Question 442439
{{{ 1/(x-1) + 1/2 = 2/(x^2 -1) }}}
First factor {{{ x^2 - 1 }}}
{{{ 1/(x-1) + 1/2 = 2 / ((x -1)*(x+1)) }}}
Multiply both sides by {{{ 2*(x-1)*(x+1) }}}
{{{ 2*(x+1) + (x-1)*(x+1) = 4 }}}
Now multiply everything out
{{{ 2x + 2 + x^2 - 1 = 4 }}}
{{{ x^2 + 2x  = 3 }}}
Solve by completing the square
{{{ x^2 + 2x + (2/2)^2 = 3 + (2/2)^2 }}}
{{{ x^2 + 2x + 1 = 3 + 1 }}}
Both sides are now a perfect square
{{{ ( x + 1 )^2 = 2^2 }}}
Take the square root of both sides
{{{x +1 = 2 }}}
{{{ x = 1 }}} 1st answer
and, also the negative square root,
{{{ x + 1 = -2 }}}
{{{x = -3 }}} 2nd answer
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check answers:
{{{ x = 1 }}}
{{{ 1/(x-1) + 1/2 = 2/(x^2 -1) }}}
{{{ 1/(1-1) + 1/2 = 2/(1^2 -1) }}}
{{{ x = 1 }}} is not allowed because {{{ 1/(1-1) = 1/0 }}}
and {{{ 2/(1^2 -1) = 2/0 }}}
and division by {{{0}}} is not allowed
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{{{x = -3}}}
{{{ 1/(-3-1) + 1/2 = 2/((-3)^2 -1) }}}
{{{ -1/4 + 1/2  = 2/8 }}}
{{{ -1/4 + 2/4 = 1/4 }}}
{{{ 1/4 = 1/4 }}}
OK
The answer is {{{ x = -3 }}}