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How would u solve
D\d+2 - 2\2-d = d+6\ d^2-4

I got to
D(d-2)(2-d) -2 (d+2)(d-2)=2d-d^2+12-6d

How would you solve? Im confused

  • algebra -

    Did you not notice that the denominator of the right side is (d-2)(d+2) ?
    I will also assume that D is really d, or else you have 2 unknowns.

    so let's rewrite ...
    d/(d+2) - 2/(2-d) = (d+6)/[(d+2)d-2)]
    d/(d+2) + 2/(d-2) = (d+6)/[(d+2)d-2)]

    now multiply each term by (d+2)(d-2)
    d(d-2) + 2(d+2) = d+6
    d^2 - 2d + 2d + 4 = d + 6
    d^2 - d - 2 = 0
    (d-2)(d+1) = 0
    d = 2 or d = -1

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