Systems of two linear equations in two variables SAT question

The correct answer is $6$. A system of two linear equations in two variables, $x$ and $y$, has no solution if the lines represented by the equations in the xy-plane are parallel and distinct. Lines represented by equations in standard form, $Ax+By=C$ and $Dx+Ey=F$, are parallel if the coefficients for $x$ and $y$ in one equation are proportional to the corresponding coefficients in the other equation, meaning $\frac{D}{A}=\frac{E}{B}$; and the lines are distinct if the constants are not proportional, meaning $\frac{F}{C}$ is not equal to $\frac{D}{A}$ or $\frac{E}{B}$. The first equation in the given system is $\frac{3}{2}y-\frac{1}{4}x=\frac{2}{3}-\frac{3}{2}y$. Multiplying each side of this equation by $12$ yields $18y-3x=8-18y$. Adding $18y$ to each side of this equation yields $36y-3x=8$, or $-3x+36y=8$. The second equation in the given system is $\frac{1}{2}x+\frac{3}{2}=py+\frac{9}{2}$. Multiplying each side of this equation by $2$ yields $x+3=2py+9$. Subtracting $2py$ from each side of this equation yields $x+3-2py=9$. Subtracting $3$ from each side of this equation yields $x-2py=6$. Therefore, the two equations in the given system, written in standard form, are $-3x+36y=8$ and$x-2py=6$. As previously stated, if this system has no solution, the lines represented by the equations in the xy-plane are parallel and distinct, meaning the proportion $\frac{1}{-3}=\frac{-2p}{36}$, or $-\frac{1}{3}=-\frac{p}{18}$, is true and the proportion $\frac{6}{8}=\frac{1}{-3}$ is not true. The proportion $\frac{6}{8}=\frac{1}{-3}$ is not true. Multiplying each side of the true proportion, $-\frac{1}{3}=-\frac{p}{18}$,  by $-18$ yields $6=p$. Therefore, if the system has no solution, then the value of $p$ is $6$.