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Question Number 50279 by Saorey last updated on 15/Dec/18

$$\{\begin{array}{l}\mathrm{x}+\mathrm{y}=6\\ \mathrm{y}+\mathrm{z}=10\end{array}(\mathrm{x},\mathrm{y},\mathrm{z}>0)$$

Answered by mr W last updated on 15/Dec/18

$$ifx,y,zareintegers,then$$ $$(x,y,z)=(1,5,5)/(2,4,6)/(3,3,7)/(4,2,8)/(5,1,9)$$ $$ifx,y,zarereal,then$$ $$x=6-a$$ $$y=a$$ $$z=10-a$$ $$with0<a<6$$

Answered by Rasheed.Sindhi last updated on 15/Dec/18

$$\mathrm{x}+\mathrm{y}=6\wedge \mathrm{x},\mathrm{y}>0\Rightarrow$$ $$0<\mathrm{x},\mathrm{y}<6\Rightarrow \mathrm{x},\mathrm{y}\in (0,6)$$ $$\mathrm{y}+\mathrm{z}=10\wedge \mathrm{y},\mathrm{z}>0\Rightarrow$$ $$0<\mathrm{y},\mathrm{z}<10\Rightarrow \mathrm{y},\mathrm{z}\in (0,10)$$ $$\mathrm{x}\in (0,6)$$ $$\mathrm{y}\in (0,6)\cap (0,10)=(0,6)$$ $$\mathrm{z}\in (4,10)$$ $$$$

Commented byRasheed.Sindhi last updated on 16/Dec/18

$$\mathrm{THANKS}\mathrm{for}\mathrm{your}\mathrm{warm}\mathrm{welcome}SIR!$$

Commented bymr W last updated on 15/Dec/18

$$niceto\mathrm{♮}see\epsilon youagainsir!$$ $$greatcomeback!$$

Commented bymr W last updated on 16/Dec/18

$$thanksforcomingbacksir!$$

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