Absolute value inequalities: | 2 + 1 / x | > 1

[captain hero]

Hello,

I need help solving absolute inequalities. I am trying to solve these two problems but I keep getting different answers than what are in the book.

Ex1

|2+1/x|>1

-1>2+1/x>1

-3>1/x>-1

-3x>1>-x or 3x<-1<x

x< -1/3 x>-1 or x<-1/3 x>-1

Did I do this right? The book says the answer is x>0 or x<-1 or -1/3<x<0

Another problem I don't understand is essentially the same:

Ex2

|3/x-2|<=4


Thanks for any help....

 

[stapel]

Take out the middle of the first line in your "solution", and you'll find that you've wriiten "-1 > 1". This isn't true.

(And "greater than" absolute-value inequalities always get split into two inequalities right at first.)

You might want to review the topic a bit more in depth before attempting the exercises....

Thank you.

Eliz.

 

[captain hero]

could someone solve one of these for me because I still don't get it. I need to see the process of how its done.

 

[pka]

We cannot actually teach lessons here. That is for your instructional provider to do. What can do is give help in the form of hints and basic guidance.

The inequality has several equivalent forms.
\(\displaystyle \begin{array}{rcl}
\left| {2 + \frac{1}{x}} \right| > 1\quad & \Rightarrow & \quad \left| {2x + 1} \right| > \left| x \right|,\quad x \not= 0 \\
& \Rightarrow & \left( {2x + 1} \right)^2 > x^2 \\
\end{array}\)

That last inequality has a solution set of \(\displaystyle x < - 1\quad or\quad \frac{{ - 1}}{3} < x .\)

However, that not the answer to your problem. We cannot allow x=0.
So you must exclude it from the solution set. The correct notation is:
\(\displaystyle \left( { - \infty , - 1} \right) \cup \left( { - \frac{1}{3},0} \right) \cup \left( {0,\infty } \right)\).

 

[captain hero]

I wasn't asking for a lesson, was I?

 

[stapel]

I wasn't asking for a lesson, was I?

Actually, it appeared that you were:

I still don't get it. I need to see the process of how its done.

Lessons are where one would get an explanation (so they "get it") and would "see the process of how its[sic] done" in the worked examples.

If you understand the topic (so you don't need lessons), then please reply showing how far you have gotten, and we'll be glad to provide hints and helps to get you going again.

Thank you.

Eliz.

 

[soroban]

Hello, captain hero!

Are you familiar with this rule for inequalities?
. . We can take the reciprocals of both sides if we reverse the inequality.

Example: \(\displaystyle 6 \:> \:2\;\;\Rightarrow\;\;\frac{1}{6}\:<\:\frac{1}{2}\)

\(\displaystyle \L|2\,+\,\frac{1}{x}| \:>\:1\)

First, we note that: \(\displaystyle \,x\:\neq\:0\) . . . we must remember that!

With an absolute value inequality with "greater than",
. . I prefer to make two inequalities and solve them separately.

[1] We have: \(\displaystyle \L\:2\,+\,\frac{1}{x}\:>\:1\)

Subtract 2: \(\displaystyle \L\:\frac{1}{x}\:>\:-1\)

Take reciprocals: \(\displaystyle \:\fbox{x\:<\:-1}\)


[2] We have: \(\displaystyle \L\:2\,+\,\frac{1}{x} \:<\:-1\)

Subtract 2: \(\displaystyle \L\:\frac{1}{x}\:< \:-3\)

Take reciprocals: \(\displaystyle \:\fbox{x \:> \:-\frac{1}{3}}\)


The solution is: \(\displaystyle \x\,<\,-1)\:\cup\:\left(x\,>\,-\frac{1}{3}\right)\) ... except \(\displaystyle x\,=\,0\)

The intervals are: \(\displaystyle \-\infty,\,-1)\:\cup\:\left(-\frac{1}{3},\,0\right)\:\cup\0,\,\infty)\)

. . That is: \(\displaystyle \:x\,<\,-1,\; -\frac{1}{3}\,<\,x\,<\,0,\;x\,>\,0\)

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Your second problem: \(\displaystyle \L\:\left|\frac{3}{x}\,-\,2\right|\:\leq\:4\)
Again: \(\displaystyle x\,\neq\,0\) . . . but this time, it doesn't matter.

So we have: \(\displaystyle \L\:-4 \;\leq \;\frac{3}{x}\,-\,2\;\leq\;4\)

Add 2: \(\displaystyle \L\:-2 \;\leq\;\frac{3}{x}\;\leq\;6\)

Take reciprocals: \(\displaystyle \L\:-\frac{1}{2}\;\geq\;\frac{x}{3}\;\geq\;\frac{1}{6}\)

Multiply by 3: \(\displaystyle \L\:-\frac{3}{2}\;\geq\;x\;\geq\;\frac{1}{2}\)


So we have: \(\displaystyle \:x\:\leq\:-\frac{3}{2}\,\) and \(\displaystyle \,x\:\geq\:\frac{1}{2}\)

The intervals are: \(\displaystyle \:\left(-\infty,\,-\frac{3}{2}\right]\:\cup\:\left[\frac{1}{2},\,\infty\right)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I highly recommend using that "Reciprocal Rule".

Multiplying by \(\displaystyle x\) is very dangerous.
We must consider whether we are multiplying by a positive or negative value.
. . Often it gives us no problems.
. . At other times, we must handle two possible cases.

The Reciprocal Rule eliminates this bothersome decision-making.

 

[pka]

Multiplying by \(\displaystyle x\) is very dangerous.
We must consider whether we are multiplying by a positive or negative value.

But of course no one is multiplying by x. The the way I suggested is multipling by |x| thus avoiding any such problem except possibly if x=0.