Math Problem Statement

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

Solution

Given that EE is between DD and FF, we can deduce that the length of segment DFDF is the sum of the lengths of segments DEDE and EFEF.

Information provided:

  • DE=x+1DE = x + 1
  • EF=x+7EF = x + 7
  • DF=4x20DF = 4x - 20

Since DF=DE+EFDF = DE + EF, we can set up the equation: DE+EF=DFDE + EF = DF Substitute the given values: (x+1)+(x+7)=4x20(x + 1) + (x + 7) = 4x - 20

Simplify the left-hand side: 2x+8=4x202x + 8 = 4x - 20

Now, solve for xx:

  1. Subtract 2x2x from both sides: 8=2x208 = 2x - 20
  2. Add 20 to both sides: 28=2x28 = 2x
  3. Divide both sides by 2: x=14x = 14

Final Answer:

x=14x = 14.

Would you like a detailed breakdown of each step?

Here are 5 related questions to deepen understanding:

  1. What is the length of DEDE when x=14x = 14?
  2. What is the length of EFEF when x=14x = 14?
  3. What is the total length of DFDF when x=14x = 14?
  4. How can you verify if the segment addition is correct for this problem?
  5. What happens if EE is not exactly between DD and FF?

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

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations
Segment Addition

Formulas

DE + EF = DF
2x + 8 = 4x - 20

Theorems

Segment Addition Postulate

Suitable Grade Level

Grades 8-10