The correct answer is $\frac{7}{2}$ . 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 distinct and parallel. Two lines represented by equations in standard form $A x + B y = C$ , where $A$ , $B$ , and $C$ are constants, are parallel if the coefficients for $x$ and $y$ in one equation are proportional to the corresponding coefficients in the other equation. The first equation in the given system, $\frac{7}{8} y - \frac{5}{8} x = \frac{4}{7} - \frac{7}{8} y$ , can be written in standard form by adding $\frac{7}{8} y$ to both sides of the equation, which yields $\frac{14}{8} y - \frac{5}{8} x = \frac{4}{7}$ , or $- \frac{5}{8} x + \frac{14}{8} y = \frac{4}{7}$ . Multiplying each term in this equation by $negative 8$ yields $5 x - 14 y = - \frac{32}{7}$ . The second equation in the given system, $\frac{5}{4} x + \frac{7}{4} = p y + \frac{15}{4}$ , can be written in standard form by subtracting $\frac{7}{4}$ and $p y$ from both sides of the equation, which yields $\frac{5}{4} x - p y = \frac{8}{4}$ . Multiplying each term in this equation by $4$ yields $5 x - 4 p y = 8$ . The coefficient of $x$ in the first equation, $5 x - 14 y = - \frac{32}{7}$ , is equal to the coefficient of $x$ in the second equation, $5 x - 4 p y = 8$ . For the lines to be parallel, and for the coefficients for $x$ and $y$ in one equation to be proportional to the corresponding coefficients in the other equation, the coefficient of $y$ in the second equation must also be equal to the coefficient of $y$ in the first equation. Therefore, $- 14 = - 4 p$ . Dividing both sides of this equation by $negative 4$ yields $\frac{- 14}{- 4} = p$ , or $p = \frac{7}{2}$ . Therefore, if the given system of equations has no solution, the value of $p$ is $\frac{7}{2}$ . Note that 7/2 and 3.5 are examples of ways to enter a correct answer.