2.решите систему двух уравнений
{у=3х-4
{3у=-х-2
{3(3х-4)= -х-2
{У=3х-4
9х-12=-х-2
9х+х=-2+12
10х=10
Х=1
{У=3×1-4
{У=-1
Х знак пренадлежит(-1;1)
3cos^2x-5+3sin^2x=3(cos^2x+sin^2x)-5=3-5=-2
3*(2х + y - 1) = 5x + 4y +2
4*(x - 2y + 1) = 2x -5y +16
6x + 3y - 3 = 5x + 4y +2
4x - 8y + 4 = 2x -5y +16
6x - 5x + 3y - 4y = 2 + 3
4x - 2x - 8y + 5y = 16 - 4
x - y = 5
2x - 3y = 12
x = 5 +y
2*(5+y) - 3y = 12
10 + 2y - 3y = 12
-y = 12 - 10
-y = 2
y = -2
x = 5 - <span>2 = 3
</span>Ответ: <span>x = </span><span>3
</span> y = -2