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

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

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

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

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

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

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

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

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## Comments

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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,

\(\frac{z-y}{z-x} = 3\) – Kartik Sharma · 2 years, 2 months ago

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3 :D – Handayani Basuki · 2 years, 2 months ago

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the answer is 3 – Dheeraj Agarwal · 2 years, 2 months ago

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3 – Farman Saifi · 2 years, 2 months ago

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3 is the answer – Sauparna Paul · 2 years, 2 months ago

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Is there a typo in eqn 1 ... should it be \(x+\frac{1}{yz}= \frac{1}{5}\) – Michael Fischer · 2 years, 2 months ago

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