To solve the system of equations:
- Equation 1: \(10x - 8y = 16\)
- Equation 2: \(5x - 3y = 1\)
We can use the method of substitution or elimination. Here, I'll use the elimination method.
First, let's simplify Equation 1 by dividing all terms by 2:
\[ 5x - 4y = 8 \quad \text{(Equation 3)} \]
Now we have the following two equations:
- \(5x - 4y = 8\) (Equation 3)
- \(5x - 3y = 1\) (Equation 2)
Next, we'll eliminate \(5x\) by subtracting Equation 2 from Equation 3:
\[ (5x - 4y) - (5x - 3y) = 8 - 1 \]
This simplifies to:
\[ -4y + 3y = 7 \]
or
\[ -y = 7 \]
Thus, we find:
\[ y = -7 \]
Now that we have \(y\), we can substitute it back into either of the original equations to find \(x\). Let's use Equation 2 for this purpose:
\[ 5x - 3(-7) = 1 \]
This simplifies to:
\[ 5x + 21 = 1 \]
Subtracting 21 from both sides gives us:
\[ 5x = 1 - 21 \] \[ 5x = -20 \]
Dividing both sides by 5:
\[ x = -4 \]
Thus, the solution to the system of equations is:
\[ \boxed{(x, y) = (-4, -7)} \]