40m + 3n = -10, 20m - 7n = -5;

To solve the system of equations:

1. \( 40m + 3n = -10 \) (Equation 1)
2. \( 20m - 7n = -5 \) (Equation 2)

We can use the substitution or elimination method. Here, we will use the elimination method.

First, let's multiply Equation 2 by 2 to align the coefficients of \( m \):

\[ 2(20m - 7n) = 2(-5) \]

This simplifies to:

\[ 40m - 14n = -10 \quad (Equation 3) \]

Now we have:

1. \( 40m + 3n = -10 \) (Equation 1)
2. \( 40m - 14n = -10 \) (Equation 3)

Now, we can subtract Equation 1 from Equation 3:

\[ (40m - 14n) - (40m + 3n) = -10 - (-10) \]

This simplifies to:

\[ -14n - 3n = 0 \]

Which gives:

\[ -17n = 0 \] \[ n = 0 \]

Now that we know \( n = 0 \), we can substitute this value back into one of the original equations to find \( m \). We'll use Equation 1:

\[ 40m + 3(0) = -10 \] \[ 40m = -10 \] \[ m = -\frac{10}{40} = -\frac{1}{4} \]

Thus, the solution to the system of equations is:

\[ m = -\frac{1}{4}, \quad n = 0 \]

If you have any further questions or need clarification, feel free to ask!