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Re: Find the solution set for the following inequality -2|3-2x| [#permalink]
14 Mar 2012, 12:03

Expert's post

NYC5648 wrote:

Hi everybody,

can please s.o. help me with this question?

Many thanks

Find the solution set for the following inequality -2|3-2x| < 14?

Is it: -2*|3-2x|<14? If yes, then -2*|3-2x|=negative*nonnegative=nonpositive, which is ALWAYS less than positive number 14. So this inequality holds true for any x. _________________

Re: Find the solution set for the following inequality -2|3-2x| [#permalink]
15 Mar 2012, 00:48

Expert's post

optimisttageja wrote:

|3-2x| can have two possible solns, either it is (3-2x) or -(3-2x) so wr can solve this ques as

-2*(3-2x)<14 => x<5 or -2*-(3-2x)<14 => x>-4

combining above two

-4<x<5 is the solution.

Posted from my mobile device

Sometimes it's a good idea to check whether your solution is correct by plug-in method. So, plug x=10 or x=-10 and see whether the inequality holds true.

Re: Find the solution set for the following inequality -2|3-2x| [#permalink]
16 Mar 2012, 03:27

Expert's post

shankar245 wrote:

Hi Buneul, Why cant we do as we do normally as in take a positive solution , then a negative solution?

x<5 x>-2

What am i doing wrong here?

If you do it properly you'll get the same answer. But you don't need that.

Consider this: -2*|3-2x|<14 --> reduce by negative -2 and flip the sign: |3-2x|>-7 --> LHS is an absolute value, which is always nonnegative, so |3-2x| will always be more than negative -7, so it'll be more for all values of x.

Re: Find the solution set for the following inequality -2|3-2x| [#permalink]
17 Mar 2012, 06:53

Expert's post

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. _________________