To solve the system of equations using the elimination method, we start with the two equations:
- \( 2x - 3y = 14 \) (Equation 1)
- \( 4x + 3y = 10 \) (Equation 2)
Step 1: Align the equations
\[ \begin{align*} (1) & \quad 2x - 3y = 14 \ (2) & \quad 4x + 3y = 10 \ \end{align*} \]
Step 2: Eliminate \(y\)
To eliminate \(y\), we can add the two equations directly, but first, we need to make the coefficients of \(y\) equal. In this case, we can multiply Equation 1 by 1 (keeping it the same) and Equation 2 by 1 (also keeping it the same) since \(3y\) in Equation 2 and \(-3y\) in Equation 1 are already opposites.
Now, we add the two equations:
\[ (2x - 3y) + (4x + 3y) = 14 + 10 \]
This simplifies to:
\[ 2x + 4x - 3y + 3y = 24 \]
Which results in:
\[ 6x = 24 \]
Step 3: Solve for \(x\)
Now we can divide both sides by 6:
\[ x = \frac{24}{6} = 4 \]
Step 4: Substitute \(x\) back into one of the original equations
Now that we have \(x = 4\), we can substitute it back into either original equation to find \(y\). Let's use Equation 1:
\[ 2(4) - 3y = 14 \]
This simplifies to:
\[ 8 - 3y = 14 \]
Step 5: Solve for \(y\)
Subtract 8 from both sides:
\[ -3y = 14 - 8 \] \[ -3y = 6 \]
Now divide both sides by -3:
\[ y = \frac{6}{-3} = -2 \]
Conclusion
The solution to the system of equations is:
\[ (x, y) = (4, -2) \]
Thus, the final answer is:
\[ \boxed{(4, -2)} \]
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