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Zorluk: ZorLinear Inequalities and Absolute Value

If xx is a real number such that ∣4−3x∣≤13|4 - 3x| \leq 13, and yy is an integer such that −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3, what is the least possible integer value of x2−yx^2 - y?

  1. A
    −8-8
  2. −7-7Cevap
  3. C
    −4-4
  4. D
    22
  5. E
    99

Cevap

The least possible integer value of x2−yx^2 - y is −7-7.
To find the minimum value of x2−yx^2 - y, we must minimize x2x^2 and maximize yy. The absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 simplifies to −3≤x≤173-3 \leq x \leq \frac{17}{3}, which contains 00, so the minimum of x2x^2 is 00. The inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 simplifies to −4≤y<8-4 \leq y < 8. Since yy is an integer, its maximum value is 77. Therefore, the minimum possible value of x2−yx^2 - y is 0−7=−70 - 7 = -7.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 for xx.
−13≤4−3x≤13  ⟹  −17≤−3x≤9  ⟹  173≥x≥−3  ⟹  −3≤x≤173-13 \leq 4 - 3x \leq 13 \implies -17 \leq -3x \leq 9 \implies \frac{17}{3} \geq x \geq -3 \implies -3 \leq x \leq \frac{17}{3}.
An absolute value inequality ∣u∣≤k|u| \leq k expands to −k≤u≤k-k \leq u \leq k. Dividing by a negative number flips the inequality signs.
2
Determine the minimum possible value of x2x^2.
Minimum x2=0x^2 = 0.
Since xx can take any real value in the interval [−3,173][-3, \frac{17}{3}], which contains 00, the minimum square of any real number in this interval is 00 at x=0x = 0.
3
Solve the double inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 for yy.
−15<1−2y≤9  ⟹  −16<−2y≤8  ⟹  8>y≥−4  ⟹  −4≤y<8-15 < 1 - 2y \leq 9 \implies -16 < -2y \leq 8 \implies 8 > y \geq -4 \implies -4 \leq y < 8.
Multiplying by 33 preserves inequalities, subtracting 11 preserves inequalities, and dividing by −2-2 reverses all inequality signs.
4
Find the maximum integer value of yy.
Maximum integer y=7y = 7.
The solution set for yy is the half-open interval [−4,8)[-4, 8). Since yy is constrained to be an integer, the largest integer strictly less than 88 is 77.
5
Minimize the expression x2−yx^2 - y.
Minimum (x2−y)=Minimum (x2)−Maximum (y)=0−7=−7\text{Minimum } (x^2 - y) = \text{Minimum } (x^2) - \text{Maximum } (y) = 0 - 7 = -7.
To minimize a difference A−BA - B, one must minimize the minuend AA and maximize the subtrahend BB.

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Linear Inequalities and Absolute Value
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Linear Inequalities and Absolute Value

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