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3x-1-x-2-3-2x-0-find-the-solution-set-




Question Number 201820 by cortano12 last updated on 13/Dec/23
   ((∣3x+1∣−∣x+2∣)/(3−∣2x∣)) ≥ 0     find the solution set.
$$\:\:\:\frac{\mid\mathrm{3x}+\mathrm{1}\mid−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{3}−\mid\mathrm{2x}\mid}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}. \\ $$
Answered by dimentri last updated on 13/Dec/23
  (((3x+1)^2 −(x+2)^2 )/(3^2 −(2x)^2 )) ≥0    (((4x+3)(2x−1))/((3+2x)(3−2x))) ≥ 0     −(3/2)<x≤−(3/4) ∨ (1/2)≤x<(3/2)
$$\:\:\frac{\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2x}\right)^{\mathrm{2}} }\:\geqslant\mathrm{0} \\ $$$$\:\:\frac{\left(\mathrm{4x}+\mathrm{3}\right)\left(\mathrm{2x}−\mathrm{1}\right)}{\left(\mathrm{3}+\mathrm{2x}\right)\left(\mathrm{3}−\mathrm{2x}\right)}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:−\frac{\mathrm{3}}{\mathrm{2}}<\mathrm{x}\leqslant−\frac{\mathrm{3}}{\mathrm{4}}\:\vee\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{x}<\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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