How do you solve #1/2abs(2x + 1 )> 2 #?

Answer 1

#x > 3/2 or x < -5/2#

in interval notation

#(-oo, -5/3) uu (3/2, oo)#

#1/2|2x+1|>2#
#|2x+1|>4#
#2x+1>4# or #2x+1<-4# so
#-4>2x+1>4#
#-5>2x>3#
#-5/2>x>3/2#
#x > 3/2 or x < -5/2#
#(-oo, -5/3)uu(3/2, oo)#
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Answer 2
To solve the inequality \( \frac{1}{2} |2x + 1| > 2 \), follow these steps: 1. Multiply both sides of the inequality by 2 to eliminate the fraction: \( |2x + 1| > 4 \). 2. Split the absolute value inequality into two separate inequalities: \( 2x + 1 > 4 \) and \( 2x + 1 < -4 \). 3. Solve each inequality separately: - For \( 2x + 1 > 4 \): - Subtract 1 from both sides: \( 2x > 3 \). - Divide both sides by 2: \( x > \frac{3}{2} \). - For \( 2x + 1 < -4 \): - Subtract 1 from both sides: \( 2x < -5 \). - Divide both sides by 2: \( x < -\frac{5}{2} \). 4. Combine the solutions: \( x < -\frac{5}{2} \) or \( x > \frac{3}{2} \).
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Answer 3
To solve \( \frac{1}{2} |2x + 1| > 2 \), follow these steps: 1. Multiply both sides by 2 to get rid of the fraction: \( |2x + 1| > 4 \) 2. Break the absolute value inequality into two cases: a) \( 2x + 1 > 4 \) b) \( 2x + 1 < -4 \) 3. Solve each case separately: a) For \( 2x + 1 > 4 \): Subtract 1 from both sides: \( 2x > 3 \) Divide both sides by 2 (since 2 is positive, inequality direction doesn't change): \( x > \frac{3}{2} \) b) For \( 2x + 1 < -4 \): Subtract 1 from both sides: \( 2x < -5 \) Divide both sides by 2 (since 2 is positive, inequality direction doesn't change): \( x < -\frac{5}{2} \) 4. Combine the solutions: \( x < -\frac{5}{2} \) or \( x > \frac{3}{2} \)
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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.

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.

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.

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.

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