Question

• q is between p and r;
• pq = 5x–10;
• qr = 3(x + 4); and
• \\(\overline{pq} \cong \overline{qr}\\).

diagram: p---q---r, with pq labeled 5x – 10, qr labeled 3(x + 4)

x = ____________

Explanation:

Step1: Set PQ equal to QR

Since \( \overline{PQ} \cong \overline{QR} \), their lengths are equal. So we set \( 5x - 10 = 3(x + 4) \).

Step2: Expand the right side

Expand \( 3(x + 4) \) to get \( 3x + 12 \). The equation becomes \( 5x - 10 = 3x + 12 \).

Step3: Subtract 3x from both sides

Subtract \( 3x \) from each side: \( 5x - 3x - 10 = 3x - 3x + 12 \), which simplifies to \( 2x - 10 = 12 \).

Step4: Add 10 to both sides

Add 10 to each side: \( 2x - 10 + 10 = 12 + 10 \), resulting in \( 2x = 22 \).

Step5: Divide by 2

Divide both sides by 2: \( \frac{2x}{2} = \frac{22}{2} \), so \( x = 11 \).

Answer:

\( 11 \)