How do you solve $1/(x+3) + 1/(x-3) = 10/ (x^2-9)$ ?

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Answer 1 · 2016-02-13T23:06:18

$x=5$

$1/(x+3)+1/(x-3)=10/(x^2-9)$

First put everything with same denominator, recalling that $(x+3)(x-3)=x^2-9:$

$(x-3)/(x^2-9)+(x+3)/(x^2-9)=10/(x^2-9)$

We can eliminate the denominators, but keeping in mind that $x!=+-3$

$x-3+x+3=10$

$2x=10$

$x=5$

How do you solve 1/(x+3) + 1/(x-3) = 10/ (x^2-9)1/(x+3) + 1/(x-3) = 10/ (x^2-9)?