Ask Question
12 August, 06:39

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

+4
Answers (2)
  1. 12 August, 06:48
    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
  2. 12 August, 06:53
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “How many integers are there in the solution set of the inequality |x-2000|+|x| ≤9999 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers