Question

1. given: de = x + 3

ef = 7x + 1
df = 20
prove: x = 2
statements

1. de=x + 3, ef = 7x+1, df = 20
2. de + ef = df
3. x + 3+7x + 1 = df

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Explanation:

Step1: Apply segment - addition postulate

Since point E lies on line segment DF, we have $DE + EF=DF$.

Step2: Substitute given expressions

Substitute $DE=x + 3$, $EF = 7x+1$, and $DF = 20$ into the equation $DE + EF=DF$. We get $(x + 3)+(7x+1)=20$.

Step3: Simplify the left - hand side

Combine like terms: $x+7x+3 + 1=20$, which simplifies to $8x+4 = 20$.

Step4: Isolate the variable term

Subtract 4 from both sides of the equation: $8x+4-4=20 - 4$, resulting in $8x=16$.

Step5: Solve for x

Divide both sides by 8: $\frac{8x}{8}=\frac{16}{8}$, so $x = 2$.

Answer:

$x = 2$