Prove that if x≠0 then if y= 3x^2+2y/x^2+2 then y=3

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Discussion Overview

The discussion centers around a mathematical problem involving the relationship between variables x and y, specifically proving that if x≠0 and a certain equation holds, then y must equal 3. The scope includes mathematical reasoning and potential homework-related queries.

Discussion Character

  • Mathematical reasoning, Homework-related, Debate/contested

Main Points Raised

  • One participant proposes the initial condition that if x≠0, then y is defined by the equation y = (3x^2 + 2y)/(x^2 + 2).
  • Another participant reformulates the equation to y(1 - 2/x^2) = 3x^2 + 2, suggesting that the expression does not appear to be constant.
  • A third participant interprets the equation differently, indicating that if y is expressed as y = (3x^4 + 2x^2)/(x^2 - 2), it does not yield a constant value.
  • A later reply clarifies the original equation and derives y = 3 through algebraic manipulation, asserting that y x^2 = 3 x^2 leads to y = 3.
  • Another participant critiques the manner of the inquiry, stating that it is not an appropriate way to seek help for homework problems and questions the forum's suitability for this discussion.

Areas of Agreement / Disagreement

Participants express differing interpretations of the original equation and its implications. There is no consensus on the validity of the claims or the appropriateness of the inquiry.

Contextual Notes

Some participants highlight potential misunderstandings in the formulation of the equation and its implications, indicating that the assumptions made may not lead to a definitive conclusion.

bean29
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suppose that x and y are real numbers. Prove that if x≠0 then if y= 3x^2+2y/x^2+2 then y=3
 
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y(1-2/x2) = 3x2 + 2

y = (3x4 + 2x2)/(x2-2)

Doesn't look constant.
 
It looks like he wanted to write
[tex]y = \frac{3x^2+2y}{x^2+2}[/tex].
Then it works out:
[tex]y x^2 + 2y = 3x^2 + 2y \Leftrightarrow y x^2 = 3 x^2 \Leftrightarrow y = 3[/tex]
 
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This is not an acceptable way to ask for help at homework problems. In addition, it is in the wrong forum.
 

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