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Ingri S
Ingri S asked in Science & MathematicsMathematics · 1 decade ago

I need help with this algebra question:?

3 / (x+3) + 1 / (x+3) = 4/ (x^2-9)

3 Answers

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  • gilvi
    Lv 7
    1 decade ago
    Favorite Answer

    3/ (x+3) + 1/ (x+3) = 4/ (x^2-9)

    4/ (x+3) = 4/ (x^2-9)

    1/ (x+3) = 1/ (x^2-9)

    L.C.M. = x^2-9

    x^2-9 =/= 0 => x =/= 3; x =/= -3

    x-3 = 1

    x = 3+1

    x = 4

    Bye

    gilvi

  • James
    Lv 6
    1 decade ago

    x= 4

  • Anonymous
    1 decade ago

    3/(x+3) + 1/(x+3) = 4/(x^2-9)

    Remember that a^2-b^2 = (a-b)(a+b) so

    3/(x+3) + 1/(x+3) = 4/(x-3)(x+3)

    common denominator (x-3)(x+3) so

    (x-3)/(x-3)*3/(x+3) + (x-3)/(x-3)*1/(x+3) = 4/(x-3)(x+3)

    3(x-3) + x-3 =4

    3x-9+x-3=4

    4x -12=4

    4x = 16

    x= 4

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