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if df = 9x - 39, find the value of ef. 47 3x + 10 d e f

Question

if df = 9x - 39, find the value of ef. 47 3x + 10 d e f

Explanation:

Step1: Set up equation based on segment - addition postulate

Since $DF=DE + EF$, we have $9x - 39=47+(3x + 10)$.

Step2: Simplify the right - hand side of the equation

$9x - 39=47 + 3x+10$, so $9x - 39=3x + 57$.

Step3: Solve for $x$

Subtract $3x$ from both sides: $9x-3x - 39=3x-3x + 57$, which gives $6x-39 = 57$. Then add 39 to both sides: $6x-39 + 39=57 + 39$, so $6x=96$. Divide both sides by 6: $x=\frac{96}{6}=16$.

Step4: Find the value of $EF$

Substitute $x = 16$ into the expression for $EF$: $EF=3x + 10$. So $EF=3\times16+10=48 + 10=61$.

Answer:

61