36. Given the system of equations [3x]+{y}+x-y=1 and [-y]-{x}-x+y=1 then

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\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

Solve the following systems of equations:

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

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

\frac{3 x + y}{x - 1} - \frac{x - y}{2 y} = 2, x - y = 4

\frac{6}{x + y} = \frac{7}{x - y} + 3, \frac{1}{2 (x + y)} = \frac{1}{3 (x - y)}

x + y \neq 0 a n d x - y \neq 0

y = 3 x - 1

\frac{1}{2} y + x = 1

(x, y)

y - x

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