Answer:
[tex]\large\boxed{A.\ 0}[/tex]
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}ax+by=c\\dx+ey=f\end{array}\right\\\\D=\left|\begin{array}{ccc}a&b\\d&e\end{array}\right|=ae-bd\\\\D_x=\left|\begin{array}{ccc}c&b\\f&e\end{array}\right|=ce-fb\\\\D_y=\left|\begin{array}{ccc}a&c\\d&f\end{array}\right|=fa-cd\\\\\text{If}\ D\neq 0\ \text{then the system of equations has one solution}\ x=\dfrac{D_x}{D}\ and\ y=\dfrac{D_y}{D}\\\\\text{If}\ D=0\ \text{and}\ D_x=0\ \text{and}\ D_y=0\ \text{then the system of equations has}\\\text{infinitely many solutions.}\\\\\text{If}\ D=0\ \text{and}\ D_x\neq0\ \text{or}\ D_y\neq0\ \text{then the system of equations has}\\\text{no solutions.}[/tex]
[tex]\text{We have}\\\\\left\{\begin{array}{ccc}3x+4y=12\\x-6y=-18\end{array}\right\\\\D=\left|\begin{array}{ccc}3&4\\1&-6\end{array}\right|=(3)(-6)+(1)(4)=-18+4=-14\\\\D_x=\left|\begin{array}{ccc}12&4\\-18&-6\end{array}\right|=(12)(-6)-(-18)(4)=-72+72=0\\\\x=\dfrac{D_x}{D}\to x=\dfrac{0}{-14}=0[/tex]
