/4x-y=2 I \Log12 x+log12 3=log12(y+1) Решите систему уравнений

Ответ:

Пошаговое объяснение:

$\displaystyle\left \{ {{4x-y=2} \atop {log_{12}3x =log_{12(y+1)} }} \right.\\\\ \\\left \{ {{4x-y=2} \atop {3x=y+1}} \right. \\\\\\$

$\displaystyle\left \{ {{4x-y=2} \atop {3x-y=1}} \right.\\ \\\\\left \{ {{-4x+y=-2} \atop {3x-y=1}} \right.$

-x= -1

x=1

4*1-y=2

4-y=2

-y=2-4

-y= -2

y=2

ответ: (1;2)