Posted by Terri on Sunday, October 27, 2013 at 4:39pm.
5/3x7≤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/(3x7) <= 2
If 3x7 > 0 (x > 7/3), then
5 <= 2(3x7)
5/2 <= 3x7
19/2 <= 3x
19/6 <= x
x >= 19/6
The original condition was x > 7/3, so x >= 19/6 works.
If 3x7 < 0 (x < 7/3),
5 >= 2(3x7)
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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