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Question

Solve for y in the following equation:
y2/32y1/3=15

A
-27 and 125
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B
-125 and 27
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C
-125 and -27
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D
None of These
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Solution

The correct option is A -27 and 125
We have,
y232y13=15
Let y13=p
So, we get,
p22p=15
p22p15=0
p25p+3p15=0
p(p5)+3(p5)=0
(p5)(p+3)=0
p=5 or p=3
y13 = 5 or y13=3
y=53=125 or y=(3)3=27
So, correct option is a.

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