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If |3x - 2 + 4x| > 7, which of the following is a possible
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28 Apr 2018, 00:45
28 Apr 2018, 00:45
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68% (01:58) correct
31% (01:51) wrong based on 29 sessions
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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. \(\frac{4}{3}\)
A. \(-\frac{2}{3}\)
B. \(-\frac{1}{2}\)
C. \(\frac{3}{4}\)
D. \(\frac{5}{4}\)
E. \(\frac{4}{3}\)
Re: If |3x - 2 + 4x| > 7, which of the following is a possible
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04 May 2018, 04:43
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.
Re: If |3x - 2 + 4x| > 7, which of the following is a possible
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09 May 2018, 23:48
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 \(\frac{9}{7}\)
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 \(\frac{9}{7}\)
