To determine which system of inequalities has a solution set that is a line, let's analyze each given system in detail:
1. \(\left\{\begin{array}{l}x+y \geq 3 \\ x+y \leq 3\end{array}\right.\)
Let's rewrite each inequality:
- \(x + y \geq 3\)
- \(x + y \leq 3\)
Combining these inequalities, we get:
[tex]\[3 \leq x + y \leq 3\][/tex]
This can be simplified:
[tex]\[x + y = 3\][/tex]
Since \(x + y = 3\) represents a line, the solution set for this system of inequalities is indeed a line.
2. \(\left\{\begin{array}{l}x+y \geq-3 \\ x+y \leq 3\end{array}\right.\)
Let's rewrite each inequality:
- \(x + y \geq -3\)
- \(x + y \leq 3\)
Combining these inequalities, we get:
[tex]\[-3 \leq x + y \leq 3\][/tex]
The solution set represents a region bounded by the lines \(x + y = -3\) and \(x + y = 3\), which is an area and not a line.
3. \(\left\{\begin{array}{l}x+y>3 \\ x+y<3\end{array}\right.\)
Let's rewrite each inequality:
- \(x + y > 3\)
- \(x + y < 3\)
Clearly, there is no overlap between \(x + y > 3\) and \(x + y < 3\), resulting in an empty set of solutions. This system does not represent a line.
4. \(\left\{\begin{array}{l}x+y>-3 \\ x+y<3\end{array}\right.\)
Let's rewrite each inequality:
- \(x + y > -3\)
- \(x + y < 3\)
Combining these inequalities, we get:
[tex]\[-3 < x + y < 3\][/tex]
The solution set represents a region between the lines \(x + y = -3\) and \(x + y = 3\), which is an area and not a line.
From this detailed analysis, we can conclude that the system of inequalities that has a solution set that is a line is:
\(\left\{\begin{array}{l}x+y \geq 3 \\ x+y \leq 3\end{array}\right.\)
Thus, the correct answer is the first system:
[tex]\[
\left\{\begin{array}{l}
x + y \geq 3 \\
x + y \leq 3
\end{array}\right.
\][/tex]
