What are the solutions to the inequality?

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Homework Help Overview

The discussion revolves around solving the inequality \(\frac{3x + 1}{2x - 6} < 3\). Participants are exploring the implications of different assumptions about the variable \(x\) and how these affect the solution set.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to manipulate the inequality by assuming different ranges for \(x\) (less than or greater than 3) and derive corresponding inequalities. Questions arise regarding how to determine the appropriate solution set based on these assumptions.

Discussion Status

Some participants have offered guidance on interpreting the results of their assumptions, suggesting that the solution regions can be simplified based on the relationships between the inequalities derived. There is an exploration of alternative methods for solving the inequality, with some participants sharing their approaches and discussing the implications of those methods.

Contextual Notes

Participants are considering the implications of the denominator being zero and the behavior of the inequality around critical points, specifically at \(x = 3\) and \(x = \frac{19}{3}\). There is an emphasis on ensuring that the assumptions made do not lead to contradictions.

Darth Frodo
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Homework Statement


[itex]\frac{3x + 1}{2x - 6}[/itex] < 3


Homework Equations





The Attempt at a Solution



[itex]\frac{3x + 1}{2(x -3}[/itex] < 3

[itex]\frac{3x +1}{x - 3}[/itex] < 6

Assume x < 3

3x + 1 > 6(x - 3)
3x + 1 > 6x - 18
3x + 1 - 6x + 18 > 0
19 > 3x
x < 19/3

No Contradiction.


Assume x > 3
3x + 1 < 6x - 18
3x - 19 > 0
3x > 19
x > 19/3

No Contradiction




How do I know which set of values to take?
 
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Darth Frodo said:

Homework Statement


[itex]\frac{3x + 1}{2x - 6}[/itex] < 3

Homework Equations


The Attempt at a Solution



[itex]\frac{3x + 1}{2(x -3}[/itex] < 3

[itex]\frac{3x +1}{x - 3}[/itex] < 6

Assume x < 3

3x + 1 > 6(x - 3)
3x + 1 > 6x - 18
3x + 1 - 6x + 18 > 0
19 > 3x
x < 19/3

No Contradiction. Assume x > 3
3x + 1 < 6x - 18
3x - 19 > 0
3x > 19
x > 19/3

No Contradiction

How do I know which set of values to take?
[

Hi Darth. You need to consider both the assumption and the consequent result in determining the solution region.

For example {x < 3 and x < 19/3} means that x must be less than 19/3, but also less than 3, but this is equivalent to simply stating x < 3.
 


For your second solution region, x must be greater than 3, but also greater than 19/3. This is equivalent to simply stating [itex]x > \ldots[/itex]. (You fill that one in. :smile:)
 


I attempted a different method

Is the correct answer

x < 3
x > 19/3
 


Yes that is correct.

BTW. What was your "different method". One alternative is to multiply both sides by the square of the denominator, which makes it into a single quadratic inequality.
 


Yeah, that was what I did
Then tested the point
 


Could also do [tex]\frac{3x+1 - 3(2x-6)}{2x-6}<0[/tex] simplify and then draw up an algebraic table to see when each of the quantities (3x-19), (2x-6) are negative, zero or positive (or undefined)for values of x less than 3, 3, 3<x<19/3, 19/3 and greater than 19/3.
 

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