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

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The discussion revolves around solving the inequality 5 < 2x - 1 ≤ 15 using interval notation. The original approach involved creating four cases, leading to an incorrect final answer of [-7, -2) ∪ (3, 8). Upon reevaluation, the correct method simplifies to solving 3 < x ≤ 8, which is confirmed as the accurate solution. The confusion may stem from mixing this problem with absolute value inequalities, which require different handling. Ultimately, the simplest solution method is preferred for clarity and efficiency.
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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:

2 &lt; |2x - 1| \le 15

where the absolute value sign creates a problem.
 

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