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Question Number 133991 by EDWIN88 last updated on 26/Feb/21

 ∣ 3x−∣ 4x+2 ∣∣ ≥ 4 > ∣ 5x+8 ∣

$$\:\mid\:\mathrm{3x}−\mid\:\mathrm{4x}+\mathrm{2}\:\mid\mid\:\geqslant\:\mathrm{4}\:>\:\mid\:\mathrm{5x}+\mathrm{8}\:\mid\: \\ $$

Answered by benjo_mathlover last updated on 26/Feb/21

 ∣5x+8∣ < 4 ≤ ∣ 3x−∣4x+2∣∣  (•) ∣5x+8∣ < 4 ⇒−4<5x+8 < 4  ⇒−12 < 5x < −4 ; −((12)/5) <x <−(4/5)  (••) ∣3x−∣4x+2∣∣ ≥ 4    3x−∣4x+2∣ ≤ −4 ∪ 3x−∣4x+2∣ ≥ 4  ∣ 4x+2 ∣ ≥ 3x+4 ∪ ∣4x+2∣ ≤ 3x−4  (i)x ≥−(1/2)⇒4x+2 ≥3x+4 ; x≥2    Solution : x≥ 2  (ii) x≤−(1/2)⇒−4x−2≥3x+4 ; x≤−(6/7)   Solution : x≤−(6/7)  (iii)∣4x+2∣ ≤ 3x−4   defined for 3x−4 >0 , x>(4/3)   then (7x−2)(x+6)≤0   No solution.  Finally solution set is   {−((12)/5) <x <−(4/5)} ∩ { x≤−(6/7) ∪ x ≥ 2 }  ≡ −−_  ^(  )  < x ≤−−_  ^

$$\:\mid\mathrm{5x}+\mathrm{8}\mid\:<\:\mathrm{4}\:\leqslant\:\mid\:\mathrm{3x}−\mid\mathrm{4x}+\mathrm{2}\mid\mid \\ $$ $$\left(\bullet\right)\:\mid\mathrm{5x}+\mathrm{8}\mid\:<\:\mathrm{4}\:\Rightarrow−\mathrm{4}<\mathrm{5x}+\mathrm{8}\:<\:\mathrm{4} \\ $$ $$\Rightarrow−\mathrm{12}\:<\:\mathrm{5x}\:<\:−\mathrm{4}\:;\:−\frac{\mathrm{12}}{\mathrm{5}}\:<\mathrm{x}\:<−\frac{\mathrm{4}}{\mathrm{5}} \\ $$ $$\left(\bullet\bullet\right)\:\mid\mathrm{3x}−\mid\mathrm{4x}+\mathrm{2}\mid\mid\:\geqslant\:\mathrm{4}\: \\ $$ $$\:\mathrm{3x}−\mid\mathrm{4x}+\mathrm{2}\mid\:\leqslant\:−\mathrm{4}\:\cup\:\mathrm{3x}−\mid\mathrm{4x}+\mathrm{2}\mid\:\geqslant\:\mathrm{4} \\ $$ $$\mid\:\mathrm{4x}+\mathrm{2}\:\mid\:\geqslant\:\mathrm{3x}+\mathrm{4}\:\cup\:\mid\mathrm{4x}+\mathrm{2}\mid\:\leqslant\:\mathrm{3x}−\mathrm{4} \\ $$ $$\left(\mathrm{i}\right)\mathrm{x}\:\geqslant−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{4x}+\mathrm{2}\:\geqslant\mathrm{3x}+\mathrm{4}\:;\:\mathrm{x}\geqslant\mathrm{2} \\ $$ $$\:\:\mathrm{Solution}\::\:\mathrm{x}\geqslant\:\mathrm{2} \\ $$ $$\left(\mathrm{ii}\right)\:\mathrm{x}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow−\mathrm{4x}−\mathrm{2}\geqslant\mathrm{3x}+\mathrm{4}\:;\:\mathrm{x}\leqslant−\frac{\mathrm{6}}{\mathrm{7}} \\ $$ $$\:\mathrm{Solution}\::\:\mathrm{x}\leqslant−\frac{\mathrm{6}}{\mathrm{7}} \\ $$ $$\left(\mathrm{iii}\right)\mid\mathrm{4x}+\mathrm{2}\mid\:\leqslant\:\mathrm{3x}−\mathrm{4} \\ $$ $$\:\mathrm{defined}\:\mathrm{for}\:\mathrm{3x}−\mathrm{4}\:>\mathrm{0}\:,\:\mathrm{x}>\frac{\mathrm{4}}{\mathrm{3}} \\ $$ $$\:\mathrm{then}\:\left(\mathrm{7x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{6}\right)\leqslant\mathrm{0} \\ $$ $$\:\mathrm{No}\:\mathrm{solution}. \\ $$ $$\mathrm{Finally}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{is}\: \\ $$ $$\left\{−\frac{\mathrm{12}}{\mathrm{5}}\:<\mathrm{x}\:<−\frac{\mathrm{4}}{\mathrm{5}}\right\}\:\cap\:\left\{\:\mathrm{x}\leqslant−\frac{\mathrm{6}}{\mathrm{7}}\:\cup\:\mathrm{x}\:\geqslant\:\mathrm{2}\:\right\} \\ $$ $$\equiv\:−\underset{ } {\overset{ } {−}}\:<\:{x}\:\leqslant−\underset{ } {\overset{ } {−}} \\ $$

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