Certainly! Let's analyze each system of equations step-by-step to determine the nature of their solutions.
### System A:
The equations provided are:
1. [tex]\(2x - y = 6\)[/tex]
2. [tex]\(2x + y = 6\)[/tex]
To determine the solution, let's add the two equations:
[tex]\[
\begin{aligned}
(2x - y) + (2x + y) &= 6 + 6 \\
4x &= 12 \\
x &= 3
\end{aligned}
\][/tex]
Now that we have [tex]\(x = 3\)[/tex], we substitute this value back into one of the original equations to find [tex]\(y\)[/tex]. Using the first equation:
[tex]\[
\begin{aligned}
2(3) - y &= 6 \\
6 - y &= 6 \\
y &= 0
\end{aligned}
\][/tex]
Thus, the system has a unique solution, which is [tex]\((x, y) = (3, 0)\)[/tex].
### System B:
The equations provided are:
1. [tex]\(4x + y = 4\)[/tex]
2. [tex]\(-4x - y + 4 = 0\)[/tex]
First, let's simplify the second equation to make it easier to compare to the first equation:
[tex]\[
\begin{aligned}
-4x - y + 4 &= 0 \\
-4x - y &= -4 \\
4x + y &= 4
\end{aligned}
\][/tex]
Notice that the second equation simplifies to exactly the same as the first equation: [tex]\(4x + y = 4\)[/tex].
When both equations are identical, it means that any solution satisfying the first equation will also satisfy the second equation. Therefore, we have infinitely many solutions parameterized by [tex]\(x\)[/tex].
We can express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex] using the equation [tex]\(4x + y = 4\)[/tex]:
[tex]\[
\begin{aligned}
4x + y &= 4 \\
y &= 4 - 4x
\end{aligned}
\][/tex]
Thus, the system has infinitely many solutions given by the equation [tex]\(y = 4 - 4x\)[/tex].
### Final Answer
[tex]\[
\begin{array}{|l|l|}
\hline
\text{System A} & \text{} \\
\begin{array}{l}
2x - y = 6 \\
2x + y = 6
\end{array} &
\begin{array}{l}
(x, y) = (3, 0) \\
\text{The system has a unique solution.}
\end{array} \\
\\ \\
\hline
\text{System B} & \text{} \\
\begin{array}{l}
4x + y = 4 \\
-4x - y + 4 = 0
\end{array} &
\begin{array}{l}
\text{The system has infinitely many solutions.} \\
\text{They must satisfy the following equation:} \\
y = 4 - 4x
\end{array} \\
\hline
\end{array}
\][/tex]
