From second: \(n_1 = m_1 + y\), plug: \(m_1 + (m_1 + y) = 2m_1 + y = S\), \(n_1 = m_1 + y\), so \(m_2 + n_2 = 2m_1 + y\).

Tue, 05 May 2026 17:34:14 +0000

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