Đặt $\frac{1}{x}$= a; $\frac{1}{y}$= b
Ta có hệ:
(1) a+b= $\frac{5}{24}$
(2) 9a+ $\frac{5}{24}$ : $\frac{6}{5}$= 1
Từ (2) <=> 9a+ $\frac{25}{144}$= 1
<=> a=$\frac{119}{1296}$
Thay vào (1): $\frac{119}{1196}$+ b= $\frac{5}{24}$ <=> b= $\frac{151}{1296}$
a= $\frac{119}{1296}$ => x= $\frac{1296}{119}$
b= $\frac{151}{1296}$ => y= $\frac{1296}{151}$
Vậy hệ có nghiệm (x;y)= ( $\frac{1296}{119} ; \frac{1296}{151}$)