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SSSSS [86.1K]
3 years ago
9

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

Mathematics
2 answers:
andriy [413]3 years ago
5 0

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

Fofino [41]3 years ago
3 0

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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