Solve |x-3| ≤ 5: Absolute Value Inequality Step-by-Step

Absolute Value Inequalities with Compound Solutions

Given:

∣x−3∣≤5 |x-3| \leq 5

Which of the following statements is necessarily true?

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Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Given:

∣x−3∣≤5 |x-3| \leq 5

Which of the following statements is necessarily true?

2

Step-by-step solution

To solve the inequality ∣x−3∣≤5 |x-3| \leq 5 , we need to consider the definition of absolute value inequalities. The inequality ∣a∣≤b |a| \leq b translates to −b≤a≤b -b \leq a \leq b .

Applying this to our expression ∣x−3∣≤5 |x-3| \leq 5 , we have:

−5≤x−3≤5 -5 \leq x-3 \leq 5 .

We add 3 to all parts of the inequality to isolate x x :

−5+3≤x−3+3≤5+3 -5 + 3 \leq x - 3 + 3 \leq 5 + 3

This simplifies to −2≤x≤8 -2 \leq x \leq 8 .

3

Final Answer

−2≤x≤8 -2 \leq x \leq 8

Key Points to Remember

Essential concepts to master this topic
  • Rule: ∣a∣≤b |a| \leq b becomes −b≤a≤b -b \leq a \leq b
  • Technique: Rewrite ∣x−3∣≤5 |x-3| \leq 5 as −5≤x−3≤5 -5 \leq x-3 \leq 5
  • Check: Test boundary values: ∣−2−3∣=5≤5 |-2-3| = 5 \leq 5 and ∣8−3∣=5≤5 |8-3| = 5 \leq 5 ✓

Common Mistakes

Avoid these frequent errors
  • Converting to two separate inequalities with OR
    Don't split ∣x−3∣≤5 |x-3| \leq 5 into x-3 ≤ 5 OR x-3 ≥ -5 = wrong solution set! This gives you all real numbers instead of the bounded interval. Always use AND with compound inequalities: −5≤x−3≤5 -5 \leq x-3 \leq 5 .

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 the absolute value inequality become a compound inequality with AND?

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Because ∣x−3∣≤5 |x-3| \leq 5 means the distance from x to 3 is at most 5 units. This creates a bounded interval where x must satisfy both conditions simultaneously!

How is this different from ∣x−3∣≥5 |x-3| \geq 5 ?

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Great question! ∣x−3∣≥5 |x-3| \geq 5 would give you two separate regions with OR: x ≤ -2 OR x ≥ 8. The ≤ keeps values between the boundaries, while ≥ takes values outside them.

What if I get confused about which direction the inequality goes?

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Think about it step by step: ∣x−3∣≤5 |x-3| \leq 5 means −5≤x−3≤5 -5 \leq x-3 \leq 5 . Then add 3 to all three parts: −5+3≤x≤5+3 -5+3 \leq x \leq 5+3 , so −2≤x≤8 -2 \leq x \leq 8 .

How can I visualize this on a number line?

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Draw a number line and mark points -2 and 8. Since we have ≤, use closed circles (filled dots) at both endpoints and shade the entire region between them. This shows all valid x values!

What happens if the right side was negative, like ∣x−3∣≤−2 |x-3| \leq -2 ?

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That would have no solution! Absolute values are always non-negative (≥ 0), so they can never be less than or equal to a negative number. Watch out for this trap!

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