[tex]\textit{difference of squares} \\\\ (a-b)(a+b) = a^2-b^2 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{2x^2+10x-28}{x^2-49}\div \cfrac{2x-4}{x+7}\implies \cfrac{2x^2+10x-28}{x^2-49}\cdot \cfrac{x+7}{2x-4} \\\\\\ \cfrac{2(x^2+5x-14)}{\underset{\textit{differenc of squares}}{x^2-7^2}}\cdot \cfrac{x+7}{2x-4}\implies \cfrac{2(x^2+5x-14)}{(x-7)~~\begin{matrix} (x+7) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{~~\begin{matrix} x+7 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{2(x-2)}[/tex]
[tex]\cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x^2+5x-14)}{x-7}\cdot \cfrac{1}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x-2)}\implies \cfrac{x^2+5x-14}{x-7}\cdot \cfrac{1}{x-2} \\\\\\ \cfrac{~~\begin{matrix} (x-2) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x+7)}{x-7}\cdot \cfrac{1}{~~\begin{matrix} x-2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\implies \cfrac{x+7}{x-7}=y[/tex]
