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Решите задачу:

$\left \{ {{ \frac{x+3}{4} - \frac{y-5}{3} = 4, } \atop { y^2 + x = 13; }} \right. \left \{ {{ 3x+9 - (4y-20) = 48, } \atop { y^2 + x = 13; }} \right. \left \{ {{ 3x - 4y = 19, } \atop { x = 13-y^2; }} \right. \left \{ {{ 3y^2 + 4y - 20 = 0, } \atop { x = 13 - y^2; }} \right. \\ D_1 = 2^2 - 3\cdot(-20) = 64 = 8^2, \\ y_1 = \frac{-2-8}{3} = - \frac{10}{3} = -3 \frac{1}{3} , \\ y_2 = \frac{-2+8}{3} = \frac{6}{3} = 2, \\ x_1 = 13 - \frac{100}{9} = \frac{17}{9} = 1 \frac{8}{9} , \\ x_2 = 13-4=9$