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This problem is solved using the **triangle inequality theorem**, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Here, the sides are \( 7 \), \( 24 \), and \( x \). The inequalities we need to solve are:

1. \( 7 + 24 > x \)
2. \( 7 + x > 24 \)
3. \( 24 + x > 7 \)

### Solving Each Inequality:
1. \( 7 + 24 > x \)  
   \( 31 > x \)  
   \( x < 31 \)

2. \( 7 + x > 24 \)  
   \( x > 24 - 7 \)  
   \( x > 17 \)

3. \( 24 + x > 7 \)  
   This inequality is always true for positive \( x \) because \( x \geq 17 \) (from inequality 2).

### Final Range for \( x \):
The possible values of \( x \) satisfy both \( x > 17 \) and \( x < 31 \), so:  
\[
17 < x < 31
\]

### Answer:
The range of possible values for \( x \) is \( \boxed{17 < x < 31} \).

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Would you like more details or further clarification?  

Here are some related questions to expand on this topic:
1. What is the triangle inequality theorem, and why is it necessary for triangles?
2. How would the range change if one of the other sides was variable?
3. Can a triangle exist if \( x = 17 \) or \( x = 31 \)? Why not?
4. What happens if all three sides are the same length in this situation?
5. How would you use the triangle inequality to check if three specific numbers can form a triangle?

**Tip:** Always check all three triangle inequalities when solving problems involving triangle side lengths!

A triangle has sides of lengths 7, 24, and x. What is the range of possible values for x?

Discover how to find the range of possible values for x in a triangle with sides 7, 24, and x using the Triangle Inequality Theorem. Detailed step-by-step solution provided. Ideal for Grades 7-9 math learners.

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