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Question Number 170868 by Tawa11 last updated on 01/Jun/22

Solve:   ∣x   −   1∣    +   ∣x    −   2∣    ≥    4

$$\mathrm{Solve}:\:\:\:\mid\mathrm{x}\:\:\:−\:\:\:\mathrm{1}\mid\:\:\:\:+\:\:\:\mid\mathrm{x}\:\:\:\:−\:\:\:\mathrm{2}\mid\:\:\:\:\geqslant\:\:\:\:\mathrm{4} \\ $$

Answered by thfchristopher last updated on 02/Jun/22

⇒∣x−1∣≥4−∣x−2∣  ⇒x−1≥4−∣x−2∣  or  x−1≤∣x−2∣−4  Case 1:  x−1≥4−∣x−2∣  ⇒∣x−2∣≥5−x  ⇒x−2≥5−x  or x−2≤x−5 (rejected)  ⇒2x≥7  ⇒x≥(7/2)  Case 2:  x−1≤∣x−2∣−4  ⇒∣x−2∣≥x+3  ⇒x−2≥x+3 (rejected) or x−2≤−x−3  ⇒2x≤−1  ⇒x≤−(1/2)  Conclusion: x≤−(1/2)  or  x≥(7/2)

$$\Rightarrow\mid{x}−\mathrm{1}\mid\geqslant\mathrm{4}−\mid{x}−\mathrm{2}\mid \\ $$$$\Rightarrow{x}−\mathrm{1}\geqslant\mathrm{4}−\mid{x}−\mathrm{2}\mid\:\:\mathrm{or}\:\:{x}−\mathrm{1}\leqslant\mid{x}−\mathrm{2}\mid−\mathrm{4} \\ $$$$\mathrm{Case}\:\mathrm{1}: \\ $$$${x}−\mathrm{1}\geqslant\mathrm{4}−\mid{x}−\mathrm{2}\mid \\ $$$$\Rightarrow\mid{x}−\mathrm{2}\mid\geqslant\mathrm{5}−{x} \\ $$$$\Rightarrow{x}−\mathrm{2}\geqslant\mathrm{5}−{x}\:\:\mathrm{or}\:{x}−\mathrm{2}\leqslant{x}−\mathrm{5}\:\left(\mathrm{rejected}\right) \\ $$$$\Rightarrow\mathrm{2}{x}\geqslant\mathrm{7} \\ $$$$\Rightarrow{x}\geqslant\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{Case}\:\mathrm{2}: \\ $$$${x}−\mathrm{1}\leqslant\mid{x}−\mathrm{2}\mid−\mathrm{4} \\ $$$$\Rightarrow\mid{x}−\mathrm{2}\mid\geqslant{x}+\mathrm{3} \\ $$$$\Rightarrow{x}−\mathrm{2}\geqslant{x}+\mathrm{3}\:\left(\mathrm{rejected}\right)\:\mathrm{or}\:{x}−\mathrm{2}\leqslant−{x}−\mathrm{3} \\ $$$$\Rightarrow\mathrm{2}{x}\leqslant−\mathrm{1} \\ $$$$\Rightarrow{x}\leqslant−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Conclusion}:\:{x}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{or}\:\:{x}\geqslant\frac{\mathrm{7}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 02/Jun/22

God bless you sir.

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

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