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The problem you've shared involves solving simultaneous linear equations. The given system of equations is:

1. \( 2(x - 12) + 3y = 0 \)
2. \( \frac{1}{2}x - 2y = 6 \)

Let's solve these step-by-step.

### Equation 1:
Simplify the first equation:

\[
2(x - 12) + 3y = 0
\]

Expand the parentheses:

\[
2x - 24 + 3y = 0
\]

Rearrange the equation:

\[
2x + 3y = 24 \quad \text{(Equation A)}
\]

### Equation 2:
The second equation is:

\[
\frac{1}{2}x - 2y = 6
\]

Multiply the whole equation by 2 to eliminate the fraction:

\[
x - 4y = 12 \quad \text{(Equation B)}
\]

### Solving the system:
Now, solve these two equations simultaneously.

1. Equation A: \( 2x + 3y = 24 \)
2. Equation B: \( x - 4y = 12 \)

From Equation B, solve for \( x \):

\[
x = 4y + 12
\]

Substitute \( x = 4y + 12 \) into Equation A:

\[
2(4y + 12) + 3y = 24
\]

Expand and simplify:

\[
8y + 24 + 3y = 24
\]
\[
11y + 24 = 24
\]
\[
11y = 0
\]
\[
y = 0
\]

Now substitute \( y = 0 \) into \( x = 4y + 12 \):

\[
x = 4(0) + 12 = 12
\]

### Solution:
The solution to the system of equations is:

\[
x = 12, \quad y = 0
\]

Let me know if you'd like further details or explanations.

### Related Questions:
1. What happens if both equations were multiplied by constants before solving?
2. How would the solution change if \( y \neq 0 \) in any part of the process?
3. Can the system be solved graphically, and how does the graph of each equation look?
4. How can you verify the solution by substitution?
5. What is the determinant of the matrix formed by these two linear equations?

#### Tip:
Always check your solution by substituting the values of \( x \) and \( y \) back into both original equations.

Solve the following simultaneous linear equations:
1. 2(x - 12) + 3y = 0
2. (1/2)x - 2y = 6

This page provides a detailed solution to the simultaneous linear equations 2(x - 12) + 3y = 0 and (1/2)x - 2y = 6. It covers algebraic methods of solving such systems, including substitution and elimination. Suitable for students in Grades 8-10.

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