Решить систему lg(x²+y²)=1+lg13lg(x+y)=lg(x-y)+lg8 Помогите решить, пожалуйста

Решите задачу:

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