Предмет: Алгебра
Автор: Аноним
Вопрос задан 6 лет назад

ДАЮ 100 БАЛЛОВ!!! Розв'яжіть системи рівнянь:
$1) \left \{ {{y=x+2} \atop {y-2x=3}} \right. 2) \left \{ {{x=3y-5} \atop {x-y=1}} \right. 3)\left \{ {{y=3+2x} \atop {y+5x=10}} \right. 4) \left \{ {{x=2+y} \atop {x+2y=8}} \right.$

Ответы

Ответ дал: Universalka

$\displaystyle\bf\\1)\\\\\left \{ {{y=x+2} \atop {y-2x=3}} \right. \\\\\\\left \{ {{y=x+2} \atop {x+2-2x=3}} \right. \\\\\\\left \{ {{y=x+2} \atop {-x=1}} \right. \\\\\\\left \{ {{x=-1} \atop {y=-1+2}} \right. \\\\\\\left \{ {{x=-1} \atop {y=1}} \right.\\\\\\Otvet: \ (-1 \ ; \ 1)$

$\displaystyle\bf\\2)\\\\\left \{ {{x=3y-5} \atop {x-y=1}} \right. \\\\\\\left \{ {{x=3y-5} \atop {3y-5-y=1}} \right. \\\\\\\left \{ {{x=3y-5} \atop {2y=6}} \right. \\\\\\\left \{ {{x=3\cdot 3-5 \atop {y=3}} \right. \\\\\\\left \{ {{x=4} \atop {y=3}} \right.\\\\\\Otvet: \ (4\ ; \ 3)$

$\displaystyle\bf\\3)\\\\\left \{ {{y=3+2x} \atop {y+5x=10}} \right. \\\\\\\left \{ {{y=3+2x} \atop {3+2x+5x=10}} \right. \\\\\\\left \{ {{y=3+2x} \atop {7x=7}} \right. \\\\\\\left \{ {{x=1} \atop {y=3+2\cdot 1}} \right. \\\\\\\left \{ {{x=1} \atop {y=5}} \right.\\\\\\Otvet: \ (1 \ ; \ 5)$

$\displaystyle\bf\\4)\\\\\left \{ {{x=2+y} \atop {x+2y=8}} \right. \\\\\\\left \{ {{x=2+y} \atop {2+y+2y=8}} \right. \\\\\\\left \{ {{x=2+y} \atop {3y=6}} \right. \\\\\\\left \{ {{x=2+2} \atop {y=2}} \right. \\\\\\\left \{ {{x=4} \atop {y=2}} \right.\\\\\\Otvet: \ (4\ ; \ 2)$