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

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

lg(x²+y²)=lg10+lg13 ⇒ x²+y²=130 lg(x+y)=lg(x-y)·8     ⇒ x+y=8x-8y  ⇒9y=7x ⇒x=9y/7 (9y/7)² +y² =130 81/49 y² +y² =130 y²(81/49+1)=130 y²=130/130/49=49 y=7 (-7 -посторонний корень) х=9у/7=9 Ответ (9;7)