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If |3x - 2 + 4x| > 7, which of the following is a possible

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amorphous
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If |3x - 2 + 4x| > 7, which of the following is a possible [#permalink] New post   28 Apr 2018, 00:45
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If \(|3x - 2 + 4x| > 7\), which of the following is a possible value for x?

A. \(-\frac{2}{3}\)

B. \(-\frac{1}{2}\)

C. \(\frac{3}{4}\)

D. \(\frac{5}{4}\)

E. \(\frac{4}{3}\)
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Re: If |3x - 2 + 4x| > 7, which of the following is a possible [#permalink] New post   04 May 2018, 04:43
Given the equation, the value of x has to be greater than 1 to satisfy the condition. So it could be 1 of the last 2 options. On plugging we would know it is the last option.
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Re: If |3x - 2 + 4x| > 7, which of the following is a possible [#permalink] New post   09 May 2018, 23:48
Since the given equation is inside a modulus.
We have to take two cases while solving the equation
1st,
3x - 2 + 4x > 7
or, 7x>9
or,\(x > \frac{9}{7}\)


2nd,
3x - 2 + 4x < -7 (when multiplying both sides by -1 the inequality sign reverses)
or, 7x< -5
or, \(x < \frac{-5}{7}\)

therefore,
only option E falls withing the range since this is greater than \(\frac{9}{7}\)
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If |3x - 2 + 4x| > 7, which of the following is a possible [#permalink] New post   09 Dec 2023, 03:24
\(|3x - 2 + 4x| > 7\)
=> |7x - 2| > 7

Theory : |X| > A
=> X > A or X < -A


|7x - 2| > 7 so we will have two cases, 7x - 2 > 7 or 7x - 2 < -7
-Case 1: 7x - 2 > 7
=> 7x > 9
=> x > \(\frac{9}{7}\) ~ 1.28		-Case 2: 7x - 2 < -7
=> 7x < -5
=> x < \(\frac{-5}{7}\) ~ -0.71


A. \(-\frac{2}{3}\) ~ 0.67 Does not fall in the range x < -0.71 or x > 1.28 => FALSE
B. \(-\frac{1}{2}\) = 0.5 Does not fall in the range x < -0.71 or x > 1.28 => FALSE
C. \(\frac{3}{4}\) = 0.75 Does not fall in the range x < -0.71 or x > 1.28 => FALSE
D. \(\frac{5}{4}\) = 1.25 Does not fall in the range x < -0.71 or x > 1.28 => FALSE
E. \(\frac{4}{3}\) = 1.3 > 1.28 => TRUE

So, Answer will be E
Hope it helps!

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If |3x - 2 + 4x| > 7, which of the following is a possible [#permalink]
09 Dec 2023, 03:24
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