We are given the system of equations:

$1) \ 2x + 3y = 6$ $2) \ 4x - 3y = 12$

Step 1: Add the two equations to eliminate $y$

We can add the two equations because the $y$-terms will cancel out due to opposite signs:

$(2x + 3y) + (4x - 3y) = 6 + 12$

Simplifying:

$6x = 18$

Step 2: Solve for $x$

Now, divide both sides by 6:

$x = \frac{18}{6} = 3$

Step 3: Substitute $x = 3$ into one of the original equations

We can substitute $x = 3$ into the first equation:

$2(3) + 3y = 6$

Simplify:

$6 + 3y = 6$

Step 4: Solve for $y$

Now subtract 6 from both sides:

$3y = 0$

Finally, divide by 3:

$y = 0$

Final Solution:

The solution to the system of equations is: $x = 3, \ y = 0$

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Would you like any further details or explanations? Here are 5 related questions you could explore:

1. How would you solve the system if the coefficients of $y$ didn't cancel out?
2. Can this system be solved using substitution? How?
3. How can graphing the two equations help verify the solution?
4. What happens if we multiply both equations by different constants and then solve?
5. How can you check if the solution $(x, y) = (3, 0)$ satisfies both original equations?

Tip: In systems of equations, always look for ways to eliminate one variable using addition or substitution.