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Question Number 16519 by tawa tawa last updated on 23/Jun/17

Solve:  ∣2 − x∣ − 2 ∣x + 1∣ < 1

$$\mathrm{Solve}: \\ $$ $$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$

Answered by ajfour last updated on 23/Jun/17

Commented byajfour last updated on 23/Jun/17

∣2−x∣−2∣x+1∣<1  whenever y=∣2−x∣ is below   y= 2∣x+1∣+1   this occurs for x<x_1  (see graph)   and  x>x_2   to obtain x_1 (wbich is <−1):   2−x_1  = 1−2(x_1 +1)   x_1 = −3   to obtain x_2   (−1<x_2  <2)   2−x_2  = 1+2(x_2 +1)   3x_2 = −1  ⇒   x_2  = −1/3    so, for given condition to  remain true             x ∉ [−3, ((−1)/3)]     that is x∈ (−∞, −3)∪(((−1)/3), ∞).

$$\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid<\mathrm{1} \\ $$ $${whenever}\:{y}=\mid\mathrm{2}−{x}\mid\:{is}\:{below} \\ $$ $$\:{y}=\:\mathrm{2}\mid{x}+\mathrm{1}\mid+\mathrm{1} \\ $$ $$\:{this}\:{occurs}\:{for}\:{x}<{x}_{\mathrm{1}} \:\left({see}\:{graph}\right) \\ $$ $$\:{and}\:\:{x}>{x}_{\mathrm{2}} \\ $$ $${to}\:{obtain}\:{x}_{\mathrm{1}} \left({wbich}\:{is}\:<−\mathrm{1}\right): \\ $$ $$\:\mathrm{2}−{x}_{\mathrm{1}} \:=\:\mathrm{1}−\mathrm{2}\left({x}_{\mathrm{1}} +\mathrm{1}\right) \\ $$ $$\:{x}_{\mathrm{1}} =\:−\mathrm{3}\: \\ $$ $${to}\:{obtain}\:{x}_{\mathrm{2}} \:\:\left(−\mathrm{1}<{x}_{\mathrm{2}} \:<\mathrm{2}\right) \\ $$ $$\:\mathrm{2}−{x}_{\mathrm{2}} \:=\:\mathrm{1}+\mathrm{2}\left({x}_{\mathrm{2}} +\mathrm{1}\right) \\ $$ $$\:\mathrm{3}{x}_{\mathrm{2}} =\:−\mathrm{1}\:\:\Rightarrow\:\:\:{x}_{\mathrm{2}} \:=\:−\mathrm{1}/\mathrm{3}\:\: \\ $$ $${so},\:{for}\:{given}\:{condition}\:{to} \\ $$ $${remain}\:{true} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:{x}\:\notin\:\left[−\mathrm{3},\:\frac{−\mathrm{1}}{\mathrm{3}}\right] \\ $$ $$\:\:\:{that}\:{is}\:{x}\in\:\left(−\infty,\:−\mathrm{3}\right)\cup\left(\frac{−\mathrm{1}}{\mathrm{3}},\:\infty\right). \\ $$

Commented bytawa tawa last updated on 23/Jun/17

God bless you sir. i really appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Jun/17

1)x≥2⇒−(2−x)−2(x+1)<1⇒  −x−4<1⇒−x<−5⇒x>5⇒x∈[5,∞)  2) −1≤x<2⇒2−x−2(x+1)<1⇒  −3x<1⇒x>−(1/3)⇒x∈[−(1/3),2)  ⇒x∈[−(1/3),2)∪[5,∞).

$$\left.\mathrm{1}\right){x}\geqslant\mathrm{2}\Rightarrow−\left(\mathrm{2}−{x}\right)−\mathrm{2}\left({x}+\mathrm{1}\right)<\mathrm{1}\Rightarrow \\ $$ $$−{x}−\mathrm{4}<\mathrm{1}\Rightarrow−{x}<−\mathrm{5}\Rightarrow{x}>\mathrm{5}\Rightarrow{x}\in\left[\mathrm{5},\infty\right) \\ $$ $$\left.\mathrm{2}\right)\:−\mathrm{1}\leqslant{x}<\mathrm{2}\Rightarrow\mathrm{2}−{x}−\mathrm{2}\left({x}+\mathrm{1}\right)<\mathrm{1}\Rightarrow \\ $$ $$−\mathrm{3}{x}<\mathrm{1}\Rightarrow{x}>−\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{2}\right) \\ $$ $$\Rightarrow{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{2}\right)\cup\left[\mathrm{5},\infty\right). \\ $$

Commented bytawa tawa last updated on 24/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented byajfour last updated on 24/Jun/17

if x=3 >2   ∣2−x∣= 1    2∣x+1∣=8    ∣2−x∣−2∣x+1∣=−7 <1  if  x=−4    ∣2−x∣= 6   2∣x+1∣ = 6  ∣2−x∣−2∣x+1∣=0 < 1   if x→−∞   ∣2−x∣−2∣x+1∣= 2−x+2(x+1)                  = x+4  = −∞+4 <1

$${if}\:{x}=\mathrm{3}\:>\mathrm{2} \\ $$ $$\:\mid\mathrm{2}−{x}\mid=\:\mathrm{1} \\ $$ $$\:\:\mathrm{2}\mid{x}+\mathrm{1}\mid=\mathrm{8} \\ $$ $$\:\:\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=−\mathrm{7}\:<\mathrm{1} \\ $$ $${if}\:\:{x}=−\mathrm{4} \\ $$ $$\:\:\mid\mathrm{2}−{x}\mid=\:\mathrm{6} \\ $$ $$\:\mathrm{2}\mid{x}+\mathrm{1}\mid\:=\:\mathrm{6} \\ $$ $$\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=\mathrm{0}\:<\:\mathrm{1} \\ $$ $$\:{if}\:{x}\rightarrow−\infty \\ $$ $$\:\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=\:\mathrm{2}−{x}+\mathrm{2}\left({x}+\mathrm{1}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}+\mathrm{4}\:\:=\:−\infty+\mathrm{4}\:<\mathrm{1}\: \\ $$

Commented bytawa tawa last updated on 24/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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