We are given a system of two linear equations:

1. \( x + 2y = 6 \)
2. \( 2x - 3y = 12 \)

To solve this system, we can use either the substitution or elimination method. Let's proceed with the **elimination method** to eliminate one variable and solve for the other.

### Step 1: Eliminate \( x \)
To eliminate \( x \), we can multiply the first equation by 2, so that the coefficients of \( x \) in both equations are the same:

\[
2(x + 2y) = 2(6) \implies 2x + 4y = 12
\]

Now the system becomes:

\[
\begin{aligned}
1. & \quad 2x + 4y = 12 \\
2. & \quad 2x - 3y = 12
\end{aligned}
\]

### Step 2: Subtract the equations
Now subtract the second equation from the first to eliminate \( x \):

\[
(2x + 4y) - (2x - 3y) = 12 - 12
\]

Simplifying:

\[
2x + 4y - 2x + 3y = 0 \implies 7y = 0
\]

### Step 3: Solve for \( y \)
\[
y = 0
\]

### Step 4: Solve for \( x \)
Substitute \( y = 0 \) into the first equation:

\[
x + 2(0) = 6 \implies x = 6
\]

### Solution:
The solution to the system of equations is \( x = 6 \) and \( y = 0 \).

Would you like further details or have any questions?

Here are some related questions you might explore:

1. How do you solve systems of equations using substitution?
2. What are the benefits of using the elimination method over substitution?
3. How can we graphically represent a system of linear equations?
4. What happens if a system of equations has no solution?
5. How can we identify a system with infinitely many solutions?

**Tip:** Always check your solution by plugging the values back into both original equations!

Discover how to solve the system of linear equations x + 2y = 6 and 2x - 3y = 12 using the elimination method. This guide explains key algebraic concepts, such as elimination and substitution methods, ideal for students in Grades 8-10.

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