Solving Inequalities: 4x² + 2x ≤ 3x + 2

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Discussion Overview

The discussion revolves around solving the inequality 4x² + 2x ≤ 3x + 2. Participants explore different methods for solving the inequality, including separating it into two parts and expressing solutions as intervals.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the inequality and seeks assistance.
  • Another participant suggests solving the two parts of the inequality separately and combining the solution sets.
  • A participant claims to have found that x is less than 0.5 for the first part and x is greater than or equal to -1 for the second part, but is unsure about the solution set for the first part.
  • A later reply corrects the interpretation of the second part's solution set, stating that if x is greater than or equal to -1, the solution set should be written as [-1, ∞).
  • There is uncertainty expressed regarding how to properly write the interval for the first part of the inequality.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solution sets for the inequalities, with some disagreement on the correct representation of the intervals.

Contextual Notes

There are unresolved aspects regarding the proper interval notation for the first part of the inequality and the conditions under which the solution sets are valid.

pinkyjoshi65
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Hey..This is a question I am having difficulty in solving.
4x[tex]\angle[/tex]2x+1[tex]\leq[/tex]3x+2

First I removed the "1"from the centre. Then I tried eliminating the X's from both sides, but that did not work. Could someone help me with this?
 
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Solve [itex]4x<2x+1[/itex] and [itex]2x+1\leq 3x+2[/itex] separately. I would express both solutions sets as intervals. Then if both inequalities must be satisfied, what would you have to do with the two solution sets you found?
 
so when i solve the 1st part i get x is less than 0.5. And when i solve the 2nd part, i got X is greater than/ equal to -1
so the solution set for the 2nd part is (-infi, -1}. I'm not sure about the solution set of the 1st part..
 
pinkyjoshi65 said:
so when i solve the 1st part i get x is less than 0.5. And when i solve the 2nd part, i got X is greater than/ equal to -1

Right.

so the solution set for the 2nd part is (-infi, -1}.

Wrong. If [itex]x\geq-1[/itex] then the solution set is [itex][-1,\infty)[/itex].

I'm not sure about the solution set of the 1st part..

But you practically have it. You already said that [itex]x<0.5[/itex]. How do you write down the interval containing all the numbers that are less than 0.5?
 

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