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March 27, 2017

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Can someone please show me the steps to this problem? I get 0,1 as proposed solutions, but I am not sure about the 1. Thanks

Please solve and check all proposed solutions

(x-2)/(x-1)+(2)/(x^2-1)= 0

  • Algebra - ,

    you're correct, x=1 and x=0

  • Algebra - ,

    So when we check the problem, would this be acceptable? Does all of this check out? Thanks!

    (x-2)/(x-1) + (2)/(x^2-1) = 0

    x-2 / x-1+2/(x-1)(x-2) = 0

    To check:

    0-2 = -2
    0-1 = -1
    -1/-2 = 0.5

    02 – 1 = -1

    -1/2 = -0.5

    0.5 + -0.5 = 0

    1-2 = -1

    1-1 = 0

    0/-1 = 0

    12 – 1 = 0

    0/2 = 0

    0+0 = 0

    x = 0, 1

  • Algebra - ,

    12 – 1 = 0 suppose to be 1^2 - 1 = 0

  • Algebra - ,

    02 – 1 = -1 suppose to be 0^2 - 1 = 0

  • Algebra - ,

    You're right in suspecting the zero of x=1.
    If you subsitiute x=1 into the equation, you will get two indefinite values, hence x=1 has to be rejected.
    Note the hint given by the question:
    "...and check all proposed solutions ".
    All you need to do is to substitute each of the proposed solution into the original equation, if they work out (i.e. LHS=RHS), then the solution is acceptable.

    See also:
    http://www.jiskha.com/display.cgi?id=1249326304

  • Algebra - ,

    Will this check work? Is this correct?

    0-2 = -2
    0-1 = -1
    -1/-2 = 0.5

    0^2 – 1 = -1

    -1/2 = -0.5

    0.5 + -0.5 = 0

    THEN

    1-2 = -1

    1-1 = 0

    0/-1 = 0

    1^2 – 1 = 0

    0/2 = 0

    0+0 = 0

    x = 0, 1 Does this all check out?

  • Algebra - ,

    The first check is OK, so 0 is admitted as a solution.
    The second check is not OK because at two places, you have inverted the numerator and denominator, namely:
    0/-1 = 0 should read -1/0 = ∞
    and
    0/2 = 0 should read 2/0 = ∞
    Since the proposed solution x=1 gives rise to infinite values, x=1 is not admissible as a solution.

  • Algebra - ,

    So 1 doesn't work in the solution of the equation at all?

  • Algebra - ,

    One (1) is rejected as a solution because when substituted into the original equation
    (x-2)/(x-1)+(2)/(x^2-1)= 0
    it gives a zero as the denominator for each term, which means that each term becomes infinite. The sum of the two terms (left-hand-side) is thus not equal to zero, so 1 is not admitted as a solution to the given equation.

  • Algebra - ,

    Thanks for the explanation!

  • Algebra - ,

    You are very welcome!

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