To solve the system of equations using the elimination method, we need to eliminate one variable by combining the two equations. The given equations are:
- \( 4x + 6y = 30 \) (Equation 1)
- \( 2x - 6y = 6 \) (Equation 2)
Let's first write down the equations:
Step 1: Align the equations.
Equation 1: \( 4x + 6y = 30 \)
Equation 2: \( 2x - 6y = 6 \)
Step 2: Prepare to eliminate one of the variables.
To eliminate \( y \), we can add the two equations directly since the coefficients of \( y \) are opposites (\(+6y\) and \(-6y\)).
Step 3: Add the two equations.
\[
(4x + 6y) + (2x - 6y) = 30 + 6
\]
Combining like terms gives us:
\[
4x + 2x + 6y - 6y = 30 + 6
\]
\[
6x = 36
\]
Step 4: Solve for \( x \).
Now we divide both sides by 6:
\[
x = \frac{36}{6} = 6
\]
Step 5: Substitute \( x \) back into one of the original equations to solve for \( y \).
We can substitute \( x = 6 \) into either Equation 1 or Equation 2. Let's use Equation 1:
\[
4x + 6y = 30
\]
Substituting \( x = 6 \):
\[
4(6) + 6y = 30
\]
\[
24 + 6y = 30
\]
Step 6: Solve for \( y \).
Now, subtract 24 from both sides:
\[
6y = 30 - 24
\]
\[
6y = 6
\]
Now divide both sides by 6:
\[
y = 1
\]
Step 7: Present the solution.
We have found \( x = 6 \) and \( y = 1 \).
The solution to the system of equations is:
\[
(x, y) = (6, 1)
\]
Final Step: Verify the solution.
Substituting \( x = 6 \) and \( y = 1 \) back into the original equations:
-
For Equation 1: \[ 4(6) + 6(1) = 24 + 6 = 30 \quad \text{(True)} \]
-
For Equation 2: \[ 2(6) - 6(1) = 12 - 6 = 6 \quad \text{(True)} \]
The solution satisfies both equations. Thus, the solution is: \[ \boxed{(6, 1)} \]