Question

Can you pls help me pls?

Answer 1

`\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}`

`\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`

Answer 2

`(16)/(7x+4)+A=(49x^2)/(7x+4)\\\\A=(49x^2)/(7x+4)-(16)/(7x+4)\\\\A=(49x^2-16)/(7x+4)\\\\A=((7x-4)(7x+4))/(7x+4)\\\\A=7x-4`