`ĐKXĐ` `:` `3x+2y` `ne 1` `;` `x+2y` `ne -3`

`@` đặt `1/(3x+2y-1)`thành `a`

`@` đặt `1/(x+2y+3)` thành `b`

`=>` ta có hệ phương trình mới

$\left \{ {{3a-2b=\frac{1}{2}} \atop {2a+b=\frac{3}{4}}} \right.$

`->` $\left \{ {{b=\frac{3a-0,5}{2}} \atop {2a+\frac{3a-0,5}{2}=0,75}} \right.$

`->`$\left \{ {{b=\frac{3a-0,5}{2}} \atop {\frac{4a+3a-0,5}{2}= 0,75}} \right.$

`->`$\left \{ {{b=\frac{3a-0,5}{2}} \atop {\frac{7a-0,5}{2}= 0,75}} \right.$

`->` $\left \{ {{b=\frac{3a-0,5}{2}} \atop {3,5a-0,25= 0,75}} \right.$

`->` $\left \{ {{b=\frac{3a-0,5}{2}} \atop {a= \frac{2}{7}}} \right.$

`->`$\left \{ {{b=\frac{5}{28}} \atop {a= \frac{2}{7}}} \right.$ 

`@` mà `a=1/(3x+2y-1)`

`**` `2/7` `=` `1/(3x+2y-1)`

`->` `7/2` `=` `3x+2y-1`

`@` `b=1/(x+2y+3)`

`**` `5/28` `=` `1/(x+2y+3)`

`->` `28/5` `=` `x+2y+3`

`->`$\left \{ {{3x+2y-1=\frac{7}{2}} \atop {x+2y+3=\frac{28}{5}}} \right.$

`->` $\left \{ {{x=\frac{19}{20}} \atop {y=\frac{7}{20}}} \right.$ 

`=>` vậy `(x;y)` `=` `(\frac{19}{20};\frac{33}{40})`