Question

segment addition
if df = 9x - 39, find ef.
if r is the mid - point of qs, find qr

Explanation:

Step1: Use segment - addition postulate for first part

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

Step2: Simplify the right - hand side of the equation

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

Step3: Isolate the variable terms

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

Step4: Solve for $x$

Add 39 to both sides: $6x-39 + 39=57 + 39$, so $6x=96$. Then divide both sides by 6: $x=\frac{96}{6}=16$.

Step5: Find $EF$

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

Step6: Use mid - point property for second part

Since $R$ is the mid - point of $QS$, then $QR=RS$. So $5x-3=21 - x$.

Step7: Solve for $x$ in the second equation

Add $x$ to both sides: $5x+x-3=21 - x+x$, which gives $6x-3 = 21$. Add 3 to both sides: $6x-3 + 3=21+3$, so $6x=24$. Divide both sides by 6: $x = 4$.

Step8: Find $QR$

Substitute $x = 4$ into the expression for $QR$: $QR=5x-3=5\times4-3=20 - 3=17$.

Answer:

$EF = 58$, $QR=17$