Let's analyze each system of linear equations to determine which one has infinitely many solutions.
### Option A:
[tex]\[
\begin{cases}
x = 2 \\
y = 4
\end{cases}
\][/tex]
This system clearly provides unique solutions for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. There is no possibility of having infinitely many solutions here since both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are explicitly defined.
### Option B:
[tex]\[
\begin{cases}
3x + y = -7 \\
2x + 2y = -14
\end{cases}
\][/tex]
First, we can rewrite the second equation in a simpler form by dividing everything by 2:
[tex]\[
2x + 2y = -14 \quad \Rightarrow \quad x + y = -7
\][/tex]
Now, let's compare the two equations:
1. [tex]\( 3x + y = -7 \)[/tex]
2. [tex]\( x + y = -7 \)[/tex]
We can subtract the second equation from the first to eliminate [tex]\( y \)[/tex]:
[tex]\[
(3x + y) - (x + y) = -7 - (-7) \\
2x = 0 \\
x = 0
\][/tex]
Substituting [tex]\( x = 0 \)[/tex] into [tex]\( x + y = -7 \)[/tex]:
[tex]\[
0 + y = -7 \\
y = -7
\][/tex]
This system has a unique solution [tex]\((x, y) = (0, -7)\)[/tex]. Hence, it does not have infinitely many solutions.
### Option C:
[tex]\[
\begin{cases}
12x - 2y = 10 \\
6x - y = 5
\end{cases}
\][/tex]
We can see that the second equation is exactly half the first equation (divide the coefficients of the first equation by 2):
[tex]\[
\frac{12x - 2y}{2} = \frac{10}{2} \quad \Rightarrow \quad 6x - y = 5
\][/tex]
Since the second equation is essentially a duplicate of the first, the two lines represented by these equations are coincident (they are the same line). Therefore, this system has infinitely many solutions as it represents the same line algebraically.
### Option D:
[tex]\[
\begin{cases}
2x + y = 9 \\
2x + y = -9
\end{cases}
\][/tex]
If we subtract one equation from the other:
[tex]\[
(2x + y) - (2x + y) = 9 - (-9) \\
0 = 18
\][/tex]
This is a contradiction, which means that there are no solutions to this system. The two lines are parallel but distinct, so they never intersect.
### Conclusion:
Only option C has infinitely many solutions since the two equations represent the same line.
So, the answer is:
(C)
