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# solve it!!!!

$$x+\dfrac{1}{zy}=\dfrac{1}{5}$$

$$y+\dfrac{1}{xz}=\dfrac{-1}{15}$$

$$z+\dfrac{1}{xy}=\dfrac{1}{3}$$

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

Note by Dheeraj Agarwal
2 years 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,

$$\frac{z-y}{z-x} = 3$$ · 2 years ago

3 :D · 2 years ago

the answer is 3 · 2 years ago

3 · 2 years ago

3 is the answer · 2 years ago

Is there a typo in eqn 1 ... should it be $$x+\frac{1}{yz}= \frac{1}{5}$$ · 2 years ago