E is between DF. if De = x +1, Ef =x +7 and Df = 4x-20 solve for x 

Given that \( E \) is between \( D \) and \( F \), we can deduce that the length of segment \( DF \) is the sum of the lengths of segments \( DE \) and \( EF \).

### Information provided:
- \( DE = x + 1 \)
- \( EF = x + 7 \)
- \( DF = 4x - 20 \)

Since \( DF = DE + EF \), we can set up the equation:
\[
DE + EF = DF
\]
Substitute the given values:
\[
(x + 1) + (x + 7) = 4x - 20
\]

Simplify the left-hand side:
\[
2x + 8 = 4x - 20
\]

Now, solve for \( x \):
1. Subtract \( 2x \) from both sides:
   \[
   8 = 2x - 20
   \]
2. Add 20 to both sides:
   \[
   28 = 2x
   \]
3. Divide both sides by 2:
   \[
   x = 14
   \]

### Final Answer:
\( x = 14 \).

Would you like a detailed breakdown of each step?

Here are 5 related questions to deepen understanding:
1. What is the length of \( DE \) when \( x = 14 \)?
2. What is the length of \( EF \) when \( x = 14 \)?
3. What is the total length of \( DF \) when \( x = 14 \)?
4. How can you verify if the segment addition is correct for this problem?
5. What happens if \( E \) is not exactly between \( D \) and \( F \)?

**Tip:** Always check if the segment addition formula applies when points are collinear.

E is between DF. If DE = x +1, EF = x +7, and DF = 4x - 20, solve for x.

Learn how to solve for x in the equation DE = x + 1, EF = x + 7, and DF = 4x - 20 using the segment addition postulate. This step-by-step solution covers essential algebraic concepts and is suitable for students in Grades 8-10.

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