Question

1. if df = 9x - 39, find ef.
2. if hj = 7x - 27, find the value of x.

Explanation:

Step1: Set up equation for 6

Since $HJ = HI+IJ$, we have $7x - 27=(3x - 5)+(x - 1)$.
$7x-27 = 3x-5+x - 1$.

Step2: Combine like - terms

$7x-27=4x - 6$.

Step3: Isolate x terms

Subtract $4x$ from both sides: $7x-4x-27=4x-4x - 6$.
$3x-27=-6$.

Step4: Isolate x

Add 27 to both sides: $3x-27 + 27=-6 + 27$.
$3x=21$.
Divide both sides by 3: $x = 7$.

Step1: Set up equation for 4

Since $DF=DE + EF$, we have $9x-39=47+(3x + 10)$.
$9x-39=3x+57$.

Step2: Isolate x terms

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

Step3: Isolate x

Add 39 to both sides: $6x-39 + 39=57 + 39$.
$6x=96$.
Divide both sides by 6: $x = 16$.
Now find $EF$ when $x = 16$.
$EF=3x + 10$.
Substitute $x = 16$ into the expression for $EF$: $EF=3\times16+10$.
$EF = 48+10$.
$EF = 58$.

Answer:

For 6: $x = 7$; For 4: $EF = 58$