To determine if the system of equations has no solutions, infinitely many solutions, or exactly one solution, we need to convert the system of equations into a form that can be easily compared.

The system of equations is:

1) 6x + y = -3
2) -28x - 5y = 13

We can multiply the first equation by 5 and the second equation by 1 to make the coefficients of y the same in both equations:

1) 30x + 5y = -15
2) -28x - 5y = 13

Now, we can add these two equations together:

30x - 28x = -15 + 13
2x = -2
x = -1

Substitute x = -1 into the first equation:

6*(-1) + y = -3
-6 + y = -3
y = -3 + 6
y = 3

So, the solution to the system of equations is x = -1 and y = 3. Therefore, the system of equations has exactly one solution.

Question

To determine if the system of equations has no solutions, infinitely many solutions, or exactly one solution, we need to convert the system of equations into a form that can be easily compared.

The system of equations is:

1) 6x + y = -3
2) -28x - 5y = 13

We can multiply the first equation by 5 and the second equation by 1 to make the coefficients of y the same in both equations:

1) 30x + 5y = -15
2) -28x - 5y = 13

Now, we can add these two equations together:

30x - 28x = -15 + 13
2x = -2
x = -1

Substitute x = -1 into the first equation:

6*(-1) + y = -3
-6 + y = -3
y = -3 + 6
y = 3

So, the solution to the system of equations is x = -1 and y = 3. Therefore, the system of equations has exactly one solution.

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