The solution of the equations `m/3+n/4=12` and `m/2-n/3=1` is

Options

A. `m=8,n=6`
B. `m=18,n=24`
C. `m=24,n=18`
D. `m=6,n=8`

Answer

To solve the simultaneous linear equations:

1. **Equation 1:** \( \frac{m}{3} + \frac{n}{4} = 12 \)
2. **Equation 2:** \( \frac{m}{2} - \frac{n}{3} = 1 \)

We will solve these equations step-by-step.

### Step 1: Eliminate the fractions in both equations.

**For Equation 1:**
Multiply through by 12 (the least common multiple of 3 and 4):

\[ 12 \left(\frac{m}{3}\right) + 12 \left(\frac{n}{4}\right) = 12 \times 12 \]

This simplifies to:

\[ 4m + 3n = 144 \quad \text{(Equation 3)} \]

**For Equation 2:**
Multiply through by 6 (the least common multiple of 2 and 3):

\[ 6 \left(\frac{m}{2}\right) - 6 \left(\frac{n}{3}\right) = 6 \times 1 \]

This simplifies to:

\[ 3m - 2n = 6 \quad \text{(Equation 4)} \]

### Step 2: Solve the new system of equations.

Now we have the following system of equations:

1. \( 4m + 3n = 144 \) (Equation 3)
2. \( 3m - 2n = 6 \) (Equation 4)

### Step 3: Eliminate \( n \).

To eliminate \( n \), we can multiply Equation 3 by 2 and Equation 4 by 3 so that the coefficients of \( n \) will be equal:

\[ 2(4m + 3n) = 2(144) \implies 8m + 6n = 288 \quad \text{(Equation 5)} \]

\[ 3(3m - 2n) = 3(6) \implies 9m - 6n = 18 \quad \text{(Equation 6)} \]

### Step 4: Add Equations 5 and 6.

Now, we can add Equation 5 and Equation 6:

\[ (8m + 6n) + (9m - 6n) = 288 + 18 \]

This simplifies to:

\[ 17m = 306 \]

### Step 5: Solve for \( m \).

Now, divide both sides by 17:

\[ m = \frac{306}{17} = 18 \]

### Step 6: Substitute \( m \) back into one of the original equations to find \( n \).

We can substitute \( m = 18 \) into Equation 2:

\[ \frac{m}{2} - \frac{n}{3} = 1 \]

Substituting \( m \):

\[ \frac{18}{2} - \frac{n}{3} = 1 \]

This simplifies to:

\[ 9 - \frac{n}{3} = 1 \]

### Step 7: Solve for \( n \).

Rearranging gives:

\[ 9 - 1 = \frac{n}{3} \]

\[ 8 = \frac{n}{3} \]

Now, multiply both sides by 3:

\[ n = 24 \]

### Final Solution:

Thus, the solution to the simultaneous equations is:

\[ m = 18, \quad n = 24 \]