Let's solve the system of linear equations using the substitution method:

Given the system:

1. $\frac{1}{2}m + \frac{1}{3}n = 3$
2. $\frac{1}{4}m - \frac{2}{3}n = -1$

Step 1: Solve one equation for one variable

Let's start with the first equation and solve for $m$ in terms of $n$:

$\frac{1}{2}m + \frac{1}{3}n = 3$

Subtract $\frac{1}{3}n$ from both sides:

$\frac{1}{2}m = 3 - \frac{1}{3}n$

Now, multiply both sides of the equation by 2 to isolate $m$:

$m = 2\left(3 - \frac{1}{3}n\right)$

Distribute the 2:

$m = 6 - \frac{2}{3}n$

Step 2: Substitute this expression for $m$ into the second equation

Now that we have $m = 6 - \frac{2}{3}n$, we substitute this into the second equation:

$\frac{1}{4}m - \frac{2}{3}n = -1$

Substitute $m = 6 - \frac{2}{3}n$:

$\frac{1}{4}(6 - \frac{2}{3}n) - \frac{2}{3}n = -1$

Now, distribute $\frac{1}{4}$ on the first term:

$\frac{1}{4}(6) - \frac{1}{4} \times \frac{2}{3}n - \frac{2}{3}n = -1$

$\frac{3}{2} - \frac{1}{6}n - \frac{2}{3}n = -1$

Step 3: Combine like terms

To combine the $n$-terms, we need a common denominator. The least common denominator between 6 and 3 is 6, so rewrite $\frac{2}{3}n$ as $\frac{4}{6}n$:

$\frac{3}{2} - \frac{1}{6}n - \frac{4}{6}n = -1$

Combine the $n$-terms:

$\frac{3}{2} - \frac{5}{6}n = -1$

Step 4: Solve for $n$

First, subtract $\frac{3}{2}$ from both sides:

$-\frac{5}{6}n = -1 - \frac{3}{2}$

Convert $-1$ to $-\frac{2}{2}$:

$-\frac{5}{6}n = -\frac{2}{2} - \frac{3}{2} = -\frac{5}{2}$

Now, multiply both sides by $-\frac{6}{5}$ to isolate $n$:

$n = \left(-\frac{5}{2}\right) \times \left(-\frac{6}{5}\right)$

$n = 3$

Step 5: Substitute $n = 3$ back into the equation for $m$

Now that we know $n = 3$, substitute this value into the expression for $m$:

$m = 6 - \frac{2}{3}(3)$

$m = 6 - 2$

$m = 4$

Final Answer:

The solution to the system of equations is:

$m = 4, \quad n = 3$

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Would you like a more detailed explanation of any step or help with another topic?

Here are 5 related questions for further exploration:

1. How can you solve a system of equations by elimination?
2. What if the system of equations had no solution or infinite solutions?
3. How does graphing relate to solving systems of linear equations?
4. How do you use substitution with a fractional coefficient?
5. What are some common mistakes to avoid when solving by substitution?

Tip: When working with fractions, always try to clear the fractions early by multiplying both sides of an equation by the least common denominator (LCD).