Question

40 БАЛЛОВ ДАЮ СРОЧНО ПОМОГИТЕ

Answer 1

$\displaystyle\bf\\ax+3y=11\\\\x=4 \ \ ; \ \ y=-3\\\\a\cdot 4+3\cdot(-3)=11\\\\4a-9=11\\\\4a=20\\\\a=5\\\\\\\left \{ {{5x+3y=11} \ |\cdot(-1)\atop {5x+2y=12}} \right. \\\\\\+\left \{ {{-5x-3y=-11} \atop {5x+2y=12}} \right. \\-----------\\-y=1\\\\y=-1\\\\5x=12-2y=12-2\cdot(-1)=12+2=14\\\\x=14:5=2,8\\\\\\Otvet: \ \boxed{a=5} \ \ ; \ \ \boxed{\Big(2,8 \ ; \ -1\Big)}$