Найдите решение систем уравнений способом подстановки : {X+y=4\3 {3x-2y=-1

Решите задачу:

$\left \{ {{x+y= \frac{4}{3} } \atop {3x-2y=-1}} \right. \\ \\ \left \{ {{x= \frac{4}{3}-y} \atop {3x-2y=-1}} \right. \\ \\ \left \{ {{x= \frac{4}{3}-y} \atop {3*(\frac{4}{3}-y)-2y=-1} \right. \\ \\ \left \{ {{x= \frac{4}{3}-y} \atop {4-3y-2y=-1}} \right. \\ \\ \left \{ {{x= \frac{4}{3}-y} \atop {-5y=-5}} \right. \\ \\ \left \{ {{x= \frac{4}{3}-y} \atop {y=1}} \right. \\ \\ \left \{ {{x= \frac{4}{3}-1} \atop {y=1}} \right. \\ \\ \left \{ {{x= \frac{1}{3} } \atop {y=1}} \right.$