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Ответ:

x=1, \ y=0, \ a=1;

x=1, \ y=0, \ a=-1;

Объяснение:

\displaystyle \left \{ {{ax+y=a} \atop {x+ay=1}} \right. \Leftrightarrow \left \{ {{y=a-ax} \atop {x+a(a-ax)=1}} \right. \Leftrightarrow \left \{ {{y=a-ax} \atop {x+a(a(1-x))=1}} \right. \Leftrightarrow

\displaystyle \Leftrightarrow \left \{ {{y=a-ax} \atop {x+a^{2}(1-x)=1}} \right. ;

x+a^{2}(1-x)=1;

a^{2}(1-x)-1+x=0;

a^{2}(1-x)-(1-x)=0;

(a^{2}-1)(1-x)=0;

a^{2}-1=0 \quad \vee \quad 1-x=0;

a^{2}=1 \quad \vee \quad x=1-0;

a=\pm 1 \quad \vee \quad x=1;

a=1, \ x=1 \Rightarrow y=1-1 \cdot 1=1-1=0;

a=-1, \ x=1 \Rightarrow y=-1-(-1) \cdot 1=-1-(-1)=0;

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