We need to find out if y > 0
1. \(3x + 2y > z + 4\)
Consider statement 1,
since we are not sure about value of z and x
we can't say anything for sure about the value of y. Insufficient.
(2) \(6x - 7 < 5y + 2z\)
This can be rewritten as \(6x - 5y < 2z +7\)
Adding one to both sides, \(6x - 5y +1 < 2z +8\)
Dividing by 2, we get \(\frac{(6x - 5y +1)}{2} < z +4\)
Hence, we can say that \(3x + \frac{(1 - 5y)}{2} < z +4\)
Again, since we have no idea about the value of z and x,
we can't say anything for sure about the value of y. Insufficient.
Now, on combining both the statements, we can get the following relation
\(3x + 2y > z + 4 > 3x + \frac{(1 - 5y)}{2}\)
Since 3x is common on both the sides, if y is negative.
\(3x + 2y < 3x + \frac{(1 - 5y)}{2}\)
Therefore, this relation can only work, when y is positive. (Option C)
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