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\(x+\dfrac{1}{zy}=\dfrac{1}{5}\)

\(y+\dfrac{1}{xz}=\dfrac{-1}{15}\)

\(z+\dfrac{1}{xy}=\dfrac{1}{3}\)

FIND THE VALUE OF-: z-y/z-x

Note by Dheeraj Agarwal 2 years, 9 months ago

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Assuming that there is a typo in the first equation,

Subtracting the 2nd eqn from the 3rd eqn,

\(z - y + \frac{1}{x}(\frac{1}{y} - \frac{1}{z}) = \frac{1}{3} - \frac{-1}{15}\)

\((z - y)(1 + \frac{1}{xyz}) = \frac{6}{15}\) ----- 4

Subtracting the 1st eqn from the 3rd eqn,

\(z - x + \frac{1}{y}(\frac{1}{x} - \frac{1}{z}) = \frac{1}{3} - \frac{1}{5}\)

\((z - x)(1 + \frac{1}{xyz}) = \frac{2}{15}\) ----- 5

Dividing 4 from 5,

\(\frac{z-y}{z-x} = 3\)

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3 :D

the answer is 3

3

3 is the answer

Is there a typo in eqn 1 ... should it be \(x+\frac{1}{yz}= \frac{1}{5}\)

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TopNewestAssuming that there is a typo in the first equation,

Subtracting the 2nd eqn from the 3rd eqn,

\(z - y + \frac{1}{x}(\frac{1}{y} - \frac{1}{z}) = \frac{1}{3} - \frac{-1}{15}\)

\((z - y)(1 + \frac{1}{xyz}) = \frac{6}{15}\) ----- 4

Subtracting the 1st eqn from the 3rd eqn,

\(z - x + \frac{1}{y}(\frac{1}{x} - \frac{1}{z}) = \frac{1}{3} - \frac{1}{5}\)

\((z - x)(1 + \frac{1}{xyz}) = \frac{2}{15}\) ----- 5

Dividing 4 from 5,

\(\frac{z-y}{z-x} = 3\)

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3 :D

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the answer is 3

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3

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3 is the answer

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Is there a typo in eqn 1 ... should it be \(x+\frac{1}{yz}= \frac{1}{5}\)

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