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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