Given that the equation `(3x + 2)k + 5 = 12x + 7` m has infinte solutions, what is the value of (k+ m) ?

Answer

To solve the equation \((3x + 2)k + 5 = 12x + 7m\) for the values of \(k\) and \(m\) such that the equation has infinite solutions, we need to ensure that the coefficients of \(x\) and the constant terms on both sides of the equation are equal.

### Step-by-step Solution:

1. **Expand the Left Side**:
Start by expanding the left-hand side of the equation:
\[ (3x + 2)k + 5 = 3kx + 2k + 5 \]
So, the equation becomes:
\[ 3kx + (2k + 5) = 12x + 7m \]

2. **Compare Coefficients**:
For the equation to have infinite solutions, the coefficients of \(x\) on both sides must be equal, and the constant terms must also be equal. Therefore, we can set up the following equations:
- Coefficient of \(x\):
\[ 3k = 12 \]
- Constant terms:
\[ 2k + 5 = 7m \]

3. **Solve for \(k\)**:
From the first equation \(3k = 12\), we can solve for \(k\):
\[ k = \frac{12}{3} = 4 \]

4. **Substitute \(k\) into the Constant Equation**:
Now substitute \(k = 4\) into the second equation \(2k + 5 = 7m\):
\[ 2(4) + 5 = 7m \]
Simplifying this gives:
\[ 8 + 5 = 7m \implies 13 = 7m \]

5. **Solve for \(m\)**:
Now, solve for \(m\):
\[ m = \frac{13}{7} \]

6. **Calculate \(k + m\)**:
Now that we have both \(k\) and \(m\), we can find \(k + m\):
\[ k + m = 4 + \frac{13}{7} \]
To add these, convert \(4\) into a fraction with a denominator of \(7\):
\[ 4 = \frac{28}{7} \]
So,
\[ k + m = \frac{28}{7} + \frac{13}{7} = \frac{41}{7} \]

### Final Answer:
The value of \(k + m\) is:
\[ \frac{41}{7} \]