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

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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?