To solve the system of equations using the elimination method, we start with the given equations:
- \( 2x + 3y = 12 \)
- \( 4x - 3y = 6 \)
We will eliminate one of the variables by adding or subtracting the equations. In this case, we can eliminate \( y \) by adding the two equations together.
Step 1: Align the equations
Here are our equations again: \[ 2x + 3y = 12 \quad \quad (1) \] \[ 4x - 3y = 6 \quad \quad (2) \]
Step 2: Add the equations
Adding both equations will eliminate \( y \):
\[ (2x + 3y) + (4x - 3y) = 12 + 6 \]
Combine like terms:
\[ (2x + 4x) + (3y - 3y) = 18 \] \[ 6x + 0y = 18 \] \[ 6x = 18 \]
Step 3: Solve for \( x \)
Now we solve for \( x \):
\[ x = \frac{18}{6} = 3 \]
Step 4: Substitute \( x \) back into one of the original equations
We will substitute \( x = 3 \) back into the first equation to find \( y \):
\[ 2x + 3y = 12 \] Substituting \( x \):
\[ 2(3) + 3y = 12 \] \[ 6 + 3y = 12 \]
Step 5: Solve for \( y \)
Now we isolate \( y \):
\[ 3y = 12 - 6 \] \[ 3y = 6 \] \[ y = \frac{6}{3} = 2 \]
Final result
Thus, the solution to the system of equations is: \[ x = 3, \quad y = 2 \]
Verification
Let's verify the solution by plugging \( x = 3 \) and \( y = 2 \) back into both original equations:
-
First equation: \[ 2(3) + 3(2) = 6 + 6 = 12 \quad \text{(true)} \]
-
Second equation: \[ 4(3) - 3(2) = 12 - 6 = 6 \quad \text{(true)} \]
Since both equations are satisfied, the solution is confirmed:
\[ \boxed{(3, 2)} \]