ПОМОГИТЕ решить систему!!!\ отдаю последние баллы!

Решите задачу:

$\left \{ {{4x-3y=1} \atop {8*2x=6*(9-5y)}} \right. \\ \\ \left \{ {{4x=1+3y} \atop {16x=54-30y}} \right. \\ \\ \left \{ {{x=\frac{1+3y}{4}} \atop {16x=54-30y}} \right. \\ \\ 16*(\frac{1+3y}{4})=54-30y \\ 4*(1+3y)=54-30y \\ 4+12y=54-30y \\ 12y+30y=54-4 \\ 42y=50 \\ y=\frac{50}{42}=\frac{25}{21} \\ \\ x=\frac{1+3*\frac{25}{21}}{4}=\frac{1+\frac{25}{7}}{4}=\frac{\frac{7}{7}+\frac{25}{7}}{4}=\frac{\frac{32}{7}}{4}= \frac{32}{7} * \frac{1}{4} = \frac{8}{7}$