Решить систему
lg(x²+y²)=1+lg13
lg(x+y)=lg(x-y)+lg8
Помогите решить, пожалуйста

$left { {{lg( x^{2} + y^{2})= lg 10 + lg 13 = textgreater x^{2} + y^{2} =130 } atop {x+y=8x-8y}} right.$
$left { {x^{2} + (frac{7}{9}x)^{2} =130 = textgreater frac{130}{81} x^{2}=130 = textgreater x=9 } atop {y= frac{7}{9}x }} right.$
$left { {x^{2} + y^{2} =130 } atop {9y-7x=0}} right.$
$left { {x^{2} + y^{2} =130 } atop {9y=7x}= textgreater y= frac{7}{9}x } right.$
$left { {x^{2} + ( frac{7}{9} x)^{2} =130= textgreater frac{130}{81} x^{2} =130 = textgreater x = 9 } atop {9y=7x}= textgreater y= frac{7}{9}x } right.$
$left { {x^{2} + ( frac{7}{9} x)^{2} =130= textgreater frac{130}{81} x^{2} =130 = textgreater x = 9 } atop { y= frac{7}{9}x } right$
$left { {x^{2} + ( frac{7}{9} x)^{2} =130= textgreater frac{130}{81} x^{2} =130 = textgreater x = 9 } atop { y= 7 } right$