The problem presents a system of linear equations:

$\frac{1}{2}x - y = -2$ $x + \frac{1}{3}y = 3$

We will solve it using the method of elimination.

Step 1: Eliminate one variable

First, we rewrite the equations in a standard form to simplify elimination. Multiply through by the denominators to eliminate fractions.

1. Multiply the first equation by 2 to clear the fraction: $x - 2y = -4$

2. Multiply the second equation by 3 to clear the fraction: $3x + y = 9$

The system now becomes: $x - 2y = -4$ $3x + y = 9$

Step 2: Eliminate $y$

To eliminate $y$, multiply the first equation by 1 (no change) and the second equation by 2: $x - 2y = -4$ $6x + 2y = 18$

Add the equations together: $(x - 2y) + (6x + 2y) = -4 + 18$ $7x = 14$ $x = 2$

Step 3: Solve for $y$

Substitute $x = 2$ into one of the original equations, e.g., $x - 2y = -4$: $2 - 2y = -4$ $-2y = -6$ $y = 3$

Final Answer:

The solution to the system is: $(x, y) = (2, 3)$

Let me know if you would like further clarification or details! Here are some related questions for further practice:

1. How would the solution change if one equation were altered slightly?
2. What is the geometric interpretation of solving a system of linear equations?
3. How does the elimination method compare to the substitution method?
4. Can you verify the solution graphically?
5. What happens if the system is inconsistent or dependent?

Tip: Always check your solution by substituting the values of $x$ and $y$ back into the original equations!