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(3)/( x^(2) +3x)- (5)/( x^(2) -9)

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\bf \cfrac{3x}{ x^(2) +3x}- \cfrac{5}{ x^(2)-9}\\\\ -----------------------------\\\\ \textit{recall your }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\qquad thus\\\\\\ x^2-9\implies x^2-3^2\implies (x-3)(x+3)\\\\ -----------------------------\\\\ \cfrac{3\boxed{x}}{\boxed{x}(x+3)}- \cfrac{5}{(x-3)(x+3) }\implies \cfrac{3}{(x+3)}- \cfrac{5}{(x-3)(x+3) }


\bf \textit{so it looks like our LCD will be then }(x-3)(x+3)\qquad thus \\\\\\ \cfrac{3(x-3)-5}{(x-3)(x+3)}\implies \cfrac{3x-9-5}{(x-3)(x+3)}\implies \cfrac{3x-14}{(x-3)(x+3)}
User Mark Elder
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