Absolute Value Inequality: How to Solve and Graph

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Homework Help Overview

The discussion revolves around solving and graphing an absolute value inequality involving the expression |[3/(x-1)] - 5| < 4. Participants are exploring the implications of manipulating the inequality and the challenges they face in obtaining valid solutions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to manipulate the inequality by adding and multiplying terms, leading to confusion about the results. Some express uncertainty about the validity of their steps and results, while others suggest breaking the problem into separate cases based on the sign of (x-1).

Discussion Status

There is an ongoing exploration of different methods to approach the inequality. Some participants have identified errors in their calculations and are seeking clarification on the correct approach. Guidance has been offered regarding the necessity of considering separate cases and the importance of accurately reflecting inequalities when multiplying.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a recognition of the need to sketch graphs to better understand the behavior of the function involved.

lovemake1
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Homework Statement



l [3/(x-1)] - 5l < 4

Homework Equations


The Attempt at a Solution



My 1st step was to make the inequality like this. -4 < 3/(x-1) - 5 < 4
and then i multiplied (x-1) to both left and right side and as well as to the 5.
but in the end, my result turns out to be really wrong.
I got 9/5 < x < 4/5
which is not possible.

please help.
 
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lovemake1 said:

Homework Statement



l [3/(x-1)] - 5l < 4


Homework Equations





The Attempt at a Solution



My 1st step was to make the inequality like this. -4 < 3/(x-1) - 5 < 4
and then i multiplied (x-1) to both left and right side and as well as to the 5.
Instead of multiplying first, add 5 to all three members. That will give you 3/(x - 1) in the middle of the inequality. Remember, you can always add any amount to both (or all) sides of an inequality.

You can multiply both (or all) members of an inequality by a positive number, without changing the direction of the inequality signs. If you multiply the members by a negative value, all of the inequality signs change direction. Remember that?

Since you don't know the sign of x - 1, you're going to have to look at two cases: one where x - 1 > 0, and the other where x - 1 < 0. Each case will give you a different inequality to solve.
lovemake1 said:
but in the end, my result turns out to be really wrong.
I got 9/5 < x < 4/5
which is not possible.

please help.
 
even if i add 5 to both side, i still get a very weird inequality.

i'll show my steps so you can see where i went wrong

-1 < 3(x-1) < 9
-1(x-1) < 3 < 9(x-1)
-x + 1 < 3 < 9x -9
-2 < -8x < -12
2/8 > x > 12/8

12/8 < x < 2/8

where did i possibly go wrong? please help.
 
lovemake1 said:
even if i add 5 to both side, i still get a very weird inequality.

i'll show my steps so you can see where i went wrong

-1 < 3(x-1) < 9
Here (above). -4 + 5 isn't -1.
lovemake1 said:
-1(x-1) < 3 < 9(x-1)
When you multiply by x - 1 you need to have two separate inequalities, as I explained earlier.
lovemake1 said:
-x + 1 < 3 < 9x -9
-2 < -8x < -12
2/8 > x > 12/8

12/8 < x < 2/8

where did i possibly go wrong? please help.
 
Two separate inequalities..
do you mean from x-1 < 3 < 4x-4 into x-1 < 3 and 4x-4 < 3
solve separately ?

using this method i got an answer that is reasonable.

x -1 < 3
x < 4

and

4x - 4 > 3
4x > 7
x > 7/4


so therefore, 7/4 < x < 4

Is this correct?
 
I expect you will clear that up - the fact that you were able to recognise a wrong answer is a good thing. I just wanted to say I think you have to watch out

Edit: some nonsense deleted.


It is always going to be useful to you from now on to sketch the graph of the function. First sketch y = ([3/(x-1)] - 5). Then reflect everything that is in the negative (y < 0) part of the graph in the positive half. Then the whole graph that is in the positive (upper) part is y = |[3/(x-1)] - 5 | .

You will then see easily what I mean if I say the inequality is true in a finest range of x, whereas if you had 6 instead of 4 on the right of the inequality it would be true for two infinite ranges of x, and the sketching habit will save you no end of trouble and confusion from now on.
 
Last edited:

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