Can you pls help me pls?

[tex]\dfrac{16}{7x+4}+A=\dfrac{49x^2}{7x+4}\\\\A=\dfrac{49x^2}{7x+4}-\dfrac{16}{7x+4}\\\\A=\dfrac{49x^2-16}{7x+4}\\\\A=\dfrac{(7x-4)(7x+4)}{7x+4}\\\\A=7x-4[/tex]
[tex]\bf \stackrel{~~~~~~~~\textit{is equivalent}}{\cfrac{16}{7x+4}+A~~=~~\cfrac{49x^2}{7x+4}}\implies \stackrel{\textit{cross-multiplying}}{\cfrac{16~~\begin{matrix} (7x+4) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 7x+4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}+A(7x+4)=49x^2} \\\\\\ 16+A(7x+4)=49x^2\implies A(7x+4)=49x^2-16\implies A=\cfrac{49x^2-16}{7x+4}[/tex]
[tex]\bf A=\cfrac{7^2x^2-4^2}{7x+4}\implies A=\cfrac{\stackrel{\textit{difference of squares}}{(7x)^2-4^2}}{7x+4}\implies A=\cfrac{(7x-4)~~\begin{matrix} (7x+4) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 7x+4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill A=7x-4~\hfill[/tex]