To solve for \( k \) in terms of \( m \) (or vice versa), let’s start by clearing the fractions. We have:

\[
\frac{2}{5}(k + 2) = \frac{3}{4}(m + 5)
\]

1. **Eliminate the Fractions**: Multiply both sides by the least common multiple of 5 and 4, which is 20.

\[
20 \times \frac{2}{5}(k + 2) = 20 \times \frac{3}{4}(m + 5)
\]

This simplifies to:

\[
4 \cdot 2(k + 2) = 5 \cdot 3(m + 5)
\]

or

\[
8(k + 2) = 15(m + 5)
\]

2. **Expand Both Sides**:

\[
8k + 16 = 15m + 75
\]

3. **Isolate \( k \)**: Subtract 16 from both sides:

\[
8k = 15m + 59
\]

Then, divide by 8:

\[
k = \frac{15m + 59}{8}
\]

---

The solution is:

\[
k = \frac{15m + 59}{8}
\]

Would you like further details or have any questions?

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### Related Questions

1. How would the solution change if the equation were \( \frac{2}{5}(k + 2) = \frac{3}{4}(m + 6) \)?
2. Can you solve for \( m \) in terms of \( k \) using the same equation?
3. What happens if you substitute a specific value for \( m \) in the equation?
4. How can you verify that this solution is correct?
5. What is the importance of clearing fractions in equations like this?

### Tip
When dealing with fractions in equations, clearing them by using the least common multiple often simplifies the problem significantly.

Discover the steps to solve the algebraic equation 2/5(k+2) = 3/4(m+5) for k in terms of m. This detailed solution involves clearing fractions, distributing, and isolating variables, perfect for students in Grades 8-10 looking to master linear equations with fractions.

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