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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 \(x\).
_________________

|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.

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 \(x\).

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.
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Re: Find the solution set for the following inequality -2|3-2x| [#permalink]

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05 Nov 2014, 08:10

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Re: Find the solution set for the following inequality -2|3-2x| [#permalink]

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26 Nov 2015, 00:13

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Re: Find the solution set for the following inequality -2|3-2x| [#permalink]

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25 Dec 2016, 08:50

Bunuel wrote:

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.

Thanx Bunuel...your solution is well understood...but i guess lot of people are asking why do we not get the same answer may be something like x=negative infinity to positive infinity if we solve it through conventional and probably lengthier process...as in the rush of things (timer etc) the "QUALITATIVE ASSESSMENT" may or may not click in mind...thanx..pl help.

Re: Find the solution set for the following inequality -2|3-2x| [#permalink]

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21 Jan 2017, 18:19

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

gmatclubot

Re: Find the solution set for the following inequality -2|3-2x|
[#permalink]
21 Jan 2017, 18:19

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