Giải hệ phương trình :
a) $\left \{ {{3x-y=1} \atop {x+y=3}} \right.$
<=>$\left \{ {{3x-y=1} \atop {x=3-y}} \right.$
<=>$\left \{ {{3(3-y)-y=1} \atop {x=3-y}} \right.$
<=> $\left \{ {{9-4y=1} \atop {x=3-y}}
<=> \right.$ $\left \{ {{4y=8} \atop {x=3-y}} \right.$
<=> $\left \{ {{y=2} \atop {x=3-2}} \right.$
<=> $\left \{ {{x=1} \atop {y=2}} \right.$
Vậy ( x ; y) = ( 1;2)
b) $\left \{ {{2x+y=2} \atop {4x-y=10}} \right.$
<=> $\left \{ {{y=2-2x} \atop {4x-(2-2x)=10}} \right.$
<=> $\left \{ {{y=2-2x} \atop {4x-2+2x=10}} \right.$
<=> $\left \{ {{y=2-2x} \atop {6x=12}} \right.$
<=> $\left \{ {{y=2-2x} \atop {x=2}} \right.$
<=> $\left \{ {{y=2-2.2} \atop {x=2}} \right.$
<=> $\left \{ {{x=2} \atop {y=-2}} \right.$
Vậy ( x ; y ) = ( 2 ; -2 ).
