Решите задание по алгебре 10-11 класс)

$\left \{ {{\left(\frac{1}{3}\right)^{2x-y}=27} \atop {5^{3x-y}=\frac{1}{25}}} \right.$
$5^{3x-y}=\frac{1}{25} \\ 5^{3x-y}=25^{-1} \\5^{3x-y}=\left(5^2\right)^{-1} \\ 5^{3x-y}=5^{2\left(-1\right)} \\ 3x-y=2\left(-1\right) \\ 3x-y=-2 \\ x=\frac{-2+y}{3}$

$\left(\frac{1}{3}\right)^{2\cdot \frac{-2+y}{3}-y}=27 \\ \\ \left(\frac{1}{3}\right)^{2\cdot \frac{-2+y}{3}-y}=\left(\frac{1}{3}\right)^{-3} \\ \\ 2\cdot \frac{-2+y}{3}-y=-3 \\ \\ \frac{2\left(-2+y\right)}{3}-y=-3 \\ \\ 2\left(-2+y\right)-3y=-9 \\ \\ -4+2y-3y=-9 \\ \\ -4-y+4=-9+4 \\ \\ -y=-5 \\ y=5 \\ \\ \\ x=\frac{-2+5}{3} \\ x=1 \\ \\ OTB:y=5,\:x=1$

$b) \\ 12x+y=2 \\ 64^{x-3y}=8 \\ \\ 12x+y=2 \\ 12x=2-y \\ \frac{12x}{12}=\frac{2}{12}-\frac{y}{12} \\ x=\frac{2-y}{12} \\ \\ \\ 64^{\frac{2-y}{12}-3y}=8 \\ \left(8^2\right)^{\frac{2-y}{12}-3y}=8 \\ 8^{2\left(\frac{2-y}{12}-3y\right)}=8 \\ 2\left(\frac{2-y}{12}-3y\right)=1 \\ \frac{2\left(\frac{2-y}{12}-3y\right)}{2}=\frac{1}{2} \\ \frac{2-37y}{12}=\frac{1}{2} \\ \frac{12\left(2-37y\right)}{12}=\frac{1\cdot \:12}{2} \\ 2-37y=6 \\-37y=4 \\ y=-\frac{4}{37} \\ \\$

$x=\frac{2-\left(-\frac{4}{37}\right)}{12}=\frac{13}{74} \\ \\ y=-\frac{4}{37},\:x=\frac{13}{74}$