Ans: D
$\begin{array}{rcl}
\dfrac{y-1}{c} & = & \dfrac{y+1}{d} \\
d(y-1) & = & c(y+1) \\
dy-d & = & cy + c \\
dy – cy & = & c+d \\
y (d-c ) & = & c+d \\
y & = & \dfrac{c+d}{d-c}
\end{array}$
$\begin{array}{rcl}
\dfrac{y-1}{c} & = & \dfrac{y+1}{d} \\
d(y-1) & = & c(y+1) \\
dy-d & = & cy + c \\
dy – cy & = & c+d \\
y (d-c ) & = & c+d \\
y & = & \dfrac{c+d}{d-c}
\end{array}$
