Let's evaluate each system of equations one by one to see if it is equivalent to the original system:
Original System:
[tex]\[
\begin{array}{l}
4x - y = -11 \\
2x + 3y = 5
\end{array}
\][/tex]
### Equivalent System 1:
[tex]\[
\begin{array}{l}
4x + 3y = 5 \\
2y = -6
\end{array}
\][/tex]
1. The first equation in Equivalent System 1:
[tex]\[
4x + 3y = 5
\][/tex]
is different from both equations in the original system, indicating this system is not equivalent.
### Equivalent System 2:
[tex]\[
\begin{array}{l}
7x - 3y = -11 \\
9x = -6
\end{array}
\][/tex]
1. The second equation:
[tex]\[
9x = -6 \quad \Rightarrow \quad x = -\frac{2}{3}
\][/tex]
Substitute [tex]\( x = -\frac{2}{3} \)[/tex] into the first equation:
[tex]\[
7\left(-\frac{2}{3}\right) - 3y = -11 \quad \Rightarrow \quad -\frac{14}{3} - 3y = -11 \quad \Rightarrow \quad -3y = -11 + \frac{14}{3} \quad \Rightarrow \quad -3y = -\frac{19}{3}
\quad \Rightarrow \quad y = \frac{19}{9}
\][/tex]
2. Plugging these solutions into the original equations shows that they do not satisfy both, indicating this system is also not equivalent.
### Equivalent System 3:
[tex]\[
\begin{array}{l}
-4x - 9y = -19 \\
-10y = -30
\end{array}
\][/tex]
1. The second equation:
[tex]\[
-10y = -30 \quad \Rightarrow \quad y = 3
\][/tex]
Substitute [tex]\( y = 3 \)[/tex] into the first equation:
[tex]\[
-4x - 9(3) = -19 \quad \Rightarrow \quad -4x - 27 = -19 \quad \Rightarrow \quad -4x = 8 \quad \Rightarrow \quad x = -2
\][/tex]
2. Check:
[tex]\[
4(-2) - 3 = -11 \quad \text{and} \quad 2(-2) + 3(3) = 5
\][/tex]
Both original equations are satisfied, indicating this system is equivalent.
### Equivalent System 4:
[tex]\[
\begin{array}{l}
12x - 3y = -33 \\
14x = -28
\end{array}
\][/tex]
1. The second equation:
[tex]\[
14x = -28 \quad \Rightarrow \quad x = -2
\][/tex]
Substitute [tex]\( x = -2 \)[/tex] into the first equation:
[tex]\[
12(-2) - 3y = -33 \quad \Rightarrow \quad -24 - 3y = -33 \quad \Rightarrow \quad -3y = -9 \quad \Rightarrow \quad y = 3
\][/tex]
2. Check:
[tex]\[
4(-2) - 3 = -11 \quad \text{and} \quad 2(-2) + 3(3) = 5
\][/tex]
Both original equations are satisfied, indicating this system is equivalent.
Answer: Both System 3 and System 4 are equivalent to the original system.
