We are given the system of equations:
- \(-x + 6y = 3\)
- \(-4x - 2y = -14\)
We can solve this system using the substitution or elimination method. Let's use the substitution method.
First, we can solve the first equation for \(x\):
\[ -x + 6y = 3 \implies x = 6y - 3 \]
Now, we can substitute this expression for \(x\) into the second equation:
\[ -4(6y - 3) - 2y = -14 \]
Expanding this gives:
\[ -24y + 12 - 2y = -14 \]
Combining like terms:
\[ -26y + 12 = -14 \]
Now, isolate \(y\):
\[ -26y = -14 - 12 \] \[ -26y = -26 \] \[ y = 1 \]
Now that we have \(y\), we can substitute back to find \(x\):
\[ x = 6(1) - 3 = 6 - 3 = 3 \]
Thus, the solution to the system of equations is:
\[ \boxed{(3, 1)} \]
To confirm, we can plug \(x\) and \(y\) back into both original equations.
For the first equation:
\[ -x + 6y = 3 \implies -3 + 6(1) = 3 \implies -3 + 6 = 3 \quad \text{(True)} \]
For the second equation:
\[ -4x - 2y = -14 \implies -4(3) - 2(1) = -14 \implies -12 - 2 = -14 \quad \text{(True)} \]
Both equations are satisfied, confirming that the solution \( (3, 1) \) is correct.
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