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Zorluk: OrtaReal Numbers, Number Line, and Absolute Value

If xx and yy are real numbers such that ∣x+2∣≤3|x + 2| \le 3 and ∣y−5∣≤2|y - 5| \le 2, which of the following could be the value of ∣x−y∣|x - y|? Select all such values.

  1. A
    1
  2. 4Cevap
  3. 8Cevap
  4. 12Cevap
  5. E
    15

Cevap

The possible values for ∣x−y∣|x - y| are 4, 8, and 12.
Solving the inequality ∣x+2∣≤3|x + 2| \le 3 gives −5≤x≤1-5 \le x \le 1, and solving ∣y−5∣≤2|y - 5| \le 2 gives 3≤y≤73 \le y \le 7. The minimum possible value of x−yx - y occurs at −5−7=−12-5 - 7 = -12, and the maximum value occurs at 1−3=−21 - 3 = -2. Thus, x−yx - y lies entirely in the interval [−12,−2][-12, -2]. Taking absolute values shows that ∣x−y∣|x - y| must lie in the interval [2,12][2, 12]. The numbers 4, 8, and 12 all fall within this interval and are valid solutions.

Adım Adım Çözüm

1
Solve the absolute value inequality for xx.
−3≤x+2≤3  ⟹  −5≤x≤1-3 \le x + 2 \le 3 \implies -5 \le x \le 1
Unpack ∣x+2∣≤3|x + 2| \le 3 into a compound inequality and isolate xx.
2
Solve the absolute value inequality for yy.
−2≤y−5≤2  ⟹  3≤y≤7-2 \le y - 5 \le 2 \implies 3 \le y \le 7
Unpack ∣y−5∣≤2|y - 5| \le 2 into a compound inequality and isolate yy.
3
Determine the minimum and maximum possible values of x−yx - y.
Minimum x−y=−5−7=−12x - y = -5 - 7 = -12; Maximum x−y=1−3=−2x - y = 1 - 3 = -2.
To minimize x−yx - y, take the smallest xx and largest yy. To maximize x−yx - y, take the largest xx and smallest yy.
4
Find the range of ∣x−y∣|x - y|.
2≤∣x−y∣≤122 \le |x - y| \le 12
Since −12≤x−y≤−2-12 \le x - y \le -2, taking the absolute value yields values in the closed interval [2,12][2, 12].
5
Evaluate the choices against the interval [2,12][2, 12].
4, 8, and 12 lie inside [2,12][2, 12], whereas 1 and 15 do not.
Any real number in [2,12][2, 12] is a achievable value for ∣x−y∣|x - y|.

Anahtar Kavram

Real Numbers, Number Line, and Absolute Value Inequalities
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Konular

Real Numbers, Number Line, and Absolute Value

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