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11 November, 19:24

How many integers are there in the solution set of the inequality |x-2000|+|x| ≤9999

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Answers (2)
  1. 11 November, 21:38
    0
    9999

    Step-by-step explanation:

    Find the greatest value of x and the least value of x that satisfy the equation:

    2x-2000 = 9999 (Yields the greatest solution)

    2x + 2000 = - 9999 (Yields the least solution)

    Solve both equations and you get:

    3999.5 (Greatest solution)

    -5999.5 (Least solution)

    That means the range of integer solutions is between 3999 and - 5999 (inclusive). That means there are 9999 integral solutions
  2. 11 November, 22:59
    0
    20,000 integers

    Step-by-step explanation:

    By triangular inequality,

    |x - 2000 + x| ≤ |x-2000|+|x|

    |x-2000|+|x| ≤ 9999

    |x - 2000 + x| ≤ 9999

    |2x - 2000| ≤ 9999

    |x - 10000| ≤ 4999.5

    x - 10000 ≤ 4999.5

    x ≤ 4999.5

    x ≤ 14999.5

    - (x - 10000) ≤ 4999.5

    -4999.5 ≤ x - 10000

    -5000.5 ≤ x

    All solutions:

    -5000.5 ≤ x ≤ 14999.5

    No. of inyegers:

    14999 - (-5000) + 1

    20,000 integers
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