Question

найдите x+y, если x+y+2z=13; x+2y+z=12; 2x+y+z=11

Answer 1

$begin{cases} x+y+2z=13\x+2y+z=12\2x+y+z=11 end{cases}\ \ begin{cases} x+y+2z=13\x+2y+z=12\z = (11-2x-y) end{cases}\ \ begin{cases} x+y+2z=13\x+2y+z=12\2x+y+z=11 end{cases}\ \ begin{cases} x+y+22-4x-2y=13\x+2y211-2x-y=12\2x+y+z=11 end{cases}\ \ begin{cases} -3x-y=9\-x-y=1\2x+y+z=11 end{cases}\ \ begin{cases} y = 9 - 3x \2x=10\2x+y+z=11 end{cases}\ \ begin{cases} x=5\y=-6\z=7 end{cases}$

 

$x + y = 5 - 6 = -1$

 

Ответ: -1