To determine which system of inequalities has a line as a solution, let's analyze each system in detail.
### System 1:
[tex]\[
\begin{cases}
2x + 4y \geq 3 \\
2x + 4y \leq 3
\end{cases}
\][/tex]
- For the inequality [tex]\(2x + 4y \geq 3\)[/tex], any point [tex]\((x, y)\)[/tex] that satisfies this inequality will lie on or above the line [tex]\(2x + 4y = 3\)[/tex].
- For the inequality [tex]\(2x + 4y \leq 3\)[/tex], any point [tex]\((x, y)\)[/tex] that satisfies this inequality will lie on or below the line [tex]\(2x + 4y = 3\)[/tex].
For a point to satisfy both inequalities simultaneously, it must lie exactly on the line [tex]\(2x + 4y = 3\)[/tex]. Therefore, this system of inequalities simplifies to the equation of the line:
[tex]\[2x + 4y = 3\][/tex]
So, system 1 has a line as its solution.
### System 2:
[tex]\[
\begin{aligned}
2x + 4y & \geq 3 \\
2x + 4y & > 3
\end{aligned}
\][/tex]
- The inequality [tex]\(2x + 4y \geq 3\)[/tex] encompasses all the points on or above the line [tex]\(2x + 4y = 3\)[/tex].
- The inequality [tex]\(2x + 4y > 3\)[/tex] encompasses all the points strictly above the line [tex]\(2x + 4y = 3\)[/tex].
There is no point [tex]\((x, y)\)[/tex] that can simultaneously satisfy both [tex]\( \geq 3\)[/tex] and [tex]\( > 3\)[/tex]. Thus, there are no common solutions, and this system does not have a line as its solution.
### System 3:
[tex]\[
\begin{cases}
2x + 4y > 3 \\
2x + 4y < 3
\end{cases}
\][/tex]
- The inequality [tex]\(2x + 4y > 3\)[/tex] encompasses all the points strictly above the line [tex]\(2x + 4y = 3\)[/tex].
- The inequality [tex]\(2x + 4y < 3\)[/tex] encompasses all the points strictly below the line [tex]\(2x + 4y = 3\)[/tex].
It is inherently impossible for a point to satisfy both [tex]\(> 3\)[/tex] and [tex]\( < 3\)[/tex] simultaneously. Thus, this system also has no solution.
Based on this analysis, system 1 is the only system where the solution is a line.
So, the answer is:
[tex]\[
\boxed{1}
\][/tex]
