Розв'язати систему рiвнянь {3y²-xy = -1, {x+3y= 5

Решите задачу:

$3 {y}^{2} - xy = - 1 \\ x + 3y = 5 \\ \\3y {}^{2} - y(5 - 3y) = - 1 \\ x = 5 - 3y \\ \\ 3y {}^{2} - y(5 - 3y) = - 1 \\ 3 {y}^{2} - 5y + 3 {y}^{2} = - 1 \\ 6 {y}^{2} - 5y + 1 = 0 \\ D = ( - 5) {}^{2} - 4 \times 6 \times 1 = 25 - 24 = 1 \: \: \: (1) \\ y {1} = \frac{5 + 1}{12} = \frac{6}{12} = \frac{1}{2} = 0.5 \\ y{2} = \frac{5 - 1}{12} = \frac{4}{12} = \frac{1}{3} \\ \\ x{1} = 5 - 3 \times 0.5 = 5 - 1.5 = 3.5 \\ x{2} = 5 - 3 \times \frac{1}{3} = 5 - 1 = 4 \\ \\ Ответ: \: (3.5 \: ; \: 0.5) \: ; \: (4 \: ; \: \frac{1}{3} )$