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Zorluk: KolayLinear Inequalities and Absolute Value

If ∣3x−6∣≤12|3x - 6| \le 12, which of the following inequality ranges represents all possible values of xx?

  1. −2≤x≤6-2 \le x \le 6Cevap
  2. B
    x≤6x \le 6
  3. C
    −6≤x≤2-6 \le x \le 2
  4. D
    −6≤x≤6-6 \le x \le 6
  5. E
    −2≤x≤2-2 \le x \le 2

Cevap

−2≤x≤6-2 \le x \le 6
The expression ∣3x−6∣≤12|3x - 6| \le 12 expands to the double inequality −12≤3x−6≤12-12 \le 3x - 6 \le 12. Adding 66 to each part produces −6≤3x≤18-6 \le 3x \le 18. Dividing all terms by 33 yields −2≤x≤6-2 \le x \le 6, which matches the correct solution range.

Adım Adım Çözüm

1
Rewrite the absolute value inequality as a compound inequality.
−12≤3x−6≤12-12 \le 3x - 6 \le 12
An absolute value inequality of the form ∣u∣≤k|u| \le k (where k≥0k \ge 0) is equivalent to −k≤u≤k-k \le u \le k.
2
Add 66 to all three parts of the inequality.
−6≤3x≤18-6 \le 3x \le 18
Isolate the term containing xx.
3
Divide all three parts of the inequality by 33.
−2≤x≤6-2 \le x \le 6
Since 3>03 > 0, dividing by 33 isolates xx without flipping the inequality signs.

Anahtar Kavram

Solving Linear Absolute Value Inequalities
Tahmini Süre:45s
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