Inequalities involving division of two absolute values

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SUMMARY

The inequality |x-2|/|x+3| > (x+2)/(x+1) requires careful consideration of cases based on the signs of the expressions involved. The discussion emphasizes the importance of enumerating cases for absolute values, specifically four intervals: x < -3, -3 < x < -1, -1 < x < 2, and x ≥ 2. Each case must be analyzed to eliminate absolute values correctly before proceeding with cross-multiplication, which necessitates reversing the inequality sign when multiplying by negative quantities. The solution process involves checking conditions such as x - 1 > 0, x - 1 = 0, and x - 1 < 0.

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  • Familiarity with case analysis in mathematical proofs
  • Basic algebraic skills for cross-multiplication
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wawar05
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Homework Statement



How to solve x for these inequality?

Homework Equations



|x-2|/|x+3|> (x+2) / (x+1)

The Attempt at a Solution



(x - 2)/(x + 3) > (x + 2) / ( x+1)

the left side holds the condition that is x >= 2

however, I wonder the next step. should I crossly multiply so I get
[(x - 2)((x + 1) / (x+3)] > (x + 2)(x +3) /(x+1) and find the value of x?

or there is another condition?
 
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wawar05 said:
should I crossly multiply so I get
[(x - 2)((x + 1) / (x+3)] > (x + 2)(x +3) /(x+1)

Whatever you did, isn't what we call "cross multiplication".

or there is another condition?

There are many other combinations of conditions. For rexample you multiply both sides of an inequality by (x-1) you must consider the cases:
Case 1) x -1 > 0
Case 2) x-1= 0
Case 3) x-1 < 0. (If you multiply both sides of an inequality by by a negative number, you must reverse in inequality sign)

When you begin the problem you should enumerate the cases that must be considerd for the absolute values signs:

Case 1) x-2 \ge 0 ; x + 3 \ge 0 implies |x-2] = x-2 and |x+3] = x + 3
Case 2) x-2 \ge 0; x + 3 &lt; 0 is impossible
Case 3) x -2 &lt; 0; x + 3 \ge 0 implies |x-2| = -(x-2) and |x+3|= x + 3
Case 4) x-2 &lt; 0; x + 3 &lt; 0 implies |x-2| = -(x-2) and |x+3| = -(x+3)

It's a pain in the neck to solve problems like this deductively. is that what your course materials want you to do? Or do they want you to solve this by graphing it?
 
To solve an inequality with absolute values, first investigate all possible cases. The sign of a factor is critical if its absolute value appears in the inequality or if it is in a denominator.. Also exclude the cases when a denominator is zero.
Here you have four intervals :x<-3, -3<x<-1,-1<x<2, and x≥2.

If x<-3: x+3<0, x-2<0, x+1<0, x+2<0,
if -3<x<-1: x+3>0, x+1<0, ...

Eliminate the absolute values in each interval, according to the signs.
Then you can cross-multiplying, but take care: the inequality turns to opposite if you multiply with a negative quantity.

ehild
 

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