Question

7     + 4x    =      2
x+3  x2 -9      x-3

Answer 1

$\dfrac{7}{x+3}+\dfrac{4x}{x^2-9}=\dfrac{2}{x-3}\Longrightarrow\dfrac{7\cdot(x-3)}{x^2-9}+\dfrac{4x}{x^2-9}=\dfrac{2\cdot(x+3)}{x^2-9}\\\\ \Longrightarrow7\cdot(x-3)+4x=2\cdot(x+3)\Longrightarrow7x-21+4x=2x+6\\\\ \Longrightarrow11x-21=2x+6\Longrightarrow9x=27\Longrightarrow x=\dfrac{27}{9}\\\\ \boxed{x=3}$