x+y=y+x,if x =-3/6 and y =1/69
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If 6xy = x^2y + 9y, what is the value of xy?
6xy=x2y+9y6xy=x2y+9y --> y(x2−6x+9)=0y(x2−6x+9)=0 --> y(x−3)2=0y(x−3)2=0 --> either x=3x=3 or y=0y=0 (or both).
(1) y – x = 3. If y=0y=0 and x=−3x=−3, then xy=0xy=0 but if x=3x=3 and y=6y=6, then xy=18xy=18. Not sufficient.
(2) x^3 < 0 --> x<0x<0 --> x≠3x≠3, then y=0y=0 and xy=0xy=0. Sufficient.
Answer: B.
6xy=x2y+9y6xy=x2y+9y --> y(x2−6x+9)=0y(x2−6x+9)=0 --> y(x−3)2=0y(x−3)2=0 --> either x=3x=3 or y=0y=0 (or both).
(1) y – x = 3. If y=0y=0 and x=−3x=−3, then xy=0xy=0 but if x=3x=3 and y=6y=6, then xy=18xy=18. Not sufficient.
(2) x^3 < 0 --> x<0x<0 --> x≠3x≠3, then y=0y=0 and xy=0xy=0. Sufficient.
Answer: B.
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