((1/2)m)+((2/3)n)-6=0 and ((-3/2)m)+((1/2)n)+3=0

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.

This page provides a step-by-step solution for solving simultaneous linear equations involving fractions, specifically the equations ((1/2)m) + ((2/3)n) - 6 = 0 and ((-3/2)m) + ((1/2)n) + 3 = 0. Learn key algebraic methods like the elimination and substitution techniques, suitable for students in Grades 9-11.

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