Question

НА ФОТО
№847(в) и №848(а)

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{\dfrac{1}{2} (x+7)-6=5y} \atop {\dfrac{2}{3}x-5y=3 }} \right. \\\\\\\left \{ {{\dfrac{1}{2} (x+7)\cdot 2-6\cdot 2=5y\cdot 2} \atop {\dfrac{2}{3}x\cdot 3-5y\cdot 3=3\cdot 3 }} \right. \\\\\\\left \{ {{x+7-12=10y} \atop {2x-15y=9}} \right. \\\\\\\left \{ {{x-10y=5 \ |\cdot (-2)} \atop {2x-15y=9}} \right. \\\\\\+\left \{ {{-2x+20y=-10} \atop {2x-15y=9}} \right. \\------------\\5y=-1\\\\y=-0,2\\\\x=5+10y=5+10\cdot(-0,2)=5-2=3\\\\Otvet \ : \ (3 \ ; -0,2)$

$\displaystyle\bf\\2)\\\\\left \{ {{5(2x-3y)=4x} \atop {6x}-4y=4(2x-1)-5} \right. \\\\\\\left \{ {{10x-15y=4x} \atop {6x-4y=8x-4-5}} \right.\\\\\\\left \{ {{10x-4x-15y=0} \atop {6x-8x-4y=-9}} \right. \\\\\\\left \{ {{6x-15y=0} \atop {-2x-4y=-9 \ |\cdot 3}} \right. \\\\\\+\left \{ {{6x-15y=0} \atop {-6x-12y=-27}} \right. \\-----------\\-27y=-27\\\\y=1\\\\2x=9-4y=9-4\cdot 1=9-4=5\\\\x=5:2=2,5\\\\Otvet \ : \ (2,5 \ ; \ 1)$