SOLUTION: I cannot differentiate between different forms of interval writing. What is interval writing and how do I know which brackets to use or what number goes where when I write out an a

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Question 340408: I cannot differentiate between different forms of interval writing. What is interval writing and how do I know which brackets to use or what number goes where when I write out an answer? For example, 0 < 5-2x.
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
When the inequality does not include an "=" sign, then the solution will not contain the interval endpoint, so then you would use a set of parentheses.
.
.
.
Example: 5-2x%3E0
-2x%3E-5
x%3C5%2F2
So the interval includes all numbers less than 5%2F2 but not 5%2F2.
In interval notation that would be,
(-infinity,5%2F2)
.
.
.
When the inequality does contain an "=" sign, then the solution would include the interval endpoint, so then you would use a bracket.
.
.
.
Example:5-2x+%3E=+0
-2x+%3E=+-5
x+%3C=+5%2F2
In interval notation,
(-infinity,5%2F2]
.
.
.
And finally for a compound inequality,
Example:0+%3C+5-2x+%3C+20
The one side you already solved,
x+%3C+5%2F2
The other side is,
5-2x%3C20
-2x%3C15
x%3E-15%2F2
Putting those two together,
(-15%2F2,5%2F2)
Then look at all of the variations with the different inequality signs,
0+%3C+5-2x+%3C+20: (-15%2F2,5%2F2)
0+%3C+5-2x+%3C=+20: [-15%2F2,5%2F2)
0+%3C=+5-2x+%3C=+20: [-15%2F2,5%2F2]
0+%3C=+5-2x+%3C+20: (-15%2F2,5%2F2]


Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If the endpoint is included, then you use [ or ]. If the endpoint is not included you use ( or ). When one of the endpoints is infinity, you ALWAYS use ( or ) for the unbounded side.



Add -5 to both sides



Multiply both sides by . Since you are multiplying by a negative, reverse the sense of the inequality.



Which is to say:



So there is no lower bound and the upper bound is

Note that your inequality symbol is strictly less than, hence the upper bound endpoint is NOT included. The interval is thus:



Contrast this with the similar situation:



which would be solve in precisely the same way, but the result WOULD include the upper bound endpoint, thus:

]


John

My calculator said it, I believe it, that settles it