Result of an inequation, different from the algebraic process.
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05-13-2024, 02:18 AM
(This post was last modified: 05-13-2024 04:56 AM by compsystems.)
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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
RE: Result of an inequation, different from the algebraic process. - Jan 11 - 05-13-2024, 05:26 AM
RE: Result of an inequation, different from the algebraic process. - parisse - 05-13-2024, 06:11 AM
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