Ans: (a) $c=4d-2$ (b) Decreased by $4$
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$\begin{array}{rcl}
3(2c+5d+4) & = & 39d \\
6c +15d + 12 & = & 39d \\
6c & = & 24d – 12 \\
c & = & 4d – 2
\end{array}$ - Let $c_0$ and $d_0$ be the original value of $c$ and $d$ respectively. Let $c_1$ and $d_1$ be the new value of $c$ and $d$ respectively. Note that $d$ is decreased by $1$, then we have $d_1 – d_0=-1$.
$\begin{array}{rcl}
c_1 – c_0 & = & (4d_1 – 2) – (4d_0 – 2) \\
& = & 4d_1 – 4d_0 \\
& = & 4(d_1 – d_0) \\
& = & 4(-1) \\
& = & -4
\end{array}$Therefore, $c$ will be decreased by $4$.
