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st bisects \\( \\overline{pr} \\) at point q. \\( pr = 7x - 15 \\) and …

Question

st bisects \\( \overline{pr} \\) at point q. \\( pr = 7x - 15 \\) and \\( pq = 7x - 25 \\). find pq.

Explanation:

Step1: Understand the Bisector

Since \( ST \) bisects \( PR \) at \( Q \), \( PQ = QR \) and \( PR = PQ + QR = 2PQ \). Also, we know \( PR = 7x - 15 \) and \( PQ = 7x - 25 \). So \( PR = 2 \times PQ \).

Step2: Set Up the Equation

Substitute the expressions for \( PR \) and \( PQ \) into \( PR = 2PQ \):

$$ 7x - 15 = 2(7x - 25) $$

Step3: Solve for \( x \)

Expand the right - hand side: \( 7x - 15 = 14x - 50 \)
Subtract \( 7x \) from both sides: \( - 15=7x - 50 \)
Add 50 to both sides: \( 7x=35 \)
Divide both sides by 7: \( x = 5 \)

Step4: Find \( PQ \)

Substitute \( x = 5 \) into the expression for \( PQ \): \( PQ=7\times5 - 25=35 - 25 = 10 \)

Answer:

\( 10 \)