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Bunuel
Official Solution:


If \(xyz < 0\), is \(x < 0\)?

\(xyz < 0\) implies that either all three unknowns are negative or one is negative and the remaining two are positive.

(1) \(x - y < 0\). This means that \(x < y\). Could \(x\) be negative? Yes, if \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(y\) are positive and \(z\) is negative. Not sufficient.

Answer: C

Hi, sorry, i keep getting confused with the concept of shifting variables from one side of the equation to the other - when to do it and when to not without affecting the inequality - please could you help me out here
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Bunuel
Official Solution:


If \(xyz < 0\), is \(x < 0\)?

\(xyz < 0\) implies that either all three unknowns are negative or one is negative and the remaining two are positive.

(1) \(x - y < 0\). This means that \(x < y\). Could \(x\) be negative? Yes, if \(x\), \(y\), and \(z\) are negative. Could \(x\) be positive? Yes, if \(x\) and \(y\) are positive and \(z\) is negative. Not sufficient.

Answer: C

Hi, sorry, i keep getting confused with the concept of shifting variables from one side of the equation to the other - when to do it and when to not without affecting the inequality - please could you help me out here

Hi,
you can always add the variables on either side of the equation and rather the correct method is to get them on one side and then move ahead..
\(xy - y > 2y +7......\)would become \(xy-y-2y>7.........y(x-3)>7......\)

Do not cross multiply or divide teh variables on two sides unless you knoe they are positive..
\(x^2>x.............. x^2-x>0 ......... x(x-1)>0..........\)now either BOTH x and x-1 are +ive or BOTH are -ive...
Youi cannot convert \(x^2>x ..............as ... x>1\)
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Bunuel
If \(xyz < 0\), is \(x < 0\)?



(1) \(x - y < 0\)

(2) \(x - z < 0\)


xyz<0 implies that odd number of variables are negative, so either 1 or all three.

(1) \(x - y < 0\)
Just tells us that x<y.
Possible that z<0<x<y or z<x<y<0.
Insufficient

(2) \(x - z < 0\)
Just tells us that x<z.
Possible that y<0<x<z or y<x<z<0.
Insufficient

Combined
We know x is less than y and less than z, so x is the least.
Now xyz<0 means only one or all three are negative.
Be it any of the two possibilities, the LOWEST will surely be negative. Thus, x<0.
Sufficient


C
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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