Respuesta :

dalendrk
[tex](*)\ x^2+y^2=1\\(**)\ y=x+1\\\\subtitute\ (**)\ to\ (*):\\\\x^2+(x+1)^2=1\ \ |use\ (a+b)^2=a^2+2ab+b^2\\\\x^2+x^2+2(x)(1)+1^2=1\\\\2x^2+2x+1=1\ \ \ |subtract\ 1\ from\ both\ sides\\\\2x^2+2x=0\ \ \ |divide\ both\ sides\ by\ 2\\\\x^2+x=0\\\\x(x+1)=0\iff x=0\ or\ x+1=0\\\\\boxed{x_1=0\ or\ x_2=-1}\\\\subtitute\ value\ of\ x\ to\ (**):\\\\y_1=0+1=1;\ y_2=-1+1=0\\\\Your\ answer:\\\boxed{(0;\ 1)\ and\ (-1;\ 0)}\to\boxed{(2)}[/tex]

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