Let's solve the given simultaneous linear equations step by step to find the values of $m$ and $n$.

Step 1: Write down the equations

The equations are:

$2n + \frac{1}{2} m - 7 = 0 \quad \text{(Equation 1)}$

$3m + \frac{1}{2} n = 5 \quad \text{(Equation 2)}$

Step 2: Solve Equation 1 for $m$

First, rearrange Equation 1 to express $m$ in terms of $n$:

$2n + \frac{1}{2} m = 7$

Multiply the entire equation by 2 to eliminate the fraction:

$4n + m = 14$

So,

$m = 14 - 4n \quad \text{(Equation 3)}$

Step 3: Substitute $m$ into Equation 2

Now substitute Equation 3 into Equation 2:

$3(14 - 4n) + \frac{1}{2} n = 5$

Expand and simplify the equation:

$42 - 12n + \frac{1}{2} n = 5$

Combine the terms involving $n$:

$42 - 11.5n = 5$

Isolate $n$ by subtracting 42 from both sides:

$-11.5n = 5 - 42$

$-11.5n = -37$

Divide both sides by -11.5 to solve for $n$:

$n = \frac{-37}{-11.5} = \frac{37}{11.5} = \frac{74}{23} \approx 3.2174$

Step 4: Find $m$

Now substitute $n = \frac{74}{23}$ back into Equation 3 to find $m$:

$m = 14 - 4\left(\frac{74}{23}\right)$

$m = 14 - \frac{296}{23} = \frac{322 - 296}{23} = \frac{26}{23} \approx 1.1304$

Step 5: Find $x$

Now use the values of $m$ and $n$ to find $x$ from the equation:

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

Substitute $m = \frac{26}{23}$ and $n = \frac{74}{23}$:

$x = \frac{1}{2} \times \frac{26}{23} + \frac{1}{3} \times \frac{74}{23}$

Simplify:

$x = \frac{13}{23} + \frac{74}{69} = \frac{13 \times 3 + 74}{69} = \frac{39 + 74}{69} = \frac{113}{69} \approx 1.6377$

Final Answer

The value of $x$ is approximately $\boxed{1.6377}$.

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Do you want more details or have any questions? Here are some related questions to consider:

1. How do you solve simultaneous linear equations using substitution?
2. What is the importance of eliminating fractions in algebraic equations?
3. How do you verify the solution of simultaneous equations?
4. What are other methods for solving simultaneous linear equations?
5. How does the value of $x$ change with different values of $m$ and $n$?

Tip: When solving simultaneous equations, substitution is often easier when one equation is already solved for a variable.