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

To solve the system of equations given by:

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

we can use the method of substitution or elimination. 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)

Next, we will subtract Equation 1 from Equation 3:

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

This simplifies to:

\[ -14n - 3n = 0 \] \[ -17n = 0 \]

From this, we find:

\[ n = 0 \]

Now substitute \(n = 0\) back into one of the original equations to solve for \(m\). We can 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:

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

In conclusion, the final answer is:

\[ (m, n) = \left(-\frac{1}{4}, 0\right) \]