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Let's go through the steps shown in your image to solve for \( x \), given that \( KM \parallel JN \) and using the side-splitter theorem.

1. **By the side-splitter theorem**: If \( JK \parallel KL \), then the ratio of the segments formed on the sides of the triangle by the parallel line will be equal. This gives us the proportion:
   \[
   \frac{x - 5}{x} = \frac{x - 3}{x + 4}
   \]

2. **Cross-multiply** to clear the fractions:
   \[
   (x - 5)(x + 4) = (x)(x - 3)
   \]

3. **Distribute** both sides:
   - Left side: \( (x - 5)(x + 4) = x(x) + x(4) - 5(x) - 5(4) = x^2 + 4x - 5x - 20 = x^2 - x - 20 \)
   - Right side: \( x(x - 3) = x^2 - 3x \)

   Now we have the equation:
   \[
   x^2 - x - 20 = x^2 - 3x
   \]

4. **Simplify** the equation:
   Subtract \( x^2 \) from both sides to eliminate the quadratic term:
   \[
   -x - 20 = -3x
   \]

5. **Solve for \( x \)**:
   Add \( 3x \) to both sides:
   \[
   2x - 20 = 0
   \]
   Add 20 to both sides:
   \[
   2x = 20
   \]
   Divide by 2:
   \[
   x = 10
   \]

So, the value of \( x \) that makes \( KM \parallel JN \) is \( \boxed{10} \).

Would you like more details or have further questions?

Here are five related questions to deepen your understanding:
1. What does the side-splitter theorem state?
2. How would the solution change if the proportion involved different side lengths?
3. Can the same theorem be used with a quadrilateral, and how?
4. How does cross-multiplication help in solving proportions?
5. What if we had different parallel lines inside the triangle? How would that affect the setup?

**Tip**: Always ensure that proportions derived from the side-splitter theorem are correctly set up based on the sides divided by the parallel line.

What value of x would make KM parallel to JN?

Learn how to solve for x to make KM parallel to JN using the side-splitter theorem. This involves solving proportions and using algebraic techniques like cross-multiplication and simplification. Suitable for students in grades 8-10.

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