Solve |x+2| < 3: Absolute Value Inequality Analysis

Absolute Value Inequalities with Compound Solutions

Given:

x+2<3 \left|x+2\right|<3

Which of the following statements is necessarily true?

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1

Understand the problem

Given:

x+2<3 \left|x+2\right|<3

Which of the following statements is necessarily true?

2

Step-by-step solution

To solve the inequality x+2<3|x + 2| < 3, we will apply the property of absolute values by rewriting it without the absolute value sign as follows:

Step 1: Transform the absolute value inequality
Using the rule A<B|A| < B implies B<A<B-B < A < B, we write

3<x+2<3-3 < x + 2 < 3.

Step 2: Solve this compound inequality. We do this by isolating xx as follows:

  • Subtract 2 from all parts: 32<x+22<32-3 - 2 < x + 2 - 2 < 3 - 2.
  • This simplifies to: 5<x<1-5 < x < 1.

Thus, the inequality x+2<3|x + 2| < 3 is solved as 5<x<1-5 < x < 1.

The correct solution is contained in choice 3: 5<x<1-5 < x < 1.

3

Final Answer

5<x<1 -5 < x < 1

Key Points to Remember

Essential concepts to master this topic
  • Rule: |A| < B becomes -B < A < B
  • Technique: From -3 < x + 2 < 3, subtract 2: -5 < x < 1
  • Check: Test x = 0: |0 + 2| = 2 < 3 ✓

Common Mistakes

Avoid these frequent errors
  • Solving |x + 2| < 3 as two separate inequalities
    Don't split into x + 2 < 3 AND x + 2 > -3 separately = missed negative case! This ignores that absolute values create compound inequalities. Always use -3 < x + 2 < 3 as one connected statement.

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 does |x + 2| < 3 become -3 < x + 2 < 3?

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The absolute value x+2 |x + 2| represents the distance from zero. When this distance is less than 3, the expression inside must be between -3 and 3!

What if I got x > -5 and x < 1 as separate answers?

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You're on the right track! But remember, both conditions must be true simultaneously. Write it as one compound inequality: 5<x<1 -5 < x < 1 .

How do I check my solution interval?

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Pick any number in your interval (like x = 0) and substitute: 0+2=2<3 |0 + 2| = 2 < 3 ✓. Also test the boundaries to make sure they're not included!

What's the difference between < and ≤ in absolute value inequalities?

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With < (strict inequality), the boundary values are not included. Use open circles on number lines. With ≤, the boundary values are included - use closed circles.

Why isn't the answer -5 < x < 2?

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That's a common error! When you have -3 < x + 2 < 3, you must subtract 2 from all three parts: -3 - 2 < x < 3 - 2, giving -5 < x < 1.

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