1

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

\begin{align} & \text{ABCD is a square,} AC \bot DB \\ & \text{AE = 2x - 6, EC = -3y + 10, DE = 5x and EB = 4y - 6} \\ \end{align}

2

Consider \$\triangle AEB\$ and \$\triangle CED\$

\begin{align} AE &= CE \\ \angle AEB &= \angle CED \\ \angle EAB &= \angle ECD \\ ∴ \triangle AEB &\cong \triangle CED \\ \end{align}

3

From \$\triangle AEB\$ and \$triangle CED\$

\begin{align} AE &= CE \\ EB &= ED \\ \end{align}

4

\begin{align} \text{Consider AE = CE} \\ \end{align}

\begin{align} && AE &= CE \\ \Rightarrow && 2x - 6 &= -3y + 10 \\ \Rightarrow && 2x + 3y &= 16 \tag{1} \label{1} \\ \end{align}

5

\begin{align} \text{Consider EB = ED} \\ \end{align}

\begin{align} && EB &= ED \\ \Rightarrow && 5x &= 4y - 6 \\ \Rightarrow && 5x - 4y &= - 6 \tag{2} \label{2} \\ \end{align}

6

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

\begin{align} 2x + 3y &= 16 \\ 5x - 4y &= -6 \\ \hline \end{align}

7

\begin{align} \text{Multiply $(1)$ with 5 and $(2)$ with 2} \end{align}

\begin{align} 10x + 15y &= 80 \\ 10x - 8y &= -12 \\ \hline \end{align}

8

\begin{align} \text{Subtract $(2)$ from $(1)$} \\ \end{align}

\begin{align} 10x + 15y &= 80 \\ - 10x + 8y &= 12 \\ \hline 23y &= 92 \\ y &= 4 \end{align}

9

\begin{align} \text{Substitute y = 4 in $(1)$ } \end{align}

\begin{align} && 2x + 3y &= 16 \\ \Rightarrow && 2x + 3(4) &= 16 \\ \Rightarrow && 2x + 12 &= 16 \\ \Rightarrow && 2x &= 16 - 12 \\ \Rightarrow && 2x &= 4 \\ \Rightarrow && x &= 2 \end{align}