Asked by Terri
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
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
Answers
Answered by
Steve
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.
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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