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

$$\mathrm{If}(x,y)\mathrm{satisfies}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}$$$$\{\begin{array}{l}\mid x\mid -x-y+2=0\\ \mid y\mid +y+5x=1\end{array}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}x+y.$$

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

$$\{\begin{array}{l}x-x-y+2=0\\ y+y+5x=1\end{array}$$$$\Rightarrow y=2,x=-\frac{3}{5}not⩾0\Rightarrow nosolution$$$$$$$$\{\begin{array}{l}x-x-y+2=0\\ -y+y+5x=1\end{array}$$$$\Rightarrow y=2not<0\Rightarrow nosolution$$$$$$$$\{\begin{array}{l}-x-x-y+2=0\Rightarrow 2x+y=2\\ y+y+5x=1\Rightarrow 5x+2y=1\end{array}$$$$\Rightarrow x=-3,y=8\Rightarrow solution$$$$$$$$\{\begin{array}{l}-x-x-y+2=0\Rightarrow 2x+y=2\\ -y+y+5x=1\end{array}$$$$\Rightarrow x=\frac{1}{5}not<0\Rightarrow nosolution$$$$$$$$solution:$$$$x=-3,y=8$$$$\Rightarrow x+y=5$$

Commented by ZiYangLee last updated on 05/Oct/20

$$\mathrm{Excellent}★★$$

Answered by bemath last updated on 05/Oct/20

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

Commented by bobhans last updated on 05/Oct/20

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

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