Question

ПОМОГИТЕ ПЖПЖПЖПЖЖПЖ ДАЮ 60 БАЛЛОВ​

Answer 1

Ответ:

Объяснение:

решение в прикрепленном фото

Answer 2

$\displaystyle\bf\\\left \{ {{\dfrac{x+3}{2} -\dfrac{y-2}{3} =4} \atop {\dfrac{x-1}{4} +\dfrac{y+1}{3} =4}} \right. \\\\\\\left \{ {{\dfrac{x+3}{2}\cdot 6 -\dfrac{y-2}{3}\cdot 6 =4\cdot 6} \atop {\dfrac{x-1}{4}\cdot 12 +\dfrac{y+1}{3}\cdot 12 =4\cdot 12}} \right. \\\\\\\left \{ {{(x+3)\cdot 3-(y-2)\cdot 2=24} \atop {(x-1)\cdot 3+(y+1)\cdot 4=48}} \right. \\\\\\\left \{ {{3x+9-2y+4=24} \atop {3x-3+4y+4=48}} \right. \\\\\\\left \{ {{3x-2y=24-4-9} \atop {3x+4y=48+3-4}} \right.$

$\displaystyle\bf\\\left \{ {{3x-2y=11|\cdot(-1) \ } \atop {3x+4y=47}} \right.\\\\\\+\left \{ {{-3x+2y=-11} \atop {3x+4y=47}} \right. \\-----------\\6y=36\\\\y=36:6=6\\\\3x=2y+11=2\cdot 6+11=12+11=23\\\\x=23:3=7\frac{2}{3} \\\\\\Otvet \ : \ (7\frac{2}{3} \ ; \ 6)$