To solve the system of equations using the elimination method, we start with the given equations:
- \( 2x - 3y = 14 \) (Equation 1)
- \( 4x + 3y = 10 \) (Equation 2)
First, we can eliminate \(y\) by adding the two equations together. To do this, we can multiply Equation 1 by 1 (no change necessary) and Equation 2 by 1 (again, no change):
- \( 2x - 3y = 14 \)
- \( 4x + 3y = 10 \)
Now, we add the two equations:
\[ (2x - 3y) + (4x + 3y) = 14 + 10 \]
This simplifies to:
\[ 6x = 24 \]
Next, we solve for \(x\):
\[ x = \frac{24}{6} = 4 \]
Now that we have \(x\), we will substitute this value back into one of the original equations to solve for \(y\). We'll use Equation 1:
\[ 2(4) - 3y = 14 \]
This simplifies to:
\[ 8 - 3y = 14 \]
Now, we can isolate \(y\):
\[ -3y = 14 - 8 \] \[ -3y = 6 \] \[ y = \frac{6}{-3} = -2 \]
Thus, the solution to the system of equations is:
\[ x = 4, \quad y = -2 \]
Final answer:
\((4, -2)\)