$#gialong209$

`{(1/x - 1/y = 1),(3/x + 4/y = 5):} (x ; y \ne 0) <=> {(4/x - 4/y = 4),(3/x + 4/y = 5):}`

`<=> {(7/x = 9),(1/x - 1/y = 1):} <=> {(x = 7/9),(y = 7/2):} (tm)`

Vậy : `(x ; y) = (7/9 ; 7/2)`