[tex]3(x - y) = 6(y + 1) \\ \frac{x}{3} - 1 \frac{1}{3} = y[/tex]

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вопрос

First, let's simplify the given equations:

[3(x - y) = 6(y + 1)]
[3x - 3y = 6y + 6]
[3x = 9y + 6]
[x = 3y + 2]

[\frac{x}{3} - 1 \frac{1}{3} = y]
[\frac{x}{3} - \frac{4}{3} = y]
[x - 4 = 3y]
[x = 3y + 4]

Now we have the following equations:
[x = 3y + 2]
[x = 3y + 4]

Since both of these equations must be true at the same time, they are actually contradicting each other. Therefore, there seems to be no solution that satisfies both given equations simultaneously.