How do you solve $(6/(x^2-1)) - (1/(x+1)) = 1/2$ ?

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Answer 1 · 2016-04-06T10:13:50

$x~~ -4.464" or "2.464$ to 3 decimal paces

$x^2-1$ can be written as $x^2-1^2$

Compare this to $a^2+b^2=(a+b)(a-b)$

We can use this to our advantage

Write the given equation as:

$6/(x^2-1^2) - 1/(x+1)=1/2$

By substitution this is

$6/((x+1)(x-1)) - 1/(x+1)=1/2$

$(6-(x-1))/((x+1)(x-1))=1/2$

Multiply both sides by 2

$(12-2(x-1))/((x+1)(x-1))=1$

$12-2(x-1)=(x+1)(x-1)$

$12-2x+2=x^2-1$

$x^2+2x-11=0$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now solve as a quadratic -

Completing the square method (skipping many steps)

$(x+1)^2 -12=0$

$x+1= +-sqrt(12)$

$x= -1+-2sqrt(3)$

$x~~ -4.464" or "2.464$ to 3 decimal paces

How do you solve (6/(x^2-1)) - (1/(x+1)) = 1/2(6/(x^2-1)) - (1/(x+1)) = 1/2?