To solve the system of equations given by:
- \( x + 5y = 15 \) (Equation 1)
- \( 4x + 7y = -5 \) (Equation 2)
We can use the substitution or elimination method. Here, let's use the substitution method.
First, we can solve Equation 1 for \( x \):
\[ x = 15 - 5y \]
Now we will substitute this expression for \( x \) into Equation 2:
\[ 4(15 - 5y) + 7y = -5 \]
Distributing \( 4 \) in the equation:
\[ 60 - 20y + 7y = -5 \]
Combine like terms:
\[ 60 - 13y = -5 \]
Next, we will isolate \( y \) by moving \( 60 \) to the right side:
\[ -13y = -5 - 60 \]
\[ -13y = -65 \]
Now, divide both sides by \( -13 \):
\[ y = \frac{-65}{-13} = 5 \]
Now that we have \( y \), we will substitute it back into the equation we derived for \( x \):
\[ x = 15 - 5(5) \]
\[ x = 15 - 25 = -10 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-10, 5) \]
To verify, let's plug these values back into the original equations:
- For \( x + 5y = 15 \):
\[ -10 + 5(5) = -10 + 25 = 15 \quad \text{(True)} \]
- For \( 4x + 7y = -5 \):
\[ 4(-10) + 7(5) = -40 + 35 = -5 \quad \text{(True)} \]
Since both equations are satisfied, the solution is confirmed:
\[ \boxed{(-10, 5)} \]