Solving an absolute value inequality

AI Thread Summary
The discussion centers on solving the absolute value inequality |x/(x-2)| < 5. The initial attempts to simplify the inequality led to incorrect conclusions about the solution set, specifically stating x < 5/3 and x > 5/2, which is logically impossible. Participants clarify that the correct approach involves rewriting the inequality as a single rational expression and considering the signs of both the numerator and denominator. Additionally, there is a focus on proper mathematical notation for clarity in forum discussions. Ultimately, the key takeaway is the importance of accurately handling inequalities and understanding their implications on solution sets.
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Homework Statement



lx/(x-2)l < 5

Homework Equations





The Attempt at a Solution


x/(x-2) < 5
x< 5x-10
10 < 4x
5/2 < x

x/(x-2) > -5
x > -5x+10
6x > 10
x > 5/3

The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?
 
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physphys said:

Homework Statement



lx/(x-2)l < 5

Homework Equations



The Attempt at a Solution


x/(x-2) < 5
x< 5x-10
10 < 4x
5/2 < x

x/(x-2) > -5
x > -5x+10
6x > 10
x > 5/3

The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?
For P a positive real number, \ \displaystyle \left| f(x) \right|&lt;P\ is equivalent to -P&lt;f(x)&lt;P\ .

More to the point, \displaystyle \ \frac{x}{x-2}&lt;5\ is not equivalent to \ x&lt;5(x-2)\ .\ In fact, if x-2 is negative, then you need to change the sense of the inequality .


You're much better off to write \displaystyle \ \frac{x}{x-2}&lt;5\ as \displaystyle \ \frac{x}{x-2}-5&lt;0\,,\ then write the left-hand side as a single rational expression by using a common denominator .
 
SammyS said:
For P a positive real number, \ \displaystyle \left| f(x) \right|&lt;P\ is equivalent to -P&lt;f(x)&lt;P\ .

More to the point, \displaystyle \ \frac{x}{x-2}&lt;5\ is not equivalent to \ x&lt;5(x-2)\ .\ In fact, if x-2 is negative, then you need to change the sense of the inequality .


You're much better off to write \displaystyle \ \frac{x}{x-2}&lt;5\ as \displaystyle \ \frac{x}{x-2}-5&lt;0\,,\ then write the left-hand side as a single rational expression by using a common denominator .

Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?
 
physphys said:
The answer is x < 5/3 and x > 5/2
so where did I go wrong on the second one?
The answer can't possibly be "x < 5/3 and x > 5/2". There aren't any numbers that are simultaneously smaller than 5/3, and larger than 5/2.

physphys said:
Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0.
How did you get this? I'm assuming you are starting from x/(x - 2) - 5 < 0, and took Sammy's advice to combine the two terms on the left into a single rational expression.

Also, if 2(3x-5) / x-2 is supposed to mean this:
$$ \frac{2(3x - 5)}{x - 2}$$
then you need to write it with parentheses around the entire denominator, like so:
2(3x-5) / (x-2)
Otherwise, we would interpret this to mean this:
$$ \frac{2(3x - 5)}{x} - 2$$
physphys said:
So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?
 
physphys said:
Okay so I simplified that farther to achieve this, 2([STRIKE] 3 [/STRIKE] 2x-5) / ( x-2 ) > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?

So you have one expression, 2(2x-5), divided by another, (x-2), and the result is positive. How does the sign of the numerator compare to the sign of the denominator ?
 
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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
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