Решение систем уравнений второй степени

$\left \{ {{m_{1} + m_{2} = m_{3}} \atop {m_{1}*w_{1} + m_{2}*w_{2} = (m_{1} + m_{2})*w_{3}}} \right. \\ m_{3} = 50\\ w_{1} = 20\\ w_{2} = 40\\ w_{3} = 28\\ \left \{ {{m_{1} + m_{2} = 50} \atop {20*m_{1} + 40 * m_{2} = 1400}} \right.\\ m_{1} = 50 - m_{2}\\ 20*(50 - m_{2}) + 40 * m_{2} = 1400\\ 1000 - 20*m_{2} + 40*m_{2} = 1400\\ 20*m_{2} = 400\\ m_{2} = 20\\ m_{1} = 50 - m_{2} = 30\\$
Ответ:30г