Question

допоможіть з 7 завданням!!!

Answer 1

Ответ:

$\displaystyle\left \{ {{5(x-3y)-6=2x+1} \atop {3(x+6y)+4=9y+19}} \right.$

$\displaystyle\left \{ {{5x-15y-6=2x+1} \atop {3x+18y+4=9y+19}} \right.$

$\displaystyle\left \{ {{3x-15y=7} \atop {3x+9y=15}} \right.$   /*-1

$\displaystyle\left \{ {{3x-15y=7} \atop {-3x-9y=-15}} \right.$

Додамо  обидва рівняння:

3x-15y-3x-9y=7-15

-24у=-8

у=8/24

$3x-15*\dfrac{8}{24} =7\\3x-5=7\\3x=12\\x=4$

Відповідь: (4; 8/24)

Answer 2

$\displaystyle\bf\\\left \{ {{5(x-3y)}-6=2x+1 \atop {3(x+6y)+4=9y+19}} \right. \\\\\\\left \{ {{5x-15y-6=2x+1} \atop {3x+18y+4=9y+19}} \right. \\\\\\\left \{ {{5x-2x-15y=1+6} \atop {3x+18y-9y=19-4}} \right. \\\\\\\left \{ {{3x-15y=7} \ |\cdot(-1)\atop {3x+9y=15}} \right.\\\\\\+\left \{ {{-3x+15y=-7} \atop {3x+9y=15}} \right.\\ -----------\\24y=8\\\\y=\frac{1}{3} \\\\3x+9y=15\\\\x+3y=5\\\\x=5-3y=5-3\cdot\frac{1}{3} =5-1=4\\\\Otvet \ :\Big(4 \ ; \ \frac{1}{3} \Big)$