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How do you solve this absolute-value inequality #6abs(2x + 5 )> 66#?

Answer 1

Savannah Anderson

Answer: #x < -8 uu x > 3#

Solution: #6|2x + 5|> 66#

#=> |2x + 5|> 11#

By definition, #|x| > k# where #k# is a constant, means: # x< -k uu x > k# This implies,

#2x + 5 < -11 uu 2x + 5 > 11#

Case1: 2x + 5 < -11

#=> 2x < -16 => x< -8#

Case2: 2x + 5 > 11

#=> 2x > 6 => x > 3#

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Answer 2

Avery Alexander

To solve the absolute-value inequality \(6|2x + 5| > 66\), you can follow these steps: 1. First, divide both sides of the inequality by 6 to isolate the absolute value term: \(|2x + 5| > 11\). 2. Next, you split the inequality into two cases: \(2x + 5 > 11\) and \(2x + 5 < -11\). 3. For the first case, subtract 5 from both sides: \(2x > 6\), then divide both sides by 2: \(x > 3\). 4. For the second case, subtract 5 from both sides: \(2x < -16\), then divide both sides by 2: \(x < -8\). 5. Therefore, the solution to the absolute-value inequality \(6|2x + 5| > 66\) is \(x < -8\) or \(x > 3\).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve this absolute-value inequality #6abs(2x + 5 )&gt; 66#?

How do you solve this absolute-value inequality #6abs(2x + 5 )> 66#?