Question

What is the sum? 3/x^2 - 9 + 5/x+3 A.) 8/x^2 + x - 6 B.) 5x - 12 / x-3 C.) -5x/ (x + 3)(x-3) D.) 5x-12/ (x+3)(x-3)

Answer 1

`\boxed{\boxed{\huge \sf D. \ (5x - 12)/((x + 3)(x - 3))}}`

Step-by-step explanation:

`\sf Summation \: of \: following : \\ \sf \implies \frac{3}{ {x}^(2) - 9} + (5)/(x + 3) \\ \\ \sf {x}^(2) - 9 = {x}^(2) - {3}^(2) : \\ \sf \implies \frac{3}{ {x}^(2) - \boxed{ \sf {3}^(2) }} + (5)/(x + 3) \\ \\ \sf Factor \: the \: difference \: of \: two \: squares. \\ \sf {x}^(2) - {3}^(2) = (x + 3)(x - 3) : \\ \sf \implies \frac{3}{ \boxed{ \sf (x + 3)(x - 3)}} + (5)/(x + 3) \\ \\ \sf Put \: each \: term \: in \: (3)/((x + 3)(x - 3)) + (5)/(x + 3) over \\ \sf the \: common \: denominator \: (x + 3)(x - 3) : \\ \sf \implies (3)/((x + 3)(x - 3)) + (5(x - 3))/((x + 3)(x - 3)) \\ \\ \sf (3)/((x + 3)(x - 3)) + (5(x - 3))/((x + 3)(x - 3)) = (5(x - 3) + 3)/((x + 3)(x - 3)) : \\ \sf \implies (5(x - 3) + 3)/((x + 3)(x - 3))`

`\sf 5(x - 3) = 5x - 15 : \\ \sf \implies \frac{ \boxed{ \sf 5x - 15} + 3}{(x + 3)(x - 3)} \\ \\ \sf Grouping \: like \: terms, \: 5x - 15 + 3 = 5x + (3 - 15) : \\ \sf \implies \frac{ \boxed{ \sf 5x + (3 - 15)}}{(x + 3)(x - 3)} \\ \\ \sf 3 - 15 = - 12 : \\ \sf \implies (5x - 12)/((x + 3)(x - 3))`