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5. pq = 3x + 3, qr = 4x + 4, rs = 5x - 15, qs = 100. solve for x. find …

Question

  1. pq = 3x + 3, qr = 4x + 4, rs = 5x - 15, qs = 100. solve for x. find pq, qr, and rs.

Explanation:

Step1: Analyze the segment addition

From the number line, \( QS = QR + RS \). Substitute the given expressions: \( 100=(4x + 4)+(5x + 15) \)

Step2: Simplify the equation

Combine like terms: \( 100 = 4x+4 + 5x+15=9x + 19 \)

Step3: Solve for \( x \)

Subtract 19 from both sides: \( 9x=100 - 19 = 81 \), then divide by 9: \( x=\frac{81}{9}=9 \)

Step4: Find \( PQ \)

Substitute \( x = 9 \) into \( PQ = 3x + 3 \): \( PQ=3(9)+3 = 27 + 3=30 \)

Step5: Find \( QR \)

Substitute \( x = 9 \) into \( QR = 4x + 4 \): \( QR=4(9)+4 = 36 + 4=40 \)

Step6: Find \( RS \)

Substitute \( x = 9 \) into \( RS = 5x + 15 \): \( RS=5(9)+15 = 45 + 15=60 \)

Answer:

\( x = 9 \), \( PQ = 30 \), \( QR = 40 \), \( RS = 60 \)