We are given two simultaneous linear equations:

1. $\frac{1}{2}m + \frac{2}{3}n - 6 = 0$
2. $-\frac{3}{2}m + \frac{1}{2}n + 3 = 0$

To solve for $m$ and $n$, let's follow these steps:

Step 1: Eliminate fractions by multiplying the equations by suitable numbers.

First equation:

$\frac{1}{2}m + \frac{2}{3}n - 6 = 0$ Multiply the entire equation by 6 (LCM of 2 and 3) to clear the fractions: $3m + 4n - 36 = 0$ $3m + 4n = 36 \quad \text{(Equation 3)}$

Second equation:

$-\frac{3}{2}m + \frac{1}{2}n + 3 = 0$ Multiply the entire equation by 2 to clear the fractions: $-3m + n + 6 = 0$ $-3m + n = -6 \quad \text{(Equation 4)}$

Step 2: Solve the system of linear equations.

We now have two equations:

1. $3m + 4n = 36$ (Equation 3)
2. $-3m + n = -6$ (Equation 4)

Add the two equations to eliminate $m$:

$(3m - 3m) + (4n + n) = 36 - 6$ $5n = 30$ $n = 6$

Step 3: Substitute $n = 6$ into one of the original equations to solve for $m$.

Substitute into Equation 4: $-3m + 6 = -6$ $-3m = -12$ $m = 4$

Final Answer:

The solution is: $m = 4, \quad n = 6$

Would you like more details or have any questions?

Here are five related questions for further practice:

1. Solve the system of equations $2x + 3y = 10$ and $x - y = 2$.
2. Find the value of $x$ and $y$ for the system $x + 4y = 12$ and $3x - 2y = 8$.
3. If $a = 3x + 2y$ and $b = 4x - y$, find $x$ and $y$ when $a = 10$ and $b = 6$.
4. Solve for $p$ and $q$ in the system: $5p - 3q = 7$ and $2p + q = 4$.
5. Solve $x - 5y = 20$ and $3x + 4y = 12$.

Tip: Always multiply the entire equation by the least common multiple (LCM) of the denominators to eliminate fractions before solving.