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Question Number 171025 by Mastermind last updated on 06/Jun/22

Answered by greougoury555 last updated on 06/Jun/22

(1) x≥−3⇒((2x+3−x−2)/(x+2)) >0 ⇒((x+1)/(x+2)) >0 ; x<−2 ∪ x>−1 ∴ solution : [−3,−2) ∪(−1,∞) (2) x<−3⇒((−3−x−2)/(x+2)) >0 ⇒ ((−x−5)/(x+2)) >0 ; ((x+5)/(x+2)) <0 ⇒−5<x<−2 ∴ solution : (−5,−3) ⇔∴ final solution ≡ (−5,−2) ∪(−1,∞)

$$\left(\mathrm{1}\right)\:{x}\geqslant−\mathrm{3}\Rightarrow\frac{\mathrm{2}{x}+\mathrm{3}−{x}−\mathrm{2}}{{x}+\mathrm{2}}\:>\mathrm{0} \\ $$$$\:\Rightarrow\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\:>\mathrm{0}\:;\:{x}<−\mathrm{2}\:\cup\:{x}>−\mathrm{1}\: \\ $$$$\:\therefore\:{solution}\::\:\left[−\mathrm{3},−\mathrm{2}\right)\:\cup\left(−\mathrm{1},\infty\right) \\ $$$$\left(\mathrm{2}\right)\:{x}<−\mathrm{3}\Rightarrow\frac{−\mathrm{3}−{x}−\mathrm{2}}{{x}+\mathrm{2}}\:>\mathrm{0} \\ $$$$\Rightarrow\:\frac{−{x}−\mathrm{5}}{{x}+\mathrm{2}}\:>\mathrm{0}\:;\:\frac{{x}+\mathrm{5}}{{x}+\mathrm{2}}\:<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{5}<{x}<−\mathrm{2}\: \\ $$$$\therefore\:{solution}\::\:\left(−\mathrm{5},−\mathrm{3}\right) \\ $$$$\Leftrightarrow\therefore\:{final}\:{solution}\:\equiv\:\left(−\mathrm{5},−\mathrm{2}\right)\:\cup\left(−\mathrm{1},\infty\right) \\ $$

Commented by Tawa11 last updated on 07/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by thfchristopher last updated on 07/Jun/22

⇒(((∣x+3∣+x)/(x+2)))(((x+2)/(x+2)))>1 ⇒(x+2)∣x+3∣+x^2 +2x>x^2 +4x+4 ⇒(x+2)∣x+3∣−2(x+2)>0 ⇒(x+2)(∣x+3∣−2)>0 Case 1: ∣x+3∣−2>0 and x+2>0⇒x>−2 ⇒∣x+3∣>2 ⇒x+3<−2 or x+3>2 ⇒x<−5(rejected) or x>−1 ∴ x>−1 Case 2: ∣x+3∣−2<0 and x+2<0⇒x<−2 ⇒∣x+3∣<2 ⇒−2<x+3<2 ⇒−5<x<−1 ∴ −5<x<−2 Conclusion: −5<x<−2 or x>−1

$$\Rightarrow\left(\frac{\mid{x}+\mathrm{3}\mid+{x}}{{x}+\mathrm{2}}\right)\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{2}}\right)>\mathrm{1} \\ $$$$\Rightarrow\left({x}+\mathrm{2}\right)\mid{x}+\mathrm{3}\mid+{x}^{\mathrm{2}} +\mathrm{2}{x}>{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4} \\ $$$$\Rightarrow\left({x}+\mathrm{2}\right)\mid{x}+\mathrm{3}\mid−\mathrm{2}\left({x}+\mathrm{2}\right)>\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{2}\right)\left(\mid{x}+\mathrm{3}\mid−\mathrm{2}\right)>\mathrm{0} \\ $$$$\mathrm{Case}\:\mathrm{1}: \\ $$$$\mid{x}+\mathrm{3}\mid−\mathrm{2}>\mathrm{0}\:\mathrm{and}\:{x}+\mathrm{2}>\mathrm{0}\Rightarrow{x}>−\mathrm{2} \\ $$$$\Rightarrow\mid{x}+\mathrm{3}\mid>\mathrm{2} \\ $$$$\Rightarrow{x}+\mathrm{3}<−\mathrm{2}\:\:\mathrm{or}\:\:{x}+\mathrm{3}>\mathrm{2} \\ $$$$\Rightarrow{x}<−\mathrm{5}\left(\mathrm{rejected}\right)\:\mathrm{or}\:{x}>−\mathrm{1} \\ $$$$\therefore\:{x}>−\mathrm{1} \\ $$$$\mathrm{Case}\:\mathrm{2}: \\ $$$$\mid{x}+\mathrm{3}\mid−\mathrm{2}<\mathrm{0}\:\mathrm{and}\:{x}+\mathrm{2}<\mathrm{0}\Rightarrow{x}<−\mathrm{2} \\ $$$$\Rightarrow\mid{x}+\mathrm{3}\mid<\mathrm{2} \\ $$$$\Rightarrow−\mathrm{2}<{x}+\mathrm{3}<\mathrm{2} \\ $$$$\Rightarrow−\mathrm{5}<{x}<−\mathrm{1} \\ $$$$\therefore\:−\mathrm{5}<{x}<−\mathrm{2} \\ $$$$\mathrm{Conclusion}:\:−\mathrm{5}<{x}<−\mathrm{2}\:\:\mathrm{or}\:\:{x}>−\mathrm{1} \\ $$

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