Algebra

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5/3x-7≤2
Solve. Note in interval notation
So I got 19/6≤x
the other is suppose to be
x less than 7/3 but I can't get that
How do you get that

  • Algebra -

    5/(3x-7) <= 2
    If 3x-7 > 0 (x > 7/3), then
    5 <= 2(3x-7)
    5/2 <= 3x-7
    19/2 <= 3x
    19/6 <= x
    x >= 19/6
    The original condition was x > 7/3, so x >= 19/6 works.

    If 3x-7 < 0 (x < 7/3),
    5 >= 2(3x-7)
    19/6 >= x
    x <= 19/6
    The original condition was x < 7/3, but 7/3 < 19/6, so only the x < 7/3 part is a solution.

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