Posted by **Terri** on Sunday, October 27, 2013 at 4:39pm.

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 -
**Steve**, Sunday, October 27, 2013 at 5:09pm
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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