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Question Number 199608 by mathlove last updated on 06/Nov/23

1) 3<∣2x−1∣<7 find Σx ;x∈Z 2) 4≤∣x−2∣<5 find Σx ;x∈Z

$$\left.\mathrm{1}\right)\:\:\:\mathrm{3}<\mid\mathrm{2}{x}−\mathrm{1}\mid<\mathrm{7}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$\left.\mathrm{2}\right)\:\:\:\mathrm{4}\leqslant\mid{x}−\mathrm{2}\mid<\mathrm{5}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$ \\ $$

Answered by mr W last updated on 06/Nov/23

1) 2x−1= −5, 5 ⇒x=−2, 3 ⇒Σx=1 2) x−2=−4, 4 ⇒x=−2, 6 ⇒Σx=4

$$\left.\mathrm{1}\right) \\ $$$$\mathrm{2}{x}−\mathrm{1}=\:−\mathrm{5},\:\mathrm{5} \\ $$$$\Rightarrow{x}=−\mathrm{2},\:\mathrm{3}\:\Rightarrow\Sigma{x}=\mathrm{1} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$${x}−\mathrm{2}=−\mathrm{4},\:\mathrm{4} \\ $$$$\Rightarrow{x}=−\mathrm{2},\:\mathrm{6}\:\Rightarrow\Sigma{x}=\mathrm{4} \\ $$

Commented by mathlove last updated on 06/Nov/23

thanks

$${thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 06/Nov/23

1) 3<∣2x−1∣<7 find Σx ;x∈Z 3<2x−1<7 ∣ 3<−(2x−1)<7 4<2x<8 ∣ 3<−2x+1<7 2<x<4 ∣ 2<−2x<6 x=3 ∣ −3<x<−1 ∣ x=−2 Σx=3−2=1 2) 4≤∣x−2∣<5 find Σx ;x∈Z 4≤x−2<5 ∣ 4≤−(x−2)<5 6≤x<7 ∣ 4≤−x+2<5 x=6 ∣ 2≤−x<3 ∣ −3<x≤−2 x=−2 Σx=6−2=4

$$\left.\mathrm{1}\right)\:\:\:\mathrm{3}<\mid\mathrm{2}{x}−\mathrm{1}\mid<\mathrm{7}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$\:\:\:\:\mathrm{3}<\mathrm{2}{x}−\mathrm{1}<\mathrm{7}\:\mid\:\mathrm{3}<−\left(\mathrm{2}{x}−\mathrm{1}\right)<\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{4}<\mathrm{2}{x}<\mathrm{8}\:\:\:\:\:\:\:\mid\:\:\:\mathrm{3}<−\mathrm{2}{x}+\mathrm{1}<\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}<{x}<\mathrm{4}\:\:\:\:\:\:\:\mid\:\:\:\mathrm{2}<−\mathrm{2}{x}<\mathrm{6} \\ $$$$\:\:\:\:\:{x}=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:−\mathrm{3}<{x}<−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:{x}=−\mathrm{2} \\ $$$$\Sigma{x}=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:\:\:\mathrm{4}\leqslant\mid{x}−\mathrm{2}\mid<\mathrm{5}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$\:\:\:\:\:\:\mathrm{4}\leqslant{x}−\mathrm{2}<\mathrm{5}\:\mid\:\mathrm{4}\leqslant−\left({x}−\mathrm{2}\right)<\mathrm{5} \\ $$$$\:\:\:\:\:\:\mathrm{6}\leqslant{x}<\mathrm{7}\:\mid\:\mathrm{4}\leqslant−{x}+\mathrm{2}<\mathrm{5} \\ $$$$\:\:\:\:\:\:{x}=\mathrm{6}\:\mid\:\mathrm{2}\leqslant−{x}<\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:−\mathrm{3}<{x}\leqslant−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=−\mathrm{2} \\ $$$$\Sigma{x}=\mathrm{6}−\mathrm{2}=\mathrm{4} \\ $$

Commented by mathlove last updated on 06/Nov/23

thanks

$${thanks} \\ $$

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