[tex]\displaystyle\bf\\\left \{ {{ {x}^{2} + 2y = 5 } \atop {y - x = 1 }} \right. \\ \displaystyle\bf\\\left \{ {{x {}^{2} + 2(x + 1) = 5 } \atop {y = x + 1 }} \right. \\ \\ {x}^{2} + 2x + 2 - 5 = 0 \\ {x}^{2} + 2x - 3 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1} + x_{2} = - 2 \\ x_{1} x_{2} = - 3\\ x_{1} = - 3 \\ x_{2} = 1 \\ \\ y_{1} = - 3 + 1 = - 2 \\y _{2} = 1 + 1 = 2[/tex]

Ответ: ( - 3 ; - 2 ) и ( 1 ; 2 )