Question Number 110294 by bemath last updated on 28/Aug/20
$$\mid\mathrm{2}{x}+\mathrm{1}\mid−\mid{x}−\mathrm{2}\mid\:<\:\mathrm{4}\: \\ $$$${find}\:{the}\:{solution}\:{set} \\ $$
Answered by Rio Michael last updated on 28/Aug/20
![∣2x + 1∣ = { ((2x + 1 , x ≥ −(1/2))),((−(2x + 1), x < −(1/2))) :} ∣x−2∣ = { ((x−2, x ≥ 2)),((−(x−2) , x < 2)) :} for x < −(1/2), −(2x + 1) −[−(x−2)] < 4 ⇒ −2x−1 + x−2 < 4 ⇒ −x −3 < 4 or x > −7 for −(1/2) ≤ x < 2, 2x + 1 + x−2 < 4 ⇒ 3x−1 < 4 or x < (5/3) for x ≥ 2, 2x + 1 − (x−2) <4 ⇒ 2x + 1 −x + 2 <4 x < 1 x > −7, x < (5/3) and x< 1 ⇒ −7 < x < (5/3)](https://www.tinkutara.com/question/Q110295.png)
$$\mid\mathrm{2}{x}\:+\:\mathrm{1}\mid\:=\:\begin{cases}{\mathrm{2}{x}\:+\:\mathrm{1}\:,\:\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\left(\mathrm{2}{x}\:+\:\mathrm{1}\right),\:{x}\:<\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\mid{x}−\mathrm{2}\mid\:=\:\begin{cases}{{x}−\mathrm{2},\:\:{x}\:\geqslant\:\mathrm{2}}\\{−\left({x}−\mathrm{2}\right)\:,\:{x}\:<\:\mathrm{2}}\end{cases} \\ $$$$\mathrm{for}\:{x}\:<\:−\frac{\mathrm{1}}{\mathrm{2}},\:\:−\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)\:−\left[−\left({x}−\mathrm{2}\right)\right]\:<\:\mathrm{4} \\ $$$$\Rightarrow\:\:−\mathrm{2}{x}−\mathrm{1}\:+\:{x}−\mathrm{2}\:<\:\mathrm{4} \\ $$$$\Rightarrow\:\:−{x}\:−\mathrm{3}\:<\:\mathrm{4}\:\mathrm{or}\:{x}\:>\:−\mathrm{7} \\ $$$$\mathrm{for}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:{x}\:<\:\mathrm{2},\:\mathrm{2}{x}\:+\:\mathrm{1}\:+\:{x}−\mathrm{2}\:<\:\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{3}{x}−\mathrm{1}\:<\:\mathrm{4}\:\mathrm{or}\:\:{x}\:<\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{for}\:{x}\:\geqslant\:\mathrm{2},\:\:\mathrm{2}{x}\:+\:\mathrm{1}\:−\:\left({x}−\mathrm{2}\right)\:<\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2}{x}\:+\:\mathrm{1}\:−{x}\:+\:\mathrm{2}\:<\mathrm{4} \\ $$$$\:\:\:\:\:\:{x}\:<\:\mathrm{1} \\ $$$$\:\:{x}\:>\:−\mathrm{7},\:{x}\:<\:\frac{\mathrm{5}}{\mathrm{3}}\:\mathrm{and}\:{x}<\:\mathrm{1} \\ $$$$\Rightarrow\:−\mathrm{7}\:<\:{x}\:<\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Commented by bemath last updated on 28/Aug/20
