Answer :
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]
### 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]
