Question

use the diagram below to answer questions

1. if pq = 7 and pr = 32, find qr.
2. if qr = 58 and pq = 13, find pr.
3. if pq = x - 5, qr = 2x - 1, and pr = 21,
4. if qr = 6x + 1, pr = 14x - 13, and pq =

Explanation:

Step1: Use segment - addition postulate

Since points $P$, $Q$, and $R$ are collinear, $PR=PQ + QR$.

Step2: Solve for $QR$ in the first case

Given $PQ = 7$ and $PR=32$, we substitute into $PR=PQ + QR$. Then $32=7 + QR$. Rearranging for $QR$, we get $QR=PR - PQ$. So $QR=32 - 7=25$.

Step3: Solve for $PR$ in the second case

Given $QR = 58$ and $PQ = 13$, using $PR=PQ + QR$, we substitute the values: $PR=13 + 58=71$.

Step4: Solve for $x$ in the third case

Given $PQ=x - 5$, $QR = 2x-1$, and $PR = 21$. By the segment - addition postulate $PR=PQ + QR$, so $21=(x - 5)+(2x-1)$.
First, simplify the right - hand side: $21=x - 5+2x-1=3x-6$.
Then, add 6 to both sides: $21 + 6=3x$, so $27 = 3x$.
Divide both sides by 3: $x = 9$.

Step5: Solve for $x$ in the fourth case

Let $PQ = 3x-1$. Given $QR=6x + 1$ and $PR=14x-13$. By the segment - addition postulate $PR=PQ + QR$.
So $14x-13=(3x - 1)+(6x + 1)$.
Simplify the right - hand side: $(3x - 1)+(6x + 1)=3x-1+6x + 1=9x$.
Then we have the equation $14x-13=9x$.
Subtract $9x$ from both sides: $14x-9x-13=0$, so $5x-13 = 0$.
Add 13 to both sides: $5x=13$, and $x=\frac{13}{5}=2.6$.

Answer:

1. $QR = 25$
2. $PR = 71$
3. $x = 9$
4. $x = 2.6$