Why Can't We Cancel (x-2) in the Inequality 2(x+3)(x-2)/(x-2)≥0?

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Discussion Overview

The discussion centers around the inequality 2(x+3)(x-2)/(x-2)≥0 and the question of why the factor (x-2) cannot be canceled from the numerator and denominator. Participants explore the implications of this operation in the context of inequalities, particularly focusing on the restrictions imposed by the denominator.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the validity of canceling (x-2) in the inequality, suggesting it may relate to the nature of inequalities.
  • Another participant explains that the inequality does not include the solution x=2 because the denominator cannot be zero, emphasizing the need to consider the original inequality's solutions.
  • A different participant clarifies that what is perceived as factoring is actually division by (x-2), which is only permissible if x is not equal to 2.
  • Several participants acknowledge the importance of not dividing by the denominator without considering its value, particularly regarding the potential need to swap the inequality sign.
  • There is a discussion about the implications of dividing by a term that could be zero and how it affects the inequality.
  • One participant expresses confusion about whether they understand the implications of the inequality and the division involved.

Areas of Agreement / Disagreement

Participants generally agree on the principle that dividing by (x-2) is problematic when x=2, but there is some uncertainty about the implications of this operation on the inequality itself. The discussion remains somewhat unresolved regarding the nuances of handling inequalities with potential zero denominators.

Contextual Notes

Participants highlight the importance of understanding the conditions under which division by a variable is valid, particularly in the context of inequalities. There is a recognition of the need to consider the original solutions of the inequality when performing algebraic manipulations.

Who May Find This Useful

This discussion may be useful for students learning about inequalities, particularly those grappling with the implications of manipulating expressions that involve variables in denominators.

LogarithmLuke
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So given the problem 2(x+3)(x-2)/(x-2)≥0

How come we can't factor x-2 in the numerator against x-2 in the denominator? I mean theyre all just factors right? Does it have something to do about it being an inequality? We got this problem on a test yesterday, and the correct way to solve it was to leave it as i originally wrote it, and then put it on a number line. Why is it wrong to just factor and solve the inequality from there?
 
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It's because that the inequality ##\frac{2(x+3)(x-2)}{x-2}\ge 0## doesn't have the solution ##x=2## for the denominator shouldn't be ##0.## But, the inequality ##2(x+3)\ge 0## does.
So, if you want to divide something in numerator and denominator simultaneously, you should confirm it would not matter its original solution, that is, think clearly.
 
What you call factoring is in fact division by x-2. This can only be done if x is not equal to 2 (you can not divide by 0, not even when you do it both on top and on the bottom). If you solve x+3 > 0 you get x > -3 and have missed excluding this x=2 for the set of x that satisfies the original inequality.

[edit] slow typist. Good thing the replies agree.
 
Oh i see now. I am aware that you can't divide or multiply by the denominator on both sides of the inequality sign, because you can't know if you have to swap the inequality sign or not. I just wasnt aware that i was doing that here.
 
LogarithmLuke said:
Oh i see now. I am aware that you can't divide or multiply by the denominator on both sides of the inequality sign, because you can't know if you have to swap the inequality sign or not. I just wasnt aware that i was doing that here.
Do I understand this ? you want to change ##2(x+3)(x-2)/(x-2)\ge 0## into ##2(x+3)(x-2) \ge 0 \; (x-2) ## and worry about the ##\ge## ?
 
No, i meant that dividing x-2 by x-2 would leave uncertainty about whether or not you have to swap the inequality sign, like you guys said.
 
It leaves the inequality sign intact. But you can't do it if x-2 is 0, that's the crux.
 
Oh, i was thinking about the fact that if you mulitply or divide by something negative on both sides, you have to swap the inequality sign, but i see now that that is not relevant here because you only divide on one side.
 
OK, I'm reassured now. Well done.
 

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