Solve the Absolute Value Inequality: |2x + 4| > 3x + 1

Absolute Value Inequalities with Multiple Cases

Given:

2x+4>3x+1 \left|2x + 4\right| > 3x + 1

Which of the following statements is necessarily true?

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Step-by-step written solution

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1

Understand the problem

Given:

2x+4>3x+1 \left|2x + 4\right| > 3x + 1

Which of the following statements is necessarily true?

2

Step-by-step solution

Firstly, let's analyze the inequality: 2x+4>3x+1 \left|2x + 4\right| > 3x + 1 .

We split it into two separate inequalities:

(1) 2x+4>3x+1 2x + 4 > 3x + 1 and (2) 2x+4<(3x+1) 2x + 4 < -(3x + 1) .

For inequality (1):

2x+4>3x+1 2x + 4 > 3x + 1

Subtract 3x 3x from both sides:

2x+43x>1 2x + 4 - 3x > 1

x+4>1 -x + 4 > 1

Subtract 4 from both sides:

x>3 -x > -3

Divide both sides by 1 -1 and flip the inequality:

x<3 x < 3

For inequality (2):

2x+4<3x1 2x + 4 < -3x - 1

Add 3x 3x to both sides:

5x+4<1 5x + 4 < -1

Subtract 4 from both sides:

5x<5 5x < -5

Divide both sides by 5:

x<1 x < -1

When we combine the solutions, we see that the answer is x<3 x < 3

3

Final Answer

x<3 x < 3

Key Points to Remember

Essential concepts to master this topic
  • Definition: Split |expression| > value into two separate inequality cases
  • Technique: Case 1: 2x + 4 > 3x + 1, Case 2: 2x + 4 < -(3x + 1)
  • Check: Test x = 0: |2(0) + 4| = 4 > 3(0) + 1 = 1 ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to consider both positive and negative cases
    Don't solve just |2x + 4| > 3x + 1 as one inequality = missing half the solution! The absolute value creates two scenarios: when the expression inside is positive AND when it's negative. Always split into both cases: expression > value AND expression < -value.

Practice Quiz

Test your knowledge with interactive questions

Given:

\( \left|2x-1\right|>-10 \)

Which of the following statements is necessarily true?

FAQ

Everything you need to know about this question

Why do I need to consider two cases for absolute value inequalities?

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The absolute value represents distance from zero, so |2x + 4| could equal either (2x + 4) or -(2x + 4) depending on whether 2x + 4 is positive or negative. You must check both possibilities!

How do I combine the solutions from both cases?

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For absolute value inequalities with '>' or '≥', use OR to combine solutions. For '<' or '≤', use AND. Since this problem uses '>', we take the union of both solution sets.

What if one case gives no solution?

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That's normal! Sometimes one case might be impossible or give no valid solutions. Just use the solution from the case that does work. Always check your final answer in the original inequality.

Why is the answer x < 3 and not just x < -1?

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From Case 1 we get x < 3, and from Case 2 we get x < -1. Since we need either condition to be true (OR), the solution includes all values where x < 3, which contains the x < -1 region too.

How can I verify my final answer?

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Pick test values! Try x = 0 (should work since 0 < 3): |2(0) + 4| = 4 and 3(0) + 1 = 1, so 4 > 1 ✓. Try x = 4 (shouldn't work since 4 > 3): |2(4) + 4| = 12 and 3(4) + 1 = 13, so 12 > 13 is false

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