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Question Number 116601 by ZiYangLee last updated on 05/Oct/20

If (x,y) satisfies the system of equations { ((∣x∣−x−y+2=0)),((∣y∣+y+5x=1)) :} find the value of x+y.

$$\mathrm{If}\:\left({x},{y}\right)\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$$\begin{cases}{\mid{x}\mid−{x}−{y}+\mathrm{2}=\mathrm{0}}\\{\mid{y}\mid+{y}+\mathrm{5}{x}=\mathrm{1}}\end{cases}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}+{y}.\: \\ $$

Answered by mr W last updated on 05/Oct/20

{ ((x−x−y+2=0)),((y+y+5x=1)) :} ⇒y=2, x=−(3/5) not ≥0 ⇒no solution { ((x−x−y+2=0)),((−y+y+5x=1)) :} ⇒y=2 not <0 ⇒ no solution { ((−x−x−y+2=0 ⇒2x+y=2)),((y+y+5x=1 ⇒5x+2y=1)) :} ⇒x=−3, y=8 ⇒solution { ((−x−x−y+2=0 ⇒2x+y=2)),((−y+y+5x=1)) :} ⇒x=(1/5) not <0 ⇒no solution solution: x=−3, y=8 ⇒x+y=5

$$\begin{cases}{{x}−{x}−{y}+\mathrm{2}=\mathrm{0}}\\{{y}+{y}+\mathrm{5}{x}=\mathrm{1}}\end{cases}\: \\ $$$$\Rightarrow{y}=\mathrm{2},\:{x}=−\frac{\mathrm{3}}{\mathrm{5}}\:{not}\:\geqslant\mathrm{0}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$$\begin{cases}{{x}−{x}−{y}+\mathrm{2}=\mathrm{0}}\\{−{y}+{y}+\mathrm{5}{x}=\mathrm{1}}\end{cases}\: \\ $$$$\Rightarrow{y}=\mathrm{2}\:{not}\:<\mathrm{0}\:\Rightarrow\:{no}\:{solution} \\ $$$$ \\ $$$$\begin{cases}{−{x}−{x}−{y}+\mathrm{2}=\mathrm{0}\:\Rightarrow\mathrm{2}{x}+{y}=\mathrm{2}}\\{{y}+{y}+\mathrm{5}{x}=\mathrm{1}\:\Rightarrow\mathrm{5}{x}+\mathrm{2}{y}=\mathrm{1}}\end{cases}\: \\ $$$$\Rightarrow{x}=−\mathrm{3},\:{y}=\mathrm{8}\:\Rightarrow{solution} \\ $$$$ \\ $$$$\begin{cases}{−{x}−{x}−{y}+\mathrm{2}=\mathrm{0}\:\Rightarrow\mathrm{2}{x}+{y}=\mathrm{2}}\\{−{y}+{y}+\mathrm{5}{x}=\mathrm{1}}\end{cases}\: \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{5}}\:{not}\:<\mathrm{0}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${solution}: \\ $$$${x}=−\mathrm{3},\:{y}=\mathrm{8} \\ $$$$\Rightarrow{x}+{y}=\mathrm{5} \\ $$

Commented by ZiYangLee last updated on 05/Oct/20

Excellent★★

$$\mathrm{Excellent}\bigstar\bigstar \\ $$

Answered by bemath last updated on 05/Oct/20

{ ((∣x∣=x+y−2)),((∣y∣=1−5x−y)) :} →∣x∣+∣y∣=−4x−1...(iii) case(1) → { ((x≥0)),((y≥0)) :} ⇒x+y=−4x−1 y=−5x−1 ∧ 2y=1−5x substitute ⇒2(−5x−1)=1−5x ⇒−10x−2=1−5x ⇒−3=5x ; x=−(3/5) ←rejected case(2)→ { ((x<0)),((y≥0)) :}⇒−x+y=−4x−1 y=−3x−1 ∧ 2y=1−5x ⇒−6x−2=1−5x ; −x=3 , x=−3←acceptable then y = 9−1=8 therefore x+y = 5

$$\begin{cases}{\mid\mathrm{x}\mid=\mathrm{x}+\mathrm{y}−\mathrm{2}}\\{\mid\mathrm{y}\mid=\mathrm{1}−\mathrm{5x}−\mathrm{y}}\end{cases} \\ $$$$\rightarrow\mid\mathrm{x}\mid+\mid\mathrm{y}\mid=−\mathrm{4x}−\mathrm{1}...\left(\mathrm{iii}\right) \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\rightarrow\begin{cases}{\mathrm{x}\geqslant\mathrm{0}}\\{\mathrm{y}\geqslant\mathrm{0}}\end{cases}\:\Rightarrow\mathrm{x}+\mathrm{y}=−\mathrm{4x}−\mathrm{1} \\ $$$$\mathrm{y}=−\mathrm{5x}−\mathrm{1}\:\wedge\:\mathrm{2y}=\mathrm{1}−\mathrm{5x} \\ $$$$\mathrm{substitute}\:\Rightarrow\mathrm{2}\left(−\mathrm{5x}−\mathrm{1}\right)=\mathrm{1}−\mathrm{5x} \\ $$$$\Rightarrow−\mathrm{10x}−\mathrm{2}=\mathrm{1}−\mathrm{5x} \\ $$$$\Rightarrow−\mathrm{3}=\mathrm{5x}\:;\:\mathrm{x}=−\frac{\mathrm{3}}{\mathrm{5}}\:\leftarrow\mathrm{rejected} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\rightarrow\begin{cases}{\mathrm{x}<\mathrm{0}}\\{\mathrm{y}\geqslant\mathrm{0}}\end{cases}\Rightarrow−\mathrm{x}+\mathrm{y}=−\mathrm{4x}−\mathrm{1} \\ $$$$\mathrm{y}=−\mathrm{3x}−\mathrm{1}\:\wedge\:\mathrm{2y}=\mathrm{1}−\mathrm{5x} \\ $$$$\Rightarrow−\mathrm{6x}−\mathrm{2}=\mathrm{1}−\mathrm{5x}\:;\:−\mathrm{x}=\mathrm{3}\:,\:\mathrm{x}=−\mathrm{3}\leftarrow\mathrm{acceptable} \\ $$$$\mathrm{then}\:\mathrm{y}\:=\:\mathrm{9}−\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{therefore}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{5} \\ $$

Commented by bobhans last updated on 05/Oct/20

waw.....funtastic..

$$\mathrm{waw}.....\mathrm{funtastic}.. \\ $$

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