Помогите решить систему, пожалуйста

$\frac{1}{y} - \frac{1}{x} = 3 \\ 0.8 = \frac{1}{x + y} \\ \\ \frac{ x- y}{xy} = 3 \\ x + y = \frac{1}{0.8} \\ \\ x - y = 3xy \\ x + y = 1.25 \\ \\ x = 1.25 - y \\ 1.25 - y - y = 3y(1.25 - y) \\ \\ x = 1.25 - y \\ 1.25 - 2y = 3.75y - 3 {y}^{2} \\ \\ x = 1.25 - y \\ 3 {y}^{2} - 5.75y + 1.25 = 0 \\ \\ 12 {y}^{2} - 23y + 5 = 0 \\ d = {b}^{2} - 4ac = 529 - 4 \times 12 \times 5 = 289 \\ y1 = \frac{23 + 17}{2 \times 12} = \frac{40}{24} = \frac{5}{3} \\ y2 = \frac{23 - 17}{2 \times 12} = \frac{6}{24} = \frac{1}{4} \\ x1 = 1.25 - \frac{5}{3} = \frac{5}{4} - \frac{5}{3} = \frac{15}{12} - \frac{20}{12} = - \frac{5}{12} \\ x2 = 1.25 - \frac{1}{4} = 1$
Ответ: (-5/12; 5/3) и (1; 1/4)