1

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

\begin{align} & \text{ABC and DBC are two triangles, AB = DB and AC = DC. } & \\ & \text{AB = y + 9, AC = 8 - x, DC = 2x and BD = 2x} \\ \end{align}

2

Consider \$\triangle ABC\$ and \$\triangle DBC\$

\begin{align} AB &= DB \\ AC &= DC \\ BC &= BC \\ ∴ \triangle ABC &\cong \triangle DBC \tag{by SSS congruency} \\ \end{align}

3

From \$\triangle ABC\$ and \$\triangle DBC\$

\begin{align} && AC &= DC \\ \Rightarrow && 8 - x &= 3y \\ \Rightarrow && x + 3y &= 8 \tag{1} \\ && \text{and} \\ && AB &= DB \\ \Rightarrow && y + 9 &= 2x \\ \Rightarrow && 2x - y &= 9 \tag{2} \\ \end{align}

4

\begin{align} \text{Solving (1) and (2) for x and y} \\ \end{align}

\begin{align} x + 3y = 8 \\ 2x - y = 9 \\ \hline \end{align}

5

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

\begin{align} 2x + 6y &= 16 \\ -2x + y &= -9 \\ \hline 7y &= 7 \\ \end{align}

\begin{align} && 7y &= 7 \\ \Rightarrow && y &= 1 \\ \end{align}

6

\begin{align} \text{Substitute y = 1 in (2)} \\ \end{align}

\begin{align} && 2x - y &= 9 \\ \Rightarrow && 2x - (1) &= 9 \\ \Rightarrow && 2x &= 9 + 1 \\ \Rightarrow && 2x &= 10 \\ \Rightarrow && x &= 5 \end{align}