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Question Number 162715 by tounghoungko last updated on 31/Dec/21

∣∣x−1∣−5∣ ≥ 2 has solution set is a ≤x≤b or x≤ c ∪ x≥d . Find ((a+d)/(b+c)) .

$$\:\:\:\mid\mid{x}−\mathrm{1}\mid−\mathrm{5}\mid\:\geqslant\:\mathrm{2}\:\:{has}\:{solution}\:{set} \\ $$$$\:{is}\:{a}\:\leqslant{x}\leqslant{b}\:{or}\:{x}\leqslant\:{c}\:\cup\:{x}\geqslant{d}\:. \\ $$$$\:{Find}\:\frac{{a}+{d}}{{b}+{c}}\:. \\ $$

Answered by bobhans last updated on 01/Jan/22

∣∣x−1∣−5∣ ≥2 ⇒ (∣x−1∣−7)(∣x−1∣−3)≥0 ⇒ ∣x−1∣ ≤3 ∪ ∣x−1∣ ≥7 ⇒ −2≤x≤4 ∪ x≤−6 ∪ x≥ 8 { ((a=−2 ∧ b=4 )),((c=−6 ∧ d=8)) :} ⇒((a+d)/(b+c)) = (6/(−2))= −3

$$\:\mid\mid\mathrm{x}−\mathrm{1}\mid−\mathrm{5}\mid\:\geqslant\mathrm{2}\:\Rightarrow\:\left(\mid\mathrm{x}−\mathrm{1}\mid−\mathrm{7}\right)\left(\mid\mathrm{x}−\mathrm{1}\mid−\mathrm{3}\right)\geqslant\mathrm{0} \\ $$$$\:\Rightarrow\:\mid\mathrm{x}−\mathrm{1}\mid\:\leqslant\mathrm{3}\:\cup\:\mid\mathrm{x}−\mathrm{1}\mid\:\geqslant\mathrm{7} \\ $$$$\:\Rightarrow\:−\mathrm{2}\leqslant\mathrm{x}\leqslant\mathrm{4}\:\cup\:\mathrm{x}\leqslant−\mathrm{6}\:\cup\:\mathrm{x}\geqslant\:\mathrm{8} \\ $$$$\:\begin{cases}{\mathrm{a}=−\mathrm{2}\:\wedge\:\mathrm{b}=\mathrm{4}\:}\\{\mathrm{c}=−\mathrm{6}\:\wedge\:\mathrm{d}=\mathrm{8}}\end{cases}\:\Rightarrow\frac{\mathrm{a}+\mathrm{d}}{\mathrm{b}+\mathrm{c}}\:=\:\frac{\mathrm{6}}{−\mathrm{2}}=\:−\mathrm{3} \\ $$

Answered by MJS_new last updated on 31/Dec/21

∣∣x−1∣−5∣≥2 (1) x≥1 ⇒ ∣x−6∣≥2 (1.1) x≥6 ⇒ x−6≥2 ⇒ x≥8 (1.2) 1≤x<6 ⇒ 6−x≥2 ⇒ 1≤x≤4 (2) x<1 ⇒ ∣x+4∣≥2 (2.1) −4<x<1 ⇒ x+4≥2 ⇒ −2≤x<1 (2.2) x≤−4 ⇒ −x−4≥2 ⇒ x≤−6 ===== x≤−6∨−2≤x≤4∨8≤x a=−2, b=4, c=−6, d=8 ⇒ ((a+d)/(b+c))=−3

$$\mid\mid{x}−\mathrm{1}\mid−\mathrm{5}\mid\geqslant\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\:{x}\geqslant\mathrm{1}\:\Rightarrow\:\mid{x}−\mathrm{6}\mid\geqslant\mathrm{2} \\ $$$$\:\:\:\:\:\left(\mathrm{1}.\mathrm{1}\right)\:{x}\geqslant\mathrm{6}\:\Rightarrow\:{x}−\mathrm{6}\geqslant\mathrm{2}\:\Rightarrow\:{x}\geqslant\mathrm{8} \\ $$$$\:\:\:\:\:\left(\mathrm{1}.\mathrm{2}\right)\:\mathrm{1}\leqslant{x}<\mathrm{6}\:\Rightarrow\:\mathrm{6}−{x}\geqslant\mathrm{2}\:\Rightarrow\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:{x}<\mathrm{1}\:\Rightarrow\:\mid{x}+\mathrm{4}\mid\geqslant\mathrm{2} \\ $$$$\:\:\:\:\:\left(\mathrm{2}.\mathrm{1}\right)\:−\mathrm{4}<{x}<\mathrm{1}\:\Rightarrow\:{x}+\mathrm{4}\geqslant\mathrm{2}\:\Rightarrow\:−\mathrm{2}\leqslant{x}<\mathrm{1} \\ $$$$\:\:\:\:\:\left(\mathrm{2}.\mathrm{2}\right)\:{x}\leqslant−\mathrm{4}\:\Rightarrow\:−{x}−\mathrm{4}\geqslant\mathrm{2}\:\Rightarrow\:{x}\leqslant−\mathrm{6} \\ $$$$===== \\ $$$${x}\leqslant−\mathrm{6}\vee−\mathrm{2}\leqslant{x}\leqslant\mathrm{4}\vee\mathrm{8}\leqslant{x} \\ $$$${a}=−\mathrm{2},\:{b}=\mathrm{4},\:{c}=−\mathrm{6},\:{d}=\mathrm{8}\:\Rightarrow\:\frac{{a}+{d}}{{b}+{c}}=−\mathrm{3} \\ $$

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