If DE = x − 4, EF = 12, and DF = 2x − 11, what is DE?

To solve for \( DE \), we need to use the fact that \( D \), \( E \), and \( F \) are on the same line, so the sum of the lengths \( DE \) and \( EF \) must equal the total length \( DF \).

We are given:
- \( DE = x - 4 \)
- \( EF = 12 \)
- \( DF = 2x - 11 \)

From this, we know:
\[
DE + EF = DF
\]
Substituting the known values into the equation:
\[
(x - 4) + 12 = 2x - 11
\]

Now, let's solve for \( x \):

\[
x - 4 + 12 = 2x - 11
\]
\[
x + 8 = 2x - 11
\]
Subtract \( x \) from both sides:
\[
8 = x - 11
\]
Add 11 to both sides:
\[
x = 19
\]

Now that we know \( x = 19 \), substitute this back into the expression for \( DE \):

\[
DE = x - 4 = 19 - 4 = 15
\]

Thus, \( DE = 15 \).

Would you like further details or have any other questions?

Here are 5 related questions to expand your understanding:
1. What is the length of \( DF \) when \( x = 19 \)?
2. How would the solution change if \( EF \) was doubled?
3. What if \( DF \) was expressed as \( 3x - 15 \), how would the equation adjust?
4. If \( EF = x \), how would that affect the final solution for \( DE \)?
5. Can you verify the solution using a different method, like graphing?

**Tip**: When dealing with line segments, always check if points are collinear to ensure that the segment lengths add up properly.

Learn how to solve for DE given DE = x − 4, EF = 12, and DF = 2x − 11 using linear equations and the segment addition postulate. This problem explores key algebraic and geometric concepts, suitable for students in Grades 8-10.

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