Solving Inequalities in Interval Notation: Where Did I Go Wrong?

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Homework Help Overview

The original poster attempts to solve the inequality 5 < 2x - 1 ≤ 15 using interval notation. They express confusion regarding their solution and the validity of their approach, particularly in checking the intervals derived from their cases.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of creating four cases for the inequality and question the necessity of certain conditions. Some express uncertainty about the correctness of their solutions and the reasoning behind their approaches.

Discussion Status

There is an ongoing exploration of different methods to solve the inequality. Some participants suggest alternative approaches, while others reflect on the original method and its implications. No consensus has been reached, but various interpretations are being considered.

Contextual Notes

Participants mention being taught to create multiple cases for solving inequalities, indicating a potential instructional context. There is also a reference to confusion with absolute value inequalities, suggesting a broader discussion about inequality types.

aeroengphys
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In interval notation, I was asked to solve for 5 < 2x - 1 ≤ 15

I did this by making the 4 cases...

2x-1>5 or 2x-1<-5 AND 2x-1≤ 15 and 2x-1≥-15
x > 3 or x < -2 AND x ≤ 8 and x ≥ -7

So for my final answer...i got [-7,-2),(3,8]

But when i check it, none of the numbers between -7 and -2 check...what'd I do wrong?

Lastly, is the answer actually just (3,8]?
 
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What's the use in checking if 2x-1<-5 and 2x-1>= -15?
 
honestly...I have no idea, but we were taught to make four cases for those...so that's how I've been doing it ever since.
 
Actually, I just realized that I could've solved for x by adding 1 to the left and right, and dividing by 2 to the left and right...which gives me an answer of 3 < x ≤ 8...i'm pretty sure that's right...guess i went about it the wrong way
 
Actually, I just realized that I could've solved for x by adding 1 to the left and right, and dividing by 2 to the left and right...which gives me an answer of 3 < x ≤ 8

Yup, that's probably the easiest way.
 
I have no idea, but we were taught to make four cases for those...so that's how I've been doing it ever since.

You're probably confusing it with a problem like:

[tex]2 < |2x - 1| \le 15[/tex]

where the absolute value sign creates a problem.
 

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