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# Absolute values

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Senior Manager
Joined: 10 Mar 2008
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28 Feb 2009, 16:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
SVP
Joined: 17 Jun 2008
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28 Feb 2009, 16:55
2
KUDOS
Consider three data points her
x <= -3/2
-3/2 < x <= 2/7
and
x > 2/7

For x <= -3/2
2x + 3 < 7x - 2
or, 5x > 5, or x > 1.....not possible.

for -3/2 < x <= 2/7

2x+3 > 2 - 7x
or, 9x > -1
or, x > -1/9....hence, values of x satisfying given inequality is -1/9 <x <= 2/7

For x > 2/7
2x+3 > 7x - 2
or, 5x < 5 or, x < 1

Hence, for the range of x -1/9 < x < 1 the given inequality is true.

Senior Manager
Joined: 08 Jan 2009
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01 Mar 2009, 05:48
4
KUDOS
|2x + 3| > |7x - 2|

Both the sides of the inequality are modulus and linear functions therefore we can square both sides,

(2x+3)^2 > ( 7x-2)^2

4x^2+9+12X > 49X^2 +4-28x

-45x^2 +40x +5 >0

9x^2 - 8x -1 <0

on solving (x-1)(9x+1) <0

so x lies between -1/9<X<1

Agree with C.
Re: Absolute values   [#permalink] 01 Mar 2009, 05:48
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