Ration Inequalities Rule

  • Thread starter rambo5330
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  • #1
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Homework Statement


Quick question on Rational Inequalities..

Say I have

[tex]\frac{1+x}{1-x}[/tex] [tex]\huge\geq[/tex] 1

Why can we not multiply 1 by denominator (1-x) is this because if x > 1 then (1-x) would be negative in effect changing the sign of the inequality.. but if x<1 then the inequality is preserved? therefore we subtract 1 from either side..setting the inequality equal to zero?

sorry the formatting isnt quite right on that expression but the left hand side is all one expression .. greater than or = 1
 

Answers and Replies

  • #2
rock.freak667
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Why can we not multiply 1 by denominator (1-x) is this because if x > 1 then (1-x) would be negative in effect changing the sign of the inequality.. but if x<1 then the inequality is preserved?

Yes.

therefore we subtract 1 from either side..setting the inequality equal to zero?

You can do this as well.

Alternatively, you can multiply by (1-x)2.
 
  • #3
HallsofIvy
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What you can do is think "IF 1- x> 0, then I can multiply both sides by it and get [itex]1+ x\ge 1- x[/itex]". Add x to both sides and subtract 1 from both sides and you have [itex]2x\ge 0[/itex] which is the same as [itex]x\ge 0[/itex]. Now that was assuming 1- x> 0 which is the same as 1> x. So you have [itex]0\le x< 1[/itex].

Now, IF 1- x< 0, multiplying both sides by a negative number reverses the inequality: [itex]1+ x\le 1- x[/itex]. Again, add x to both sides and subtract 1 from both sides and you have [itex]2x\le 0[/itex] or [itex]x\le 0[/itex]. The "IF" is true as long as 1< x which can't be true if [itex]x\le 0[/itex] so this gives NO new solution.

The algebra gets a bit complicated but, yes, you can subtract 1 from each side: [itex]\frac{x+1}{1-x}- 1= \frac{x+ 1- (1- x)}{1-x}= \frac{2x}{1-x}\ge 0[/itex] which is true as long as the numerator and denominator have the same sign.

That is, both 2x> 0 and 1- x> 0 which gives x> 0 and x< 1 as before, or both 2x< 0 and 1- x< 0 which gives x< 0 and x> 1 which is impossible.
 
  • #4
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thanks for the excellent explanation!
 

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