Помогите решить {6x/5+y/15=2.3 {x/10-2y/3=1.2 это система способ подстановки

$\left \{ {{\frac{6x}{5} + \frac{y}{15}=2,3} \atop {\frac{x}{10} - \frac{2y}{3}=1,2}} \right. \\\\\frac{6x}{5} + \frac{y}{15} = 2,3\\\\\frac{6x}{5} + \frac{y}{15} = \frac{23}{10}\\\\\frac{180x}{5} + \frac{30y}{15} = \frac{690}{10}\\36x + 2y = 69\\36x = 69-2y\\x = \frac{69}{36} - \frac{2y}{36}\\\\x = \frac{23}{12} - \frac{y}{18}$

Выразили x, вернёмся к системе.

$\left \{ {{x = \frac{23}{12} - \frac{y}{18}} \atop {\frac{x}{10} - \frac{2y}{3} = 1,2}} \right.\\\\\frac{\frac{23}{12} - \frac{y}{18}}{10} - \frac{2y}{3} = 1,2\\\\\frac{\frac{69-2y}{36}}{10} - \frac{2y}{3} = \frac{6}{5}\\\\\frac{69-2y}{360} - \frac{2y}{3} = \frac{6}{5}\\69-2y-240y=432\\-242y=363\\y=-\frac{3}{2}\\y = -1,5$

Нашли y.

$\left \{ {{x=\frac{23}{12}-\frac{y}{18}} \atop {y=-1,5}} \right. \\\\\left \{ {{x =\frac{23}{12} + \frac{1,5}{18}\\}} \atop {y=-1,5}} \right.\\\\\left \{ {{x=\frac{345}{180} + \frac{15}{180}} \atop {y=-1,5}} \right.\\\\\left \{ {{x=\frac{360}{180}} \atop {y=-1,5}} \right.\\\\\left \{ {{x=2} \atop {y=-1,5}} \right.\\$

Ответ: (2; -1,5)