➊ |x + 2| - x ≥ 0 ← Notice that you have a positive number minus x,
i.e. |x+2| minus x. And this must be either
positive of zero ... i.e. ≥ 0
Case 1
If x is positive, then x+2 is positive so that |x + 2| = x+2.
It follows that |x + 2| - x = x+2 - x = 2 which is definitely positive.
So, x > 0 is part of the solution.
Case 2
If x = 0, then |x + 2| - x = |0 + 2| - 0 = 2 which is definitely positive.
So, x = 0 is part of the solution.
Case 3
If x is negative, then |x + 2| - x is a positive number minus a negative number which simplifies to a positive number plus a positive number ... which is definitely positive.
So, x < 0 is part of the solution.
Therefore, from Cases 1,2, and 3, we find that the solution to |x + 2| - x ≥ 0 is all real numbers.
Answers & Comments
Verified answer
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➊ |x + 2| - x ≥ 0 ← Notice that you have a positive number minus x,
i.e. |x+2| minus x. And this must be either
positive of zero ... i.e. ≥ 0
Case 1
If x is positive, then x+2 is positive so that |x + 2| = x+2.
It follows that |x + 2| - x = x+2 - x = 2 which is definitely positive.
So, x > 0 is part of the solution.
Case 2
If x = 0, then |x + 2| - x = |0 + 2| - 0 = 2 which is definitely positive.
So, x = 0 is part of the solution.
Case 3
If x is negative, then |x + 2| - x is a positive number minus a negative number which simplifies to a positive number plus a positive number ... which is definitely positive.
So, x < 0 is part of the solution.
Therefore, from Cases 1,2, and 3, we find that the solution to |x + 2| - x ≥ 0 is all real numbers.
ANSWER
all real numbers
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x+2-xâ¥0.......or -(x+2)-xâ¥0
2â¥0..................-2x-2â¥0
........................-2xâ¥2
...........................xâ¤-1 this is the answer.
And this is the answer in:
Set Notation (-â, -1]
NOTE: -â is negative infinity.