Determine if the following equations represent linear functions. Show your work.

Subtract $x$ and 4 from both sides.

$$x - 3y + 4 = \frac{2}{3}x - 1$$ $$-3y = \frac{2}{3}x - x - 1 - 4$$ $$-3y = \left(\frac{2}{3} - \frac{3}{3}\right)x - 5$$ $$-3y = -\frac{1}{3}x - 5$$

$$y = \frac{-\frac{1}{3}x}{-3} - \frac{5}{-3}$$ $$y = \frac{1}{9}x + \frac{5}{3}$$

The equation $y = \frac{1}{9}x + \frac{5}{3}$ is in the form $y = mx + c$, where $m = \frac{1}{9}$ and $c = \frac{5}{3}$.
Therefore, it is a linear function.

$$\boxed{\text{Yes, it is a linear function.}}$$

C)

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

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

The equation contains $x^2$ and $x$ in the denominator, which means it cannot be written in the form $y = mx + c$.
Therefore, it is not a linear function.

$$\boxed{\text{No, it is not a linear function.}}$$

D)

Assuming $x-y \neq 0$, the fraction $\frac{x-y}{x-y}$ simplifies to 1.

$$\frac{x-y}{x-y} = x + y$$ $$1 = x + y$$

The equation $y = -x + 1$ is in the form $y = mx + c$, where $m = -1$ and $c = 1$.
Therefore, it is a linear function (with the condition $x \neq y$).

$$\boxed{\text{Yes, it is a linear function.}}$$

E)

$$\frac{2x+2y}{3} = \frac{3(x+2y)}{2}$$ $$6 \cdot \frac{2x+2y}{3} = 6 \cdot \frac{3(x+2y)}{2}$$ $$2(2x+2y) = 3 \cdot 3(x+2y)$$ $$4x+4y = 9(x+2y)$$

$$4y - 18y = 9x - 4x$$ $$-14y = 5x$$

The equation $y = -\frac{5}{14}x$ is in the form $y = mx + c$, where $m = -\frac{5}{14}$ and $c = 0$.
Therefore, it is a linear function.

$$\boxed{\text{Yes, it is a linear function.}}$$