How many integers are there in the solution set of the inequality |x−2000|+|x| ≤9999
2 answers:
Answer:
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
Answer:
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
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