TH1: y≥0
$\left \{ {{2x+y=4} \atop {4x-3y=1}} \right.$
⇔$\left \{ {{4x+2y=8} \atop {4x-3y=1}} \right.$
⇔$\left \{ {{5y=7} \atop {2x+y=4}} \right.$
⇔$\left \{ {{y=\frac{7}{5}} \atop {x=\frac{13}{10}}} \right.$ (nhận)
TH2: y<0
$\left \{ {{2x-y=4} \atop {4x-3y=1}} \right.$
⇔$\left \{ {{4x-2y=8} \atop {4x-3y=1}} \right.$
⇔$\left \{ {{y=7} \atop {x=\frac{11}{2}}} \right.$ (loại)
Vậy (x;y) = ($\frac{13}{10}$; $\frac{7}{5}$)
