[tex]\begin{gathered} \frac{1}{x-4}+\frac{x}{x-2}=\frac{2}{x^2-6x+8} \\ Add\text{ing} \\ \frac{1}{x-4}+\frac{x}{x-2}=\frac{(1)(x-2)+(x-4)(x)}{(x-4)(x-2)}=\frac{(x-2)+(x-4)(x)}{(x-4)(x-2)} \\ Factor\text{ing} \\ \frac{2}{x^2-6x+8}=\frac{2}{(x-4)(x-2)} \\ \frac{(x-2)+(x-4)(x)}{(x-4)(x-2)}=\frac{2}{(x-4)(x-2)} \\ (x-2)+(x-4)(x)=2 \\ x-2+x^2-4x=2 \\ x^2-3x-2=2 \\ x^2-3x-2-2=0 \\ x^2-3x-4=0 \\ (x-4)(x+1)=0 \\ x-4=0 \\ x=4 \\ x+1=0 \\ x=-1 \\ \text{The solution is x=4 and x=-1} \end{gathered}[/tex]
