y=[2x-1] = 3[x-6]. Find the possible values of [3x+y].

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\frac{5}{6} < \frac{1}{2} x - \frac{1}{2} y < \frac{3}{2}

x - y

\frac{6}{x + y} = \frac{7}{x - y} + 3

\frac{1}{2 (x + y)} = \frac{1}{3 (x - y)}

2 x + y = 9

3 x - y = 6

x

x

y

x - 3 y = 3 x - 1 = 2 x - y

\frac{6}{x + y} = \frac{7}{x - y} + 3

\frac{1}{2 (x + y)} = \frac{1}{3 (x - y)},

x + y \neq 0

x - y \neq 0

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