Absolute Value Inequality: How to Solve and Graph

AI Thread Summary
The discussion focuses on solving the absolute value inequality |[3/(x-1)] - 5| < 4. The initial approach involved incorrectly manipulating the inequality, leading to an impossible result. A suggested correction is to add 5 to all sides first, creating a clearer middle expression. It is emphasized that two cases must be considered based on the sign of (x-1) to solve the inequality accurately. Additionally, graphing the function can provide visual insight into the solution ranges for x.
lovemake1
Messages
147
Reaction score
1

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.
 
Physics news on Phys.org
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 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?
 
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:
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top