Ans: $x=\dfrac{BC}{A-4C}$
$\begin{array}{rcl}
Ax & = & (4x+B)C \\
Ax & = & 4Cx + BC \\
Ax – 4Cx & = & BC \\
x(A-4C) & = & BC \\
x & = & \dfrac{BC}{A-4C}
\end{array}$
$\begin{array}{rcl}
Ax & = & (4x+B)C \\
Ax & = & 4Cx + BC \\
Ax – 4Cx & = & BC \\
x(A-4C) & = & BC \\
x & = & \dfrac{BC}{A-4C}
\end{array}$
