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Inequality with Modulus

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Inequality with Modulus

by kartikshah » Fri Jul 27, 2012 5:59 pm
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

I don't have the OA for this question. But my working showed A to be correct.
Can someone confirm this to be the correct answer?
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by tutorphd » Fri Jul 27, 2012 6:58 pm
I get C.

One way to solve this problem is to split it in 4 cases to resolve the absolute values.

Alternative way is to square it:
|2x + 3| > |7x - 2| is equivalent to (2x + 3)^2 > (7x - 2)^2

Simplify the squared inequality to 0 > 9x^2 - 8x -1 = (9x +1)(x-1), find the roots and deduce the answer -1/9 < x < 1.
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by eagleeye » Fri Jul 27, 2012 7:02 pm
kartikshah wrote: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

I don't have the OA for this question. But my working showed A to be correct.
Can someone confirm this to be the correct answer?
tutorphd is right.
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by tisrar02 » Fri Jul 27, 2012 7:09 pm
tutorphd wrote:I get C.

One way to solve this problem is to split it in 4 cases to resolve the absolute values.

Alternative way is to square it:
|2x + 3| > |7x - 2| is equivalent to (2x + 3)^2 > (7x - 2)^2

Simplify the squared inequality to 0 > 9x^2 - 8x -1 = (9x +1)(x-1), find the roots and deduce the answer -1/9 < x < 1.
Could you please elaborate further on this question. I seem to have difficulty understanding your methodology.

Thank you
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by Birottam Dutta » Fri Jul 27, 2012 7:10 pm
|2x + 3| > |7x - 2|

=> 2x + 3 > 7x-2 or x<1------ (1)

|2x + 3| > |7x - 2|

=> 2x-3< -(7x-2) or -1/9 <x.--- (2)

Again, -(2x-3) < (7x-2) which again gives -1/9<x.

Combining (1), (2) and (3), we get,

-1/9<x<1.

My answer is C!
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by GMATGuruNY » Fri Jul 27, 2012 9:02 pm
kartikshah wrote: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

I don't have the OA for this question. But my working showed A to be correct.
Can someone confirm this to be the correct answer?
According to A, x<-1/9.
According to C, x >-1/9.
Check whether x=-1/2 satisfies |2x + 3| > |7x - 2|:
|2(-1/2) + 3| > |7(-1/2) - 2|
2 > 11/2.
Doesn't work.
Eliminate any answer choice includes x=-1/2 in its range.
Eliminate A, B and E.

According to C, x<1.
According to D, x<5.
Check whether x=2 -- a value BETWEEN 1 and 5 -- satisfies |2x + 3| > |7x - 2|:
|2*2 + 3| > |7*2 - 2|
7 > 12
Doesn't work.
Eliminate D, since it includes x=2 in its range.

The correct answer is C.
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