To solve the system of equations:
\[ \begin{align*}
- & \quad 2x - 5y = 6 \quad \text{(Equation 1)} \
- & \quad -2x + 10y = -16 \quad \text{(Equation 2)} \end{align*} \]
We can start by manipulating these equations. Notice that Equation 2 can be simplified to align its structure with Equation 1. If we divide all terms of Equation 2 by -2, we get:
\[ \frac{-2x}{-2} + \frac{10y}{-2} = \frac{-16}{-2} \implies x - 5y = 8 \quad \text{(Equation 3)} \]
Now we have a new system of equations:
\[ \begin{align*}
- & \quad 2x - 5y = 6 \
- & \quad x - 5y = 8 \end{align*} \]
Next, we can substitute Equation 3 into Equation 1. From Equation 3, we express \( x \) in terms of \( y \):
\[ x = 8 + 5y \]
Now substitute \( x \) in Equation 1:
\[ 2(8 + 5y) - 5y = 6 \]
Simplifying this gives:
\[ 16 + 10y - 5y = 6 \ 16 + 5y = 6 \]
Now we isolate \( y \):
\[ 5y = 6 - 16 \ 5y = -10 \ y = -2 \]
Now that we have \( y \), we can substitute \( y = -2 \) back into Equation 3 to find \( x \):
\[ x - 5(-2) = 8 \ x + 10 = 8 \ x = 8 - 10 \ x = -2 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-2, -2) \]
To verify, we can substitute \( x = -2 \) and \( y = -2 \) back into the original equations:
For Equation 1:
\[ 2(-2) - 5(-2) = -4 + 10 = 6 \quad \text{(True)} \]
For Equation 2:
\[ -2(-2) + 10(-2) = 4 - 20 = -16 \quad \text{(True)} \]
Both equations are satisfied, confirming that our solution is correct.
Finally, the solution is:
\[ \boxed{(-2, -2)} \]