\(2\log_5{x}+3\log_3{y}=8 \ . \ . \ . \ (1)\\6\log_5{x}+2\log_3{y}=2 \ . \ . \ . \ (2)\\ \\ (1)\times3:6\log_5{x}+9\log_3{y}=24 \ . \ . \ . \ (3) \\ \\ (3)-(2):7\log_3{y}=22\\ \log_3y=\frac{22}{7}\\3^{\log_3y}=3^{\frac{22}{7}}=31.59\\y=31.59\\ \\ From \ (1), 2\log_5{x}+3\log_3{31.59}=8\\2\log_5{x}=8-9.429=-1.429\\ \log_5{x}= -\frac{1.429}{2} =-0.7143\\5^{\log_5{x}}=5^{-0.7143}\\x=0.3168\)
Therefore, x = 0.3168 and y = 31.59
