解下列方程組

lg (x^2 + y^2) – 1 = lg 13 ①
lg (x+y) – lg(x-y) = 3lg2 ②

由①得lg(x^2+ y^2) - lg10=lg13
lg(x^2+ y^2)/10=lg13
(x^2+ y^2)/10=13
即x^2+ y^2=130 (3)

由②得 lg (x+y)/(x-y) = lg2^3
lg (x+y)/(x-y) = lg8
(x+y)/(x-y)=8
(x+y)=8*(x-y)
7x=9y

即x=9y/7 代入(3) (9y/7)^2 + y^2=130
(81 y^2)/49 + y^2=130
(81 y^2 + 49 y^2)/49=130
130 y^2 =130*49
y^2=49
即 y=7 or y=-7 (舍)

x=9*7/7 =9

so x=9, y=7