.

If we get the solutions graphically it comes out as -2 < x < 5. But it should have come for any value of x. What am i doing wrong?

The range i have got is by simplifying -2|3-2x| < 14 in the following way

|3-2x| > 7

=| -2 (x-3/2) | > 7

=2 |x-3/2| > 7

=|x-3/2 | > 7/2