{2x -5y =31 ; x² +xy =y² -31.
решаем методом постановки
{ y =(2x -31)/5 ; x² +x*(2x -31)/5 =((2x -31)/5)² - 31 .
x² +x*(2x -31)/5 =((2x -31)/5)² - 31 ;
x² +x*(2x -31)/5 =(4x² -124x +31²)/25 - 31 ;
25x² +10x² -155x =4x² -124x +31² -31*25;
31x² -31x =31(31 -25) ;
x² -x -6 =0;'
x₁ = -2 ⇒y₁ =(2(-2) -31)/5 = -7 ;
x₂ = 3 ⇒y₂ =(2*3 -31)/5 = -5.
ответ : x₁ = -2 ; .y₁ = -7 ;
x₂ = 3 ; y₂ = - 5 .
или ( -2 ; -7) б ( 3 ; -5) .
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{ y =(2x -31)/5 ; [ x = -2 ;x =3.