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Zorluk: OrtaLinear Inequalities and Absolute Value

If xx is a real number that satisfies the inequality ∣3x−3∣≤9|3x - 3| \le 9, which of the following statements must be true? Select all that apply.

  1. −4≤2x≤8-4 \le 2x \le 8Cevap
  2. ∣x−1∣≤3|x - 1| \le 3Cevap
  3. x2≤16x^2 \le 16Cevap
  4. D
    x≥0x \ge 0
  5. E
    1−x≤21 - x \le 2

Cevap

The statements that must be true are −4≤2x≤8-4 \le 2x \le 8, ∣x−1∣≤3|x - 1| \le 3, and x2≤16x^2 \le 16.
Solving ∣3x−3∣≤9|3x - 3| \le 9 gives −9≤3x−3≤9-9 \le 3x - 3 \le 9, which simplifies to −2≤x≤4-2 \le x \le 4. Multiplying this range by 2 yields −4≤2x≤8-4 \le 2x \le 8. Subtracting 1 gives −3≤x−1≤3-3 \le x - 1 \le 3, which is ∣x−1∣≤3|x - 1| \le 3. Squaring values in [−2,4][-2, 4] yields non-negative numbers up to 16, so x2≤16x^2 \le 16 is also true.

Adım Adım Çözüm

1
Unfold the absolute value inequality into a compound inequality.
−9≤3x−3≤9-9 \le 3x - 3 \le 9
By definition, ∣u∣≤k|u| \le k (where k≥0k \ge 0) is equivalent to −k≤u≤k-k \le u \le k.
2
Add 3 to all parts of the compound inequality.
−6≤3x≤12-6 \le 3x \le 12
Isolating the variable term 3x3x.
3
Divide all parts by 3 to solve for xx.
−2≤x≤4-2 \le x \le 4
Dividing by a positive constant preserves the direction of the inequality signs.
4
Test each proposed statement against the interval [−2,4][-2, 4].
−4≤2x≤8-4 \le 2x \le 8 is true; ∣x−1∣≤3|x - 1| \le 3 is true; x2≤16x^2 \le 16 is true; x≥0x \ge 0 fails for x=−1x = -1; 1−x≤21 - x \le 2 fails for x=−2x = -2.
Determining which properties hold for every real number in the solution set.

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Linear Inequalities and Absolute Value
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Linear Inequalities and Absolute Value

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