To solve the system of equations using the elimination method, we start with the two equations:
- \( x + y = 4 \)
- \( 2x - 3y = 18 \)
First, we can eliminate one variable by manipulating the equations. We can solve for \( y \) in the first equation:
\[ y = 4 - x \]
Now we can substitute \( y \) in the second equation. Substitute \( y \) into equation 2:
\[ 2x - 3(4 - x) = 18 \]
Now, distribute \( -3 \):
\[ 2x - 12 + 3x = 18 \]
Combine like terms:
\[ 5x - 12 = 18 \]
Next, add 12 to both sides:
\[ 5x = 30 \]
Now divide by 5:
\[ x = 6 \]
Now that we have \( x \), we can substitute \( x = 6 \) back into the equation we found for \( y \):
\[ y = 4 - x = 4 - 6 = -2 \]
So, the solution to the system is:
\[ (x, y) = (6, -2) \]
To verify, we can plug these values back into the original equations.
- For \( x + y = 4 \):
\[ 6 + (-2) = 4 \quad \text{(True)} \]
- For \( 2x - 3y = 18 \):
\[ 2(6) - 3(-2) = 12 + 6 = 18 \quad \text{(True)} \]
Thus, the solution \( (6, -2) \) satisfies both equations, confirming our answer is correct. Therefore, the solution is:
\[ \boxed{(6, -2)} \]