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Zorluk: OrtaLinear Inequalities and Absolute Value

If xx is an integer that satisfies both ∣2x−5∣≤9|2x - 5| \le 9 and 3−x<53 - x < 5, how many possible values of xx exist?

  1. A
    4
  2. B
    8
  3. 9Cevap
  4. D
    10
  5. E
    0

Cevap

There are 9 possible integer values for x.
Solving ∣2x−5∣≤9|2x - 5| \le 9 gives −2≤x≤7-2 \le x \le 7. Solving 3−x<53 - x < 5 gives x>−2x > -2. Taking the intersection yields −2<x≤7-2 < x \le 7. The integers in this interval are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, amounting to 9 values in total.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣2x−5∣≤9|2x - 5| \le 9
−2≤x≤7-2 \le x \le 7
Unfold ∣2x−5∣≤9|2x - 5| \le 9 as −9≤2x−5≤9-9 \le 2x - 5 \le 9, add 5 to obtain −4≤2x≤14-4 \le 2x \le 14, and divide by 2.
2
Solve the linear inequality 3−x<53 - x < 5
x>−2x > -2
Subtract 3 to get −x<2-x < 2, then divide by −1-1 and reverse the inequality symbol.
3
Determine the intersection of both inequalities
−2<x≤7-2 < x \le 7
Combine −2≤x≤7-2 \le x \le 7 and x>−2x > -2 on the real number line.
4
Count the integer values within the intersection −2<x≤7-2 < x \le 7
9 integer values
The valid integers are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, giving a total count of 7−(−1)+1=97 - (-1) + 1 = 9.

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Linear Inequalities and Absolute Value Bounds
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Linear Inequalities and Absolute Value

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