Question 103008
{{{3/((x-1))}}} - {{{6/((x^2-1))}}} = 1
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You can see that (x^2-1) is the difference of squares so factor that:
{{{3/((x-1))}}} - {{{6/((x-1)(x+1))}}} = 1
:
Common denominator would be (x-1)(x+1), multiply equation by that:
(x-1)(x+1){{{3/((x-1))}}} - (x-1)(x+1){{{6/((x-1)(x+1))}}} = 1(x-1)(x+1)
:
Cancel out the denominators and you have;
3(x+1) - 6 = (x-1)(x+1)
:
Multiply what's inside the brackets on the left, FOIL the right
3x + 3 - 6 = x^2 - 1
3x - 3 = x^2 - 1
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Arrange as a quadratic equation:
x^2 - 3x - 1 + 3 = 0
x^2 - 3x + 2 = 0
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Factors to:
(x-2)(x-1) = 0
Two solutions:
x = +2
x = +1
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However x = 1, cannot be a solution. Note that if you substitute 1 for x,
both fractions have division by 0
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x = 2 is the only solution, check by subsitution
:
{{{3/((2-1))}}} - {{{6/((2^2-1))}}} = 1
 3 - {{{6/3}}} = 1
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