Помогите решить уравнение
3х-4=2-3у
у=1 1/3-3у
Х+у=х+у
х-у+2=0
2х+3у=2х+3у+2
Х-7у+1=0
3х+5у=5(х+3у)-2(х+5у)
у-3+х=2х+(х+у-3)
3х+4у+1=(х+у-2)+(2х+3у+3)
х+у+2=у+(2+х)

Предыдущий
вопрос Следующий
вопрос

3x - 4 = 2 - 3y
3x - 4 + 4 = 2 - 3y + 4
3x = 6 - 3y
x = (6 - 3y) / 3
x = 2 - y

y = 1 1/3 - 3y
y = 4/3 - 9y
1 = 4/3 + 8y
8y = 3 - 4/3
8y = 9/3 - 4/3
8y = 5/3
y = 5/24

x + y = x + y
This is always true

x - y + 2 = 0
This is a linear equation

2x + 3y = 2x + 3y + 2
This is always true

x - 7y + 1 = 0
This is a linear equation

3x + 5y = 5(x + 3y) - 2(x + 5y)
3x + 5y = 5x + 15y - 2x - 10y
3x + 5y = 3x + 5y
This is always true

y - 3 + x = 2x + (x + y - 3)
y - 3 + x = 2x + x + y - 3
y - 3 + x = 3x + y - 3
x = 2x
This is not possible

3x + 4y + 1 = (x + y - 2) + (2x + 3y + 3)
3x + 4y + 1 = x + y - 2 + 2x + 3y + 3
3x + 4y + 1 = 3x + 4y + 2
1 = 2
This is not possible

x + y + 2 = y + (2 + x)
x = 2
This is valid.