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guddo
Asked: If 4x + 10 = 3y and y ≥ 2, which of the following intervals contains possible values for x?

3y = 4x + 10 ≥ 3*2 = 6
4x + 10 ≥ 6
4x ≥ - 4
x ≥ - 1

I. x < -1: Not feasible
II. -1 < x < 0 : Feasible
III. x > 0 : Feasible

A. I only
B. II only
C. III only
D. I and II
E. II and III
IMO E
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ShilpiAgnihotrii
­Is it really a 700 level question ?

 
­The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question. So, this is a 655-705 level question based on our statistics.
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Hi! Would x = 0 also not be a possible solution? I guess since the wording of the question is"which of the following contains possible values" it doesn't mean that the intervals shown here will capture all values of x, but we just have to pick from these ones to see which do contain the values of x?
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guddo
If 4x + 10 = 3y and y ≥ 2, which of the following intervals contains possible values for x?

I. x < -1
II. -1 < x < 0
III. x > 0

A. I only
B. II only
C. III only
D. I and II
E. II and III

Attachment:
2024-01-03_18-38-14.png
Since y ≥ 2, then 3y ≥ 6. Substituting the value of 3y gives 4x + 10 ≥ 6, which simplifies to x ≥ -1. Therefore, both -1 < x < 0 and x > 0 intervals contain possible values for x.

Answer: E.­
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I did it in a different manner.
4x+10 = 3y
=> (4x+10)/3 > or equal 2
=> 4x+10 > or equal 6
=> 4x > or equal -4
so x > or = -1
fits II and III
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