You can plug numbers. Based on the question stem, it seems pretty easy to plug and play with some numbers.

\frac{x}{y} or

\frac{x}{(y+1)}(1) x is not equal to 0 ----> I think from a quick glance, this is not sufficient because we don't know any thing about y. But just for kicks work through the process. Pick numbers: y= 2 , x=2

\frac{2}{2}=1 and

\frac{2}{2+1}=\frac{2}{3} Therefore,

\frac{x}{y} 1 > 2/3

\frac{x}{(y+1)}but if we pick y= -2 , x=-2 then

\frac{-2}{-2} = 1 and

\frac{-2}{-2+1}=\frac{-2}{-1}= 2, Therefore,

\frac{x}{y} 1 < 2 \frac{x}{(y+1)}So based on the results, we have two answer, therefore the statement N/S

(2) x > y ----> same thing for this statement. Try using a variation of the same number. Make the both positive and both negative but satisfying the parameters of the statement. Pick numbers: y= 2 , x=4

\frac{4}{2}=2 and

\frac{4}{2+1}=\frac{4}{3} or 1 \frac{1}{3}. Therefore,

\frac{x}{y} 2 >1 \frac{1}{3} \frac{x}{(y+1)}but if we pick y= -4 , x=-2 then

\frac{-2}{-4} =

\frac{1}{2}and

\frac{-2}{-4+1}=\frac{-2}{-3}= \frac{2}{3}, Therefore,

\frac{x}{y} .50 < .666 \frac{x}{(y+1)}So we get two different results from picking numbers that satisfied the parameters in Stmt 2. So this is N/S. When you put the two statements together its also I/S. From Stmt 1, we don't if x is positive or negative. That will have a bearing on the results of the numbers as noted above.

I know its been awhile since you made this post, but the time it took me rationalize this correct answer helped me understand why i got the question wrong. Hope this explanation helps someone else where an algebra equation is not intuitive.

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