1

\begin{align} \text{Given} \\ \end{align}

\$AF ∥ BC\$, \$AD \bot EF\$, \$AB \bot BC\$
\begin{align} \text{AB = 2x + 3, AC = 3y + 1, BD = x and DC = y + 1} \\ \end{align}

2

\begin{align} &\text{From the given parallell lines and} \\ &\text{perpendicular line} \\ \end{align}

\begin{align} \angle EAD &= \angle FAD = 90^\circ \\ \angle EAB &= \angle FAC &&\text{(Given)} \\ \angle EAD - \angle EAB &= \angle FAD - \angle FAC \\ ∴ \angle BAD &= \angle CAD \\ \end{align}

3

Consider \$\triangle ABD\$ and \$\triangle ACD\$

\begin{align} AD &= AD &&\text{(Common side)} \\ \angle BAD &= \angle CAD \\ \angle ADB &= \angle ADC = 90^\circ \\ ∴ \triangle ABD &\cong \triangle ACD \\ \end{align}

4

Consider \$\triangle ABD\$ and \$\triangle ACD\$

\begin{align} && ∵ \angle ABD &= \angle ACD \\ && AB &= AC \\ \text{and} && BD &= CD \\ \end{align}

5

\begin{align} \text{Consider AB = AC} \\ \end{align}

\begin{align} && AB &= AC \\ \Rightarrow && 2x + 3 &= 3y + 1 \\ \Rightarrow && 2x - 3y &= -2\tag{1} \label{1} \\ \end{align}

6

\begin{align} \text{Consider BD = CD} \\ \end{align}

\begin{align} && BD &= CD \\ \Rightarrow && x &= y + 1 \\ \Rightarrow && x - y &= 2 \tag{2} \label{2} \\ \end{align}

7

\begin{align} \text{Solving $(1)$ and $(2)$} \\ \end{align}

\begin{align} 2x - 3y &= -2 \\ x - y &= 1 \\ \hline \end{align}

8

\begin{align} \text{Multiply $(2)$ with 2 and subtract $(2)$ from $(1)$} \end{align}

\begin{align} 2x - 3y &= -2 \\ 2x - 2y &= 2 \\ \hline \end{align}

\begin{align} 2x - 3y &= -2 \\ - 2x + 2y &= -2 \\ \hline - y &= -4 \\ y &= 4 \end{align}

9

\begin{align} \text{Substitute y = 4 in $(2)$ } \end{align}

\begin{align} && x &= y + 1 \\ \Rightarrow && x &= (4) + 1 \\ \Rightarrow && x &= 5\\ \end{align}