[tex]\bf \cfrac{-x-2}{9x^2-1}+\cfrac{-5x+4}{9x^2-1}\impliedby \begin{array}{llll} \textit{since the denominators are the same,}\\ \textit{we simply add the numerators}\\ \textit{and keep the same denominator} \end{array} \\\\\\ \cfrac{-x-2~~+~~(-5x+4)}{9x^2-1}\implies \cfrac{2-6x}{9x^2-1}\implies \cfrac{2(1-3x)}{3^2x^2-1^2}\implies \cfrac{2(1-3x)}{\stackrel{\textit{difference of squares}}{(3x)^2-1^2}}[/tex]
[tex]\bf \cfrac{2(1-3x)}{(3x-1)(3x+1)}\implies \cfrac{-2(-1+3x)}{(3x-1)(3x+1)} \\\\\\ \cfrac{-2\underline{(3x-1)}}{\underline{(3x-1)}(3x+1)}\implies \cfrac{-2}{3x+1}[/tex]
recall that -2(-1+3x) is really 2(1 - 3x) in disguise.
also recall that 1² = 1³ = 1⁴ = 1¹⁰⁰⁰⁰⁰⁰⁰⁰⁰ = 1.
