Giải hệ pt 2(x-2)+3(1+y)=-2 3(x-2)-2(1+y)=-3 28/10/2021 Bởi Natalia Giải hệ pt 2(x-2)+3(1+y)=-2 3(x-2)-2(1+y)=-3
$\left \{ {{2(x-2)+3(1+y)=-2} \atop {3(x-2)-2(1+y)=-3}} \right.$ ⇔$\left \{ {{2x-4+3+3y=-2} \atop {3x-6-2-2y=-3}} \right.$ ⇔$\left \{ {{2x+3y=-1} \atop {3x-2y=5}} \right.$ ⇔$\left \{ {{6x+9y=-3} \atop {6x-4y=10}} \right.$ ⇔$\left \{ {{9y+4y=-3-10} \atop {2x+3y=-1}} \right.$ ⇔$\left \{ {{13y=-13} \atop {2x+3y=-1}} \right.$ ⇔$\left \{ {{y=-1} \atop {2x+3(-1)=-1}} \right.$ ⇔$\left \{ {{y=-1} \atop {2x=-1+3=2}} \right.$ ⇔$\left \{ {{y=-1} \atop {x=1}} \right.$ Vậy HPT có 1 cặp nghiệm duy nhất (x;y)=(1;-1) Bình luận
$\left \{ {{2(x-2)+3(1+y)=-2} \atop {3(x-2)-2(1+y)=-3}} \right.$
⇔$\left \{ {{2x-4+3+3y=-2} \atop {3x-6-2-2y=-3}} \right.$
⇔$\left \{ {{2x+3y=-1} \atop {3x-2y=5}} \right.$
⇔$\left \{ {{6x+9y=-3} \atop {6x-4y=10}} \right.$
⇔$\left \{ {{9y+4y=-3-10} \atop {2x+3y=-1}} \right.$
⇔$\left \{ {{13y=-13} \atop {2x+3y=-1}} \right.$
⇔$\left \{ {{y=-1} \atop {2x+3(-1)=-1}} \right.$
⇔$\left \{ {{y=-1} \atop {2x=-1+3=2}} \right.$
⇔$\left \{ {{y=-1} \atop {x=1}} \right.$
Vậy HPT có 1 cặp nghiệm duy nhất (x;y)=(1;-1)