How do you solve $10/(n^2-1)+(2n-5)/(n-1)=(2n+5)/(n+1)$ ?

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Answer 1 · 2017-02-22T01:08:49

$n = 5/3$

Note that $n^2 - 1 = (n + 1)(n - 1)$ .

$10/((n + 1)(n - 1)) + ((2n - 5)(n + 1))/((n + 1)(n - 1)) = ((2n + 5)(n - 1))/((n - 1)(n + 1))$

$10 + 2n^2 - 5n + 2n - 5 = 2n^2 + 5n - 2n - 5$

$10 = 6n$

$5/3 = n$

Hopefully this helps!

How do you solve 10/(n^2-1)+(2n-5)/(n-1)=(2n+5)/(n+1)10/(n^2-1)+(2n-5)/(n-1)=(2n+5)/(n+1)?