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9/x + 9/x-2= 12

  • alg/trig -

    The first thing you want to do is get rid of x in the denominator, because that simplifies things quite a bit. Think of the easiest common denominator between 9/x and 9/(x-2), which would be x(x-2).

    You know that x-2/x-2 = 1, regardless of the value of x; the same with x/x, so you therefore can multiply 9/x with (x-2)/(x-2) (which is the same thing as multiplying by one) and 9/(x-2) by x/x. This yields

    9(x-2)/[x(x-2)] and 9x/[x(x-2)], giving an equation of

    9(x-2)/[x(x-2)] + 9x/[x(x-2)] = 12

    Because you have a common denominator, you can condense the two terms on the right side of the equation to:

    [9(x-2) + 9x]/[x(x-2)] = 12.

    Multiply both sides of the equation by x(x-2), eliminating the denominator:
    9(x-2) + 9x = 12x(x-2)

    Distribute the terms:
    9x-18 + 9x = 12x^2 - 24x

    18x-18=12x^2 - 24x

    Move all terms to the left side of the equation (subtract 18x from both sides and add 18 to both sides):

    0= 12x^2 - 24x - 18x + 18


    0=12x^2 - 42x + 18

    You now have a quadratic formula and can now do one of two things: apply the quadratic formula or factor. I'll just factor it out (you can undoubtedly plug the numbers into the quadratic formula on your own time, should you so desire):

    The values that would make this equation true would be
    2x-1=0 or x-3=0, giving x-values of 0.5 and 3.

    Plug in the x-values into your equation. Do you get 12?

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