Question

Please send the solution

Answer 1

Answer:

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Answer 2

Answer:

Let 2x = 3y = 6-2 = k

= 2x = k

Therefore, k 1/x= 2

Similarly, k 1/y = 3 and k = - 1/z = 6

= k - 1/z=2 × 3

= k - 1/z = k 1/x × k 1/y

= k - 1/z = k 1/x + 1/y

= - 1/2 = 1/x + 1/y

Therefore, 1/x + 1/y + 1/z = 0

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