Let's analyze each of the given systems of equations to determine their solution sets.
### System 1
[tex]\[
\begin{array}{l}
y=6x-3 \\
y=3(2x+1)
\end{array}
\][/tex]
First, we simplify the second equation:
[tex]\[
3(2x+1) = 6x + 3
\][/tex]
Thus, the system becomes:
[tex]\[
\begin{array}{l}
y=6x-3 \\
y=6x+3
\end{array}
\][/tex]
Let's set the right sides equal to each other:
[tex]\[
6x-3 = 6x+3
\][/tex]
Subtract [tex]\(6x\)[/tex] from both sides:
[tex]\[
-3 = 3
\][/tex]
This is a contradiction, which means there are no values of [tex]\(x\)[/tex] that satisfy both equations simultaneously. Therefore, this system has zero solutions.
### System 2
[tex]\[
\begin{array}{l}
y=2x+1 \\
y=6x+3
\end{array}
\][/tex]
Set the right sides equal to each other:
[tex]\[
2x+1 = 6x+3
\][/tex]
Rearrange the terms to solve for [tex]\(x\)[/tex]:
[tex]\[
2x - 6x = 3 - 1 \\
-4x = 2 \\
x = -\frac{1}{2}
\][/tex]
Plug [tex]\(x = -\frac{1}{2}\)[/tex] back into either equation to find [tex]\(y\)[/tex]. Using the first equation:
[tex]\[
y = 2\left(-\frac{1}{2}\right) + 1 = -1 + 1 = 0
\][/tex]
So, the solution is [tex]\((- \frac{1}{2}, 0)\)[/tex]. Therefore, this system has one solution.
### Summary of Results
1. System 1:
[tex]\[
\begin{array}{l}
y=6x-3 \\
y=3(2x+1)
\end{array}
\][/tex]
- Zero Solutions: [tex]\(6x-3\)[/tex] can never be equal to [tex]\(6x+3\)[/tex].
2. System 2:
[tex]\[
\begin{array}{l}
y=2x+1 \\
y=6x+3
\end{array}
\][/tex]
- One Solution: [tex]\(2x+1 = 6x+3\)[/tex] has one solution.
Therefore, we align the statements accordingly for the systems of equations:
1. Zero Solutions: [tex]\(6x-3\)[/tex] can never be equal to [tex]\(6x+3\)[/tex].
- System:
[tex]\[
\begin{array}{l}
y=6x-3 \\
y=3(2x+1)
\end{array}
\][/tex]
2. One Solution: [tex]\(2x+1 = 6x+3\)[/tex] has one solution.
- System:
[tex]\[
\begin{array}{l}
y=2x+1 \\
y=6x+3
\end{array}
\][/tex]
