Question

Х во 2 + у во 2= 17. х-у=3 это система ​

Answer 1

$(-1; -4) \quad ; \quad (4; 1) \quad ;$

Объяснение:

$$ \displaystyle \left \{ {{x^{2}+y^{2}=17} \atop {x-y=3}} \right. \Leftrightarrow \left \{ {{x^{2}+(x-3)^{2}=17} \atop {y=x-3}} \right. \Leftrightarrow \left \{ {{x^{2}+x^{2}-6x+9-17=0} \atop {y=x-3}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{2x^{2}-6x-8=0} \atop {y=x-3}} \right. \Leftrightarrow \left \{ {{x^{2}-3x-4=0} \atop {y=x-3}} \right. \Leftrightarrow \left \{ {{x^{2}-4x+x-4=0} \atop {y=x-3}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{x(x-4)+1(x-4)=0} \atop {y=x-3}} \right. \Leftrightarrow \left \{ {{(x+1)(x-4)=0} \atop {y=x-3}} \right. \Leftrightarrow \left \{ {{x+1=0} \atop {y=x-3}} \right. \vee \left \{ {{x-4=0} \atop {y=x-3}} \right. \Leftrightarrow $$$$ \displaystyle \Leftrightarrow \left \{ {{x=-1} \atop {y=-4}} \right. \vee \left \{ {{x=4} \atop {y=1}} \right. ; $$

$(-1; -4) \quad ; \quad (4; 1) \quad ;$