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Result of an inequation, different from the algebraic process.
05-13-2024, 02:18 AM (This post was last modified: 05-13-2024 04:56 AM by compsystems.)
Post: #1
Result of an inequation, different from the algebraic process.
Hello

solve(((x-2)/(x+2))<2,x,'=') [ENTER] returns set[x<-6,x>-2] this is
(-2,+∞) union (-∞, -6)

but my calculations give me another domain.
(-2,+∞) union (-6,+∞)

Where am I going wrong?

step by step solution.

((x-2)/(x+2))<2
(x+2)>0
[(x+2)>0]+-2
(x+2-2)>(0-2)
(x+0)>-2
x>-2
(-2,+∞)

((x-2)/(x+2))<2
[((x-2)/(x+2))<2]*(x+2)
[((x-2)/(x+2))*(x+2)]<[2*(x+2)]
(x-2)<(2*x+4)
[(x-2)<(2*x+4)]+-2*x
[(x-2*x-2)<(2*x-2*x+4)]
(-x-2)<(4)
(-x-2+2)<(4+2)
(-x)<(6)
[-x<6]*-1
[-1*-x<6*-1]
(x)>(-6)
(-6,+∞)

Edit. a second case must be considered when the denominator is negative.
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Result of an inequation, different from the algebraic process. - compsystems - 05-13-2024 02:18 AM



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