Ответ: (-3;-2),    (1;2).

Объяснение:

[tex]\displaystyle\\\left \{ {{x^2+2y=5} \atop {y-x=1}} \right. \ \ \ \ \left \{ {{x^2+2(x+1)=5} \atop {y=x+1}} \right. \ \ \ \ \left \{ {{x^2+2x+2=5} \atop {y=x+1}} \right. \\\\\\\\\left \{ {{x^2+2x-3=0} \atop {y=x+1}} \right. \ \ \ \ \left \{ {{x^2+3x-x-3=0} \atop {y=x+1}} \right.\ \ \ \ \left \{ {{x(x+3)-(x+3)=0} \atop {y=x+1}} \right.[/tex]

[tex]\displaystyle\\\left \{ {{(x+3)(x-1)=0} \atop {y=x+1}} \right. \ \ \ \ \left \{ {{x_1=-3\ \ \ \ x_2=1} \atop {y_1=-2\ \ \ \ y_2=2}} \right..[/tex]