- 12 x + 14 y = 36 - 6 x + 7 y = -

Choice D is correct. A system of two linear equations in two variables, $x$ and $y$, has zero solutions if the lines representing the equations in the xy-plane are distinct and parallel. Two lines are distinct and parallel if they have the same slope but different y-intercepts. Each equation in the given system can be written in slope-intercept form $y=mx+b$, where $m$ is the slope of the line representing the equation in the xy-plane and $\left(0,b\right)$ is the y-intercept. Adding $12x$ to both sides of the first equation in the given system of equations, $-12x+14y=36$, yields $14y=12x+36$. Dividing both sides of this equation by $14$ yields $y=\frac{6}{7}x+\frac{18}{7}$. It follows that the first equation in the given system of equations has a slope of $\frac{6}{7}$ and a y-intercept of $\left(0,\frac{18}{7}\right)$. Adding $6x$ to both sides of the second equation in the given system of equations, $-6x+7y=-18$, yields $7y=6x-18$. Dividing both sides of this equation by $7$ yields $y=\frac{6}{7}x-\frac{18}{7}$. It follows that the second equation in the given system of equations has a slope of $\frac{6}{7}$ and a y-intercept of $\left(0,-\frac{18}{7}\right)$. Since the slopes of these lines are the same and the y-intercepts are different, it follows that the given system of equations has zero solutions.

Alternate approach: To solve the system by elimination, multiplying the second equation in the given system of equations, $-6x+7y=-18$, by $-2$ yields $12x-14y=36$. Adding this equation to the first equation in the given system of equations, $-12x+14y=36$, yields $(-12x+12x)+(-14y+14y)=36+36$, or $0=72$. Since this equation isn't true, the given system of equations has zero solutions.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.