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Question Number 850 by sagarwal last updated on 25/Mar/15

Solve ∣2x+4∣≥14

$$\mathrm{Solve}\:\mid\mathrm{2}{x}+\mathrm{4}\mid\geqslant\mathrm{14} \\ $$

Answered by 123456 last updated on 25/Mar/15

∣2x+4∣≥14  2x+4≤−14∨2x+4≥14  2x≤−18∨2x≥10  x≤−9∨x≥5

$$\mid\mathrm{2}{x}+\mathrm{4}\mid\geqslant\mathrm{14} \\ $$$$\mathrm{2}{x}+\mathrm{4}\leqslant−\mathrm{14}\vee\mathrm{2}{x}+\mathrm{4}\geqslant\mathrm{14} \\ $$$$\mathrm{2}{x}\leqslant−\mathrm{18}\vee\mathrm{2}{x}\geqslant\mathrm{10} \\ $$$${x}\leqslant−\mathrm{9}\vee{x}\geqslant\mathrm{5} \\ $$

Commented by sagarwal last updated on 26/Mar/15

Thanks

$$\mathrm{Thanks} \\ $$

Answered by malwaan last updated on 25/Mar/15

2x+4=0 ⇒x=−2  x≥−2⇒2x+4≥14⇒x≥5  x<−2⇒−2x−4≥14  ⇒2x+4≤−14⇒x≤−9  ∴solution set =R/]−9 . 5[   OR =]−∞ . −9]∪[5 . ∞[

$$\mathrm{2}{x}+\mathrm{4}=\mathrm{0}\:\Rightarrow{x}=−\mathrm{2} \\ $$$${x}\geqslant−\mathrm{2}\Rightarrow\mathrm{2}{x}+\mathrm{4}\geqslant\mathrm{14}\Rightarrow{x}\geqslant\mathrm{5} \\ $$$${x}<−\mathrm{2}\Rightarrow−\mathrm{2}{x}−\mathrm{4}\geqslant\mathrm{14} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{4}\leqslant−\mathrm{14}\Rightarrow{x}\leqslant−\mathrm{9} \\ $$$$\left.\therefore{solution}\:{set}\:={R}/\right]−\mathrm{9}\:.\:\mathrm{5}\left[\:\right. \\ $$$$\left.{O}\left.{R}\:=\right]−\infty\:.\:−\mathrm{9}\right]\cup\left[\mathrm{5}\:.\:\infty\left[\right.\right. \\ $$

Commented by sagarwal last updated on 26/Mar/15

Thanls

$$\mathrm{Thanls} \\ $$

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