Determining the Constant Value for a System of Equations Without Solutions
The system of equations is given by:
\[
\begin{align*}
3x + \frac{1}{2}y &= \frac{3}{2} \\
2x + \frac{3}{4}y &= 1 \\
\frac{1}{2}x + \frac{1}{4}y &= \frac{p}{2} + \frac{1}{2}
\end{align*}
\]
To determine a value of \( p \) that leads to no solution, the equations must be inconsistent. This happens when the ratios of the coefficients of \( x \) and \( y \) are equal for two equations, but the ratio of the constants is different.
First, let's make the coefficients of \( y \) similar in the first two equations:
\[
\begin{align*}
3x + \frac{1}{2}y &= \frac{3}{2} \ |\times2 \\
2x + \frac{3}{4}y &= 1 \ |\times4 \\
\end{align*}
\]
This results in:
\[
\begin{align*}
6x + y &= 3 \\
8x + 3y &= 4
\end{align*}
\]
Now, let's make the coefficient of \( x \) in the third equation similar to the first equation:
\[
\begin{align*}
\frac{1}{2}x + \frac{1}{4}y &= \frac{p}{2} + \frac{1}{2} \ |\times6 \\
\end{align*}
\]
This results in:
\[
3x + \frac{3}{2}y = 3p + 3
\]
Comparing the first and last equations, the coefficient ratio of \( x \) to \( y \) is \( 6:1 \) and \( 3:\frac{3}{2} \). Simplifying the second ratio gives us \( 2:1 \), which is the same as \( 6:3 \).
For the system to have no solution, the constant term (after the simplification) should not have the same ratio. Comparing the constants from the first and the last equations, \( 3 \) and \( 3p + 3 \) must not have the same ratio as the coefficients of \( x \) and \( y \), which is \( 2:1 \). Therefore, \( 2(3) \) should not equal \( 3p + 3 \).
Equating and solving gives us:
\[
6 \neq 3p + 3
\]
Subtracting \( 3 \) from both sides gives:
\[
3 \neq 3p
\]
Dividing by \( 3 \) we get:
\[
1 \neq p
\]
Thus, the value of \( p \) that results in no solution is any value except \( p = 1 \).
