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, $A x + B y = C$ and $D x + E y = F$ , 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 $18 y - 3 x = 8 - 18 y$ . Adding $18 y$ to each side of this equation yields $36 y - 3 x = 8$ , or $- 3 x + 36 y = 8$ . The second equation in the given system is $\frac{1}{2} x + \frac{3}{2} = p y + \frac{9}{2}$ . Multiplying each side of this equation by $2$ yields $x + 3 = 2 p y + 9$ . Subtracting $2 p y$ from each side of this equation yields $x + 3 - 2 p y = 9$ . Subtracting $3$ from each side of this equation yields $x - 2 p y = 6$ . Therefore, the two equations in the given system, written in standard form, are $- 3 x + 36 y = 8$ and $x - 2 p y = 6$ . 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 $\frac{1}{- 3} = \frac{- 2 p}{36}$ , 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 $negative 18$ yields $6 equals p$ . Therefore, if the system has no solution, then the value of $p$ is $6$ .