I. 27(p + 2) = 2p(24 – p)
II. 2q2 – 25q + 78 = 0
ATQ, From I: 27(p + 2) = 2p(24 – p) Or, 27p + 54 – 48p + 2p2 = 0 Or, 2p2 – 21p + 54 = 0 Or, 2p2 – 12p – 9p + 54 = 0 Or, 2p(p – 6) – 9(p – 6) = 0 Or, (2p – 9)(p – 6) = 0 Or, p = (9/2) or 6 From II: 2q2 – 25q + 78 = 0 Or, 2q2 – 12q – 13q + 78 = 0 Or, 2q(q – 6) – 13(q – 6) = 0 Or, (2q – 13)(q – 6) = 0 Or, q = (13/2) or 6 Therefore, q ≥ p
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