Question

Помогите пожалуйста решить системы

Answer 1

1) Сделаем замену :

$\frac{1}{2x-3y}=m\\\\\frac{1}{3x-2y}=n\\\\\left \{ {{m+2n=\frac{3}{4}}|*(-2) \atop {3m+4n=\frac{7}{4} }} \right.\\\\\\+\left \{ {{-2m-4n=-\frac{6}{4}} \atop {3m+4n=\frac{7}{4} }} \right.\\--------\\m=\frac{1}{4}\\\\2n=\frac{3}{4}-m=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\\\\n=\frac{1}{2}:2=\frac{1}{4}$

$1)\frac{1}{2x-3y}=\frac{1}{4}\\\\2x-3y=4\\\\2)\frac{1}{3x-2y}=\frac{1}{4}\\\\3x-2y=4\\\\\+\left \{ {{2x-3y=4}|*(-2) \atop {3x-2y=4}|*3} \right.\\\\\\+\left \{ {{-4x+6y=-8} \atop {9x-6y=12}} \right.\\ --------\\5x=4\\\\x=0,8\\\\2y=3x-4=3*0,8-4=2,4-4=-1,6\\\\y=-0,8\\\\Otvet:\boxed{(0,8;-0,8)}$

$2)\left \{ {{\frac{x}{4}*12+\frac{y}{6}*12=\frac{1}{2}*12} \atop {(6x+4y)(x+\frac{1}{2}y)=-12 }} \right.\\\\\\\left \{ {{3x+2y=6} \atop {(6x+4y)(x+\frac{1}{2}y)=-12 }} \right.\\\\\\\left \{ {{6x+4y=12} \atop {12(x+\frac{1}{2}y)=-12 }} \right.\\\\\\\left \{ {{3x+2y=6} \atop {x+\frac{1}{2}y =-1}|*-4} \right. \\\\\\+\left \{ {{3x+2y=6} \atop {-4x-2y=4}} \right. \\-------\\-x=10\\\\x=-10\\\\2y=6-3x=6-3*(-10)=6+30=36\\\\y=18\\\\Otvet:\boxed{(-10;18)}$