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Question
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
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Solution
Let 2x = 3y = 6-z = k
`rArr2=k^(1/x),` `3=k^(1/y),` `6=k^(1/-z)`
Now,
`6 = 2xx3=k^(1/-z)`
`rArrk^(1/x)xxk^(1/y)=k^(1/-z)`
`rArrk^(1/x+1/y)=k^(1/-z)`
`rArr1/x+1/y=1/-z`
`rArr1/x+1/y+1/z=0`
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