Решите методом подстановки систему 2x^2 - y^2 = 2 3x + 2y = 1

Решите задачу:

$\left \{ {{2x^2 - y^2 = 2,} \atop {3x + 2y = 1;}} \right. \left \{ {{2x^2 - \frac{1}{4}(1-3x)^2 = 2,} \atop {y = \frac{1}{2}(1-3x);}} \right. \left \{ {{8x^2 - (1-6x+9x^2) = 8,} \atop {y = \frac{1}{2}(1-3x);}} \right. \left \{ {{-x^2+6x-9 = 0,} \atop {y = \frac{1}{2}(1-3x);}} \right. \\ \left \{ {{-(x-3)^2 = 0,} \atop {y = \frac{1}{2}(1-3x);}} \right. \left \{ {{x-3 = 0,} \atop {y = \frac{1}{2}(1-3x);}} \right. \left \{ {{x = 3,} \atop {y = -4.}} \right.$