Absolute Value Inequality

[car0le_la]

My work: ( I solved all the options through the absolute rule:if |u| <a,a> 0 then -a<u<a
|x-7|<3
-3<x-7<3
x-7>-3 and x-7<3
x>4
x<10
solved for x interval: 4<x<10

|x-5|<2
-2<x-5<2
x-5>-2 and x-5<2
x>3
x<7
solved for x interval: 3<x<7

|x-4|<3
-3<x-4<3
x-4>-3 and x-4<3
x>1
x<7
solved for x interval: 1<x<7

|x-3|<4
-4<x-3<4
x-3>-4 and x-3<4
x>-1
x<7
solved for x interval: -1<x<7

|x-2|<5
-5<x-2<5
x-2>-5 and x-2<5
x>-3
x<7
solved for x interval: -3<x<7

But my results don't match up with 3<x<4, I feel like I made an error somehow, or that I am not understanding the concept of the problem. Is there any step that needs to be done after solving for the x interval?

NOTE:
The answer key says it's supposed to be b, if that helps.

 

[car0le_la]

I noticed that for the range of b, 3<x<7, the number of 4 is the range between 3 and 7 and that would somehow connect back to the 3<x<4, but that would not make sense at all and it feels like I'm just stretching it to make it seem like B is the answer.

 

[Dr.Peterson]

It would be nice if there were instructions; but it seems to be asking, if you know that 3<x<4, then which of the following do you know to be true? It apparently is not asking for an equivalent inequality, but for a one-way implication.

So, which of the five intervals (all of which you found correctly) include the given interval? Answer: all except (a).

That is, if x is between 3 and 4, then it is also (b) within 2 units of 5; (c) within 3 units of 4; (d) within 4 units of 3; and (e) within 5 units of 2.

I can't see any interpretation of the problem for which only (b) would be correct (unless, of course, there is a typo somewhere). Was there any further explanation of what they are asking?

 

[MarkFL]

I would begin with:

[MATH]\left|x-\frac{7}{2}\right|<\frac{1}{2}[/MATH]
Multiply by 2:

[MATH]|2x-7|<1[/MATH]
None of the given choices work. (b) would work for:

[MATH]3<x<7[/MATH]

 

[pka]

View attachment 19825

This was one of my favorite to teach.
If \(a<b\) then the set \(\{x: a<x<b\}\) is written as:
Let \(c=\frac{a+b}{2}~\&~r=\frac{b-a}{2}\) where \(c\) is for centre and \(r\) is for radius.
\(a<x<b\) is \(|x-c|<r\)
So the answer is \(\left|x-\frac{7}{2}\right|<\frac{1}{2}\).
So none of the given distractors is correct

 

[pka]

View attachment 19825

Let's look at each of these distractors.
\(a)~|x-7|<3 \leftrightarrow 4<x<10\\b)~|x-5|<2 \leftrightarrow 3<x<7\\c)~|x-4|<3 \leftrightarrow 1<x<7\\d)~|x-3|<4 \leftrightarrow -1<x<7\\e)~|x-2|<5 \leftrightarrow -3<x<7\)
Are any of those subsets of \(3<x<4~?\)

 

[car0le_la]

It would be nice if there were instructions; but it seems to be asking, if you know that 3<x<4, then which of the following do you know to be true? It apparently is not asking for an equivalent inequality, but for a one-way implication.

So, which of the five intervals (all of which you found correctly) include the given interval? Answer: all except (a).

That is, if x is between 3 and 4, then it is also (b) within 2 units of 5; (c) within 3 units of 4; (d) within 4 units of 3; and (e) within 5 units of 2.

I can't see any interpretation of the problem for which only (b) would be correct (unless, of course, there is a typo somewhere). Was there any further explanation of what they are asking?

There wasn't any further explanation or interpretation of the problem. It is most likely that there was a type somewhere.

 

[car0le_la]

Let's look at each of these distractors.
\(a)~|x-7|<3 \leftrightarrow 4<x<10\\b)~|x-5|<2 \leftrightarrow 3<x<7\\c)~|x-4|<3 \leftrightarrow 1<x<7\\d)~|x-3|<4 \leftrightarrow -1<x<7\\e)~|x-2|<5 \leftrightarrow -3<x<7\)
Are any of those subsets of \(3<x<4~?\)

No, none of those are subsets of 3<x<4. So logically none of the answers are possible.

 

[car0le_la]

This was one of my favorite to teach.
If \(a<b\) then the set \(\{x: a<x<b\}\) is written as:
Let \(c=\frac{a+b}{2}~\&~r=\frac{b-a}{2}\) where \(c\) is for centre and \(r\) is for radius.
\(a<x<b\) is \(|x-c|<r\)
So the answer is \(\left|x-\frac{7}{2}\right|<\frac{1}{2}\).
So none of the given distractors is correct

I'm actually curious, what is this concept called? I wasn't taught something like that when learning to resolve these types of questions and it seems like it would be more helpful/efficient instead of having to solve through each multiple choice like I did.

 

[Dr.Peterson]

No, none of those are subsets of 3<x<4. So logically none of the answers are possible.

I think possibly pka meant to ask whether (3,4) is a subset of any of those sets; it's a subset of all but (a). This is what I said in post #3: if x is in (3,4), then it is in all the others. That is what the problem is literally asking for, though it's likely that's not what they intended. One way or another, they messed up the problem.

I'm actually curious, what is this concept called? I wasn't taught something like that when learning to resolve these types of questions and it seems like it would be more helpful/efficient instead of having to solve through each multiple choice like I did.

What pka did (assuming the problem is asking which of the five is equivalent) is to reverse the problem by writing the given interval in absolute value form, so it can be directly compared to the choices.

The idea is that the absolute value form, |x-c|<r, says "x is within r units of c" (which I also mentioned in post #3): c is the center, the midpoint of (a,b), and r is the radius, the distance from c to either a or b. This is an extremely useful way to think of these, and saves a good bit of work.

 

[Steven G]

I am a bit lost here.

If I had two choices to pick from, namely

if 3 < x < 4 then 2 < x < 5 and
if 3 < x < 4 then 3.2 < x < 3.8

I would pick the 1st one.

Assume e>0. If a<x<b then a-e<x<b+e.

In this case all 4 choices are valid.

No, none of those are subsets of 3<x<4. So logically none of the answers are possible.

Not true that there are no answers. You want the given inequality to be contained in one (or more ) of the choices to consider that choice to be correct.