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22 Nov 2019, 03:26
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E
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How many values of x are there such that x is an integer and \(|3x — 2| < 8\)?
A. One
B. Two
C. Three
D. Four
E. Five
A. One
B. Two
C. Three
D. Four
E. Five
22 Nov 2019, 07:48
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Bunuel
How many values of x are there such that x is an integer and \(|3x — 2| < 8\)?
A. One
B. Two
C. Three
D. Four
E. Five
A. One
B. Two
C. Three
D. Four
E. Five
---------ASIDE---------------------
When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive
------------------------------------
Take: \(|3x — 2| < 8\)
Applying Rule #1, we can write: \(-8 < 3x — 2 < 8\)
Add 2 to all sides to get: \(-6 < 3x < 10\)
Divide all sides by 3 to get: \(-2 < x < \frac{10}{3}\)
In other words: \(-2 < x < 3 \frac{1}{3}\)
So, the INTEGER values of \(x\) that satisfy the above inequality are: \(x = -1, 0, 1, 2\) and \(3\)
There are FIVE such values.
Answer: E
Cheers,
Brent
General Discussion
Updated on: 19 Jan 2025, 05:09
Originally posted by BrushMyQuant on 21 May 2022, 13:23.
Last edited by BrushMyQuant on 19 Jan 2025, 05:09, edited 2 times in total.
Last edited by BrushMyQuant on 19 Jan 2025, 05:09, edited 2 times in total.
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧
\(|3x — 2| \) is an absolute value of 3x-2
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
\(|3x — 2| \) is an absolute value of 3x-2
| Whenever there is an absolute value we need to consider two cases | |
| -Case 1: Whatever is inside the modulus is >= 0 3x-2 >= 0 If the value inside the modulus is non-negative then the modulus can be removed directly As 3x-2 >= 0 which means 3x >=2 or \(x >= \frac{2}{3}\) so \(|3x — 2| \) = 3x - 2 => \(3x — 2 < 8\) => 3x < 8+2 => 3x < 10 => x < \(\frac{10}{3}\) Now we have two conditions for x \(\frac{2}{3} <= x < \frac{10}{3}\) => So, x = 1,2,3 is a SOLUTION | -Case 2: Whatever is inside the modulus is < 0 3x-2 < 0 If the value inside the modulus is negative then the modulus can be removed after multiplying the terms inside the modulus by -1 As 3x-2 >= 0 which means 3x < 2 or \(x < \frac{2}{3}\) so \(|3x — 2| \) = -(3x - 2) => \(-(3x — 2) < 8\) => 3x > 2 - 8 => 3x > -6 => x > \(\frac{-6}{3}\) => x > -2 Now we have two conditions for x \(-2 < x < \frac{2}{3}\) => So, x = -1,0 is a SOLUTION |
So, Answer will be E
Hope it helps!
Watch the following video to learn how to Solve Inequality + Absolute value Problems
