Ans: (a) $n=\dfrac{7-5m}{2}$ (b) decreased by $5$
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$\begin{array}{rcl}
2(3m + n) & = & m + 7 \\
6m + 2n & = & m + 7 \\
2n & = & -5m + 7\\
n & = & \dfrac{-5m + 7}{2}
\end{array}$ - Let $m_0$ and $n_0$ be the original values of $m$ and $n$ respectively. Hence we have
$\begin{array}{rcl}
n_0 & = & \dfrac{-5m_0 + 7}{2}
\end{array}$Let $n_1$ be the new value of $n$. Hence, we have
$\begin{array}{rcl}
n_1 & = & \dfrac{-5(m_0 + 2) + 7}{2} \\
& = & \dfrac{-5m_0 – 10 + 7}{2} \\
& = & \dfrac{-5m_0 + 7}{2} – \dfrac{10}{2} \\
& = & n_0 – 5
\end{array}$Therefore, the decrease of $n$ is $5$.
