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Question Number 11764 by Peter last updated on 31/Mar/17

if x y and z are solution of

\frac{x + xy}{x + y + 1} = 2

\frac{2 x + xz}{x + z + 2} = 3

\frac{2 + 2 y + z + yz}{y + z + 3} = 4

so, the value of

\frac{1}{x} + \frac{1}{y + 1} + \frac{1}{z + 2} = \dots . ?

Answered by ajfour last updated on 31/Mar/17

\frac{x (y + 1)}{x + (y + 1)} = 2; \frac{x (z + 2)}{x + (z + 2)} = 3;

\frac{(y + 1) (z + 2)}{(y + 1) + (z + 2)} = 4

\Rightarrow 2 (\frac{1}{x} + \frac{1}{y + 1} + \frac{1}{z + 2}) = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}

\Rightarrow \frac{1}{x} + \frac{1}{y + 1} + \frac{1}{z + 2} = \frac{13}{24} .

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