SOLUTION: 1/(3x) + 1/(2y) + 1/(2z) = -1
1/(5x) - 1/(3y) + 1/(4z) = -1/3
1/(2x) + 1/y + 1/(2z) = -1/2
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Question 1016859: 1/(3x) + 1/(2y) + 1/(2z) = -1
1/(5x) - 1/(3y) + 1/(4z) = -1/3
1/(2x) + 1/y + 1/(2z) = -1/2
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
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Is it correct?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
1/(3x) + 1/(2y) + 1/(2z) = -1
1/(5x) - 1/(3y) + 1/(4z) = -1/3
1/(2x) + 1/y + 1/(2z) = -1/2
====================
Sub r = 1/x, s = 1/y & t = 1/z
-------
r/3 + s/2 + t/2 = -1
r/5 - s/3 + t/4 = -1/3
r/2 + s + t/2 = -1/2
------
Eliminate fractions
2r + 3s + 3t = -6
12r - 20s + 15t = -20
r + 2s + t = -1
==================
Eliminate r
2r + 3s + 3t = -6
2r + 4s + 2t = -2
-------------------------- Subtract
-s + t = -4 Eqn A
==============
12r + 18s + 18t = -36
12r - 20s + 15t = -20
--------------------------- Subtract
38s + 3t = -16
-3s + 3t = -12 Eqn A times 3
----------------------------- Subtract
41s = -4
s = -4/41 --> y = -41/4
===========================
-s + t = -4 Eqn A
t = s-4
t = -168/41 --> z = -41/168
===========================
r + 2s + t = -1
r = -1 - 2s - t = 1 +8/41 + 168/41
r = 217/41 ---> x = 41/217
==========================================
Check it carefully, mistakes are possible, but that's the method.
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