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Question Number 122081 by bobhans last updated on 14/Nov/20

solve { ((∣x−1∣+∣y−1∣=1)),((∣x−1∣−y=−5)) :}

$$\:{solve}\:\begin{cases}{\mid{x}−\mathrm{1}\mid+\mid{y}−\mathrm{1}\mid=\mathrm{1}}\\{\mid{x}−\mathrm{1}\mid−{y}=−\mathrm{5}}\end{cases} \\ $$

Answered by mathmax by abdo last updated on 14/Nov/20

⇒ { ((∣x−1∣=y−5)),((∣y−1∣+y =6 let solve ∣y−1∣+y =6 withy≥5)) :} ⇒∣y−1∣+y−6=0 if y≥1 e⇒y−1+y−6=0 ⇒2y−7=0 ⇒y=(7/2) (not solution) if y≤1 e⇒1−y+y−6 =0 ⇒−5=0 impossible ⇒no solution for this system

$$\Rightarrow\begin{cases}{\mid\mathrm{x}−\mathrm{1}\mid=\mathrm{y}−\mathrm{5}}\\{\mid\mathrm{y}−\mathrm{1}\mid+\mathrm{y}\:=\mathrm{6}\:\:\:\:\:\mathrm{let}\:\mathrm{solve}\:\mid\mathrm{y}−\mathrm{1}\mid+\mathrm{y}\:=\mathrm{6}\:\:\mathrm{withy}\geqslant\mathrm{5}}\end{cases} \\ $$$$\Rightarrow\mid\mathrm{y}−\mathrm{1}\mid+\mathrm{y}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{y}\geqslant\mathrm{1}\:\:\mathrm{e}\Rightarrow\mathrm{y}−\mathrm{1}+\mathrm{y}−\mathrm{6}=\mathrm{0}\:\Rightarrow\mathrm{2y}−\mathrm{7}=\mathrm{0}\:\Rightarrow\mathrm{y}=\frac{\mathrm{7}}{\mathrm{2}}\:\left(\mathrm{not}\:\mathrm{solution}\right) \\ $$$$\mathrm{if}\:\mathrm{y}\leqslant\mathrm{1}\:\:\:\mathrm{e}\Rightarrow\mathrm{1}−\mathrm{y}+\mathrm{y}−\mathrm{6}\:=\mathrm{0}\:\Rightarrow−\mathrm{5}=\mathrm{0}\:\mathrm{impossible}\:\Rightarrow\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{this} \\ $$$$\mathrm{system} \\ $$

Answered by MJS_new last updated on 14/Nov/20

∣x−1∣−y=−5 ⇒ y=∣x−1∣+5 insert in (1) ∣x−1∣+∣∣x−1∣+5−1∣=1 ∣x−1∣+∣∣x−1∣+4∣=1 ∣x−1∣+∣x−1∣+4=1 2∣x−1∣=−3 no solution

$$\mid{x}−\mathrm{1}\mid−{y}=−\mathrm{5}\:\Rightarrow\:{y}=\mid{x}−\mathrm{1}\mid+\mathrm{5} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right) \\ $$$$\mid{x}−\mathrm{1}\mid+\mid\mid{x}−\mathrm{1}\mid+\mathrm{5}−\mathrm{1}\mid=\mathrm{1} \\ $$$$\mid{x}−\mathrm{1}\mid+\mid\mid{x}−\mathrm{1}\mid+\mathrm{4}\mid=\mathrm{1} \\ $$$$\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{1}\mid+\mathrm{4}=\mathrm{1} \\ $$$$\mathrm{2}\mid{x}−\mathrm{1}\mid=−\mathrm{3} \\ $$$$\mathrm{no}\:\mathrm{solution} \\ $$

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