How do you solve $1/(x+3) + 1/(x-3) = 1/(x^2-9)$ ?

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Answer 1 · 2015-05-27T03:33:33

$color(blue)(a^2 - b^2 = (a+b)(a-b)$

$(x^2-9) = (x+3)(x-3)$

the expression given is:
$1/(x+3) + 1/(x-3) = 1/(x^2-9)$

$1/(x+3) + 1/(x-3) = 1/((x+3)(x-3)$

$(1.(x-3))/((x+3)(x-3)) + (1.(x+3))/((x-3)(x+3)) = 1/((x+3)(x-3)$

$((x-3)+ (x+3))/((x+3)(x-3)) = 1/((x+3)(x-3))$

$(2x)/cancel(x^2-9) = 1/cancel(x^2-9)$

$2x = 1$

$x =color(blue)( 1/2$ is the solution for the expression.

How do you solve 1/(x+3) + 1/(x-3) = 1/(x^2-9)1/(x+3) + 1/(x-3) = 1/(x^2-9)?