To solve the system of equations:
- \( -10x + 2y = 6 \)
- \( 10x - 3y = 1 \)
We can use either substitution or elimination. Here, I'll use the elimination method.
First, let's multiply the first equation by \( 1 \) (which doesn't change it) and the second equation by \( 1 \):
- \( -10x + 2y = 6 \) (Equation 1)
- \( 10x - 3y = 1 \) (Equation 2)
Now, if we add Equation 1 and Equation 2, we eliminate the \( x \):
\[ (-10x + 2y) + (10x - 3y) = 6 + 1 \] \[ 0x - y = 7 \] \[ -y = 7 \] \[ y = -7 \]
Now that we have \( y \), we can substitute \( y = -7 \) back into one of the original equations to solve for \( x \). Let's substitute into Equation 1:
\[ -10x + 2(-7) = 6 \] \[ -10x - 14 = 6 \] \[ -10x = 6 + 14 \] \[ -10x = 20 \] \[ x = -2 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-2, -7) \]
So the answer is \((-2, -7)\).