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

Question

  1. if df = 9x - 39, find ef.

d
47
e
3x + 10
f

Explanation:

Step1: Set up equation

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

Step2: Simplify the right - hand side

$9x-39=47 + 3x+10$, which simplifies to $9x-39=3x + 57$.

Step3: Move terms with $x$ to one side

Subtract $3x$ from both sides: $9x-3x-39=3x-3x + 57$, resulting in $6x-39 = 57$.

Step4: Isolate the term with $x$

Add 39 to both sides: $6x-39 + 39=57+39$, so $6x=96$.

Step5: Solve for $x$

Divide both sides by 6: $x=\frac{96}{6}=16$.

Step6: Find $EF$

Substitute $x = 16$ into the expression for $EF$. $EF=3x + 10=3\times16+10=48 + 10=58$.

Answer:

58