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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{∣x−1∣+∣y−1∣=1∣x−1∣−y=−5
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
⇒{∣x−1∣=y−5∣y−1∣+y=6letsolve∣y−1∣+y=6withy⩾5⇒∣y−1∣+y−6=0ify⩾1e⇒y−1+y−6=0⇒2y−7=0⇒y=72(notsolution)ify⩽1e⇒1−y+y−6=0⇒−5=0impossible⇒nosolutionforthissystem
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
∣x−1∣−y=−5⇒y=∣x−1∣+5insertin(1)∣x−1∣+∣∣x−1∣+5−1∣=1∣x−1∣+∣∣x−1∣+4∣=1∣x−1∣+∣x−1∣+4=12∣x−1∣=−3nosolution

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