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Re: If |–3x + 1| < 7, then which of the following represents all
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12 Nov 2016, 16:58
Expert Reply
Explanation
You can Plug In or solve on this problem. To Plug In, choose a value that fits one of the answer choices, such as x = 2, which would fit in the range for choice (C). If x = 2, then \(|-3x + 1| = 5\), which is true, so we can eliminate any answer choice that doesn’t include x = 2: choices (A), (D), and (E).
Logically, it doesn’t make sense that an inequality with a < sign would have a ≤ sign when it’s been solved, but to be sure, check x = −2. In that case, \(|-3x + 1| = 7,\) and is not < 7, so the answer must be choice (B). If you solve this problem, remember that you have to solve both \(-3x + 1 < 7\), and \(-3x + 1 > 7\).
Also remember that you must flip the sign any time you multiply or divide both sides of an inequality by a negative number.
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If |3x + 1| < 7, then which of the following represents all
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17 Nov 2016, 12:58
1
sandy wrote:
If |-3x + 1| < 7, then which of the following represents all possible values of x?
A. -2 < x B. -2 < x < 8/3 C. -2 < x < 8/3 D. x < -2 or x > 8/3 E. x < -2 or x < 8/3
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
Given: |-3x + 1| < 7 In this case, we'll apply Rule #1 to get: -7 < -3x + 1 < 7 Subtract 1 from all 3 sides to get: -8 < -3x < 6 Divide all 3 sides by -3 to get: 8/3 > x > -2 [we REVERSE the inequalities when we divide by a NEGATIVE value] Rewrite as -2 < x < 8/3
If |3x + 1| < 7, then which of the following represents all
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30 Aug 2022, 22:20
1
Given that |-3x + 1| < 7 and we need to find the range for all possible values of x
Let's solve the problem using two methods
Method 1: Substitution
We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)
A. -2 < x
Lets take x = 5 (which falls in this range of -2 < x ) and substitute in the equation |-3x + 1| < 7 => |-3*5 + 1| < 7 => |-14| < 7 => 14 < 7 which is FALSE
B. -2 < x < \(\frac{8}{3}\)
Lets take x = 0 (which falls in this range of -2 < x < \(\frac{8}{3}\) ) and substitute in the equation |-3x + 1| < 7 => |-3*0 + 1| < 7 => |1| < 7 => 1 < 7 which is TRUE In test, we don't need to solve further. But I am solving to completed the solution.
C. \(-2 \leq x \leq \frac{8}{3}\)
Lets take x = 5 (which falls in this range of \(-2 \leq x \leq \frac{8}{3}\)) and substitute in the equation |-3x + 1| < 7 => |-3*\(\frac{8}{3}\) + 1| < 7 => |-8 + 1| < 7 => |-7| < 7 => 7 < 7 which is FALSE
D. \(x < -2\) or \(x > \frac{8}{3}\)
Lets take x = -3 (which falls in this range of \(x < -2\) or \(x > \frac{8}{3}\)) and substitute in the equation |-3x + 1| < 7 => |-3*-3 + 1| < 7 => |-9 + 1| < 7 => |-8| < 7 => 8 < 7 which is FALSE
E. \(x \leq -2\) or \(x \geq \frac{8}{3}\)
We can again take x = -3 to prove this one FALSE
So, Answer will be B
Method 2: Algebra
Now, we know that |A| < B can be opened as (Watch this video to know about the Basics of Absolute Value) A < B for A ≥ 0 and -A < B for A < 0
=> |-3x + 1| < 7 can be written as
Case 1: -3x + 1 ≥ 0 or x ≤ \(\frac{1}{3}\) => |-3x + 1| = -3x + 1 => -3x + 1 < 7 => 3x > 1 - 7 => 3x > -6 => x > \(\frac{-6}{3}\) => x > -2 And the condition was x ≤ \(\frac{1}{3}\), so answer will be the range common in x ≤ \(\frac{1}{3}\) and x > -2 => -2 < x ≤ \(\frac{1}{3}\) is the solution
Attachment:
-2 to 1by3.JPG [ 17.37 KiB | Viewed 1319 times ]
Case 2: -3x + 1 < 0 or x > \(\frac{1}{3}\) => |-3x + 1| = -(-3x + 1) = 3x - 1 => 3x - 1 < 7 => 3x < 7 + 1 => 3x < 8 => x < \(\frac{8}{3}\) And the condition was x > \(\frac{1}{3}\), so answer will be the range common in x > \(\frac{1}{3}\) and x < \(\frac{8}{3}\) => \(\frac{1}{3}\) < x < \(\frac{8}{3}\) is the solution
Attachment:
1by3 to 8by3.JPG [ 15.92 KiB | Viewed 1300 times ]
Final answer will be a combination of the two answers -2 < x ≤ \(\frac{1}{3}\) and \(\frac{1}{3}\) < x < \(\frac{8}{3}\) => -2 < x < \(\frac{8}{3}\)
So, Answer will be B Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems