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Absolute Value Inequality

  • Thread starter lovemake1
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
33,634
5,291

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.
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.
 
  • #3
150
0
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.
 
  • #4
33,634
5,291
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.
-1(x-1) < 3 < 9(x-1)
When you multiply by x - 1 you need to have two separate inequalities, as I explained earlier.
-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.
 
  • #5
150
0
Two seperate inequalities..
do you mean from x-1 < 3 < 4x-4 into x-1 < 3 and 4x-4 < 3
solve seperately ?

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?
 
  • #6
epenguin
Homework Helper
Gold Member
3,724
765
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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