Giải bằng cách đặt ẩn phụ

$\left \{ {{\frac{1}{x}+\frac{1}{y}=\frac{3}{4}} \atop {\frac{1}{6x}+\frac{1}{5y}=\frac{2}{15}}} \right.$   
$\left \{ {{\frac{1}{x}+\frac{1}{y}=\frac{3}{4}} \atop {\frac{1}{6}\frac{1}{x}+\frac{1}{5}\frac{1}{y}=\frac{2}{15}}} \right.$    
đặt $\frac{1}{x}=a; \frac{1}{y}=b$  
ta có:$\left \{ {{a+b=\frac{3}{4}} \atop {\frac{1}{6}a+\frac{1}{5}b=\frac{2}{15}}} \right.$  
$\left \{ {{a+b=\frac{3}{4}} \atop {\frac{5}{6}a+b=\frac{2}{3}}} \right.$   
$\left \{ {{\frac{1}{6}a=\frac{1}{12}} \atop {a+b=\frac{3}{4}}} \right.$   
$\left \{ {{a=\frac{1}{2}} \atop {\frac{1}2+b=\frac{3}4}} \right.$   
$\left \{ {{a=\frac{1}2} \atop {b=\frac{1}4}} \right.$   
thay `1/x`=`a`; `1/y`=`b` ta có:  
$\left \{ {{\frac{1}x=\frac{1}2} \atop {\frac{1}y=\frac{1}4}} \right.$   
$\left \{ {{y=4} \atop {x=2}} \right.$