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? asked in Science & MathematicsMathematics · 5 years ago

4/(t+3)= 1 - 1/(t-3)?

So far I got to the point where I have 0=t^2-5t. They want a large x value and a small x value. I don't know what to do next.

2 Answers

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  • DWRead
    Lv 7
    5 years ago

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

    4(t-3) = (t+3)(t-3) - (t+3)

    4t-12 = t²- 9 - t - 3

    t² -5t = 0

    t(t-5) = 0

    t = 0,5

  • rmm
    Lv 7
    5 years ago

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

    multiply by (t+3)(t-3)

    4(t - 3) = (t+3)(t-3) - (t + 3)

    4t - 12 = t^2 - 9 - t - 3

    0 = t^2 - 5t

    0 = t(t - 5)

    t = 0, t = 5

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