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$ \frac{a_1}{x} + \frac{b_1}{y} + \frac{c_1}{z} = d_1 \\ \frac{a_2}{x} + \frac{b_2}{y} + \frac{c_2}{z} = d_2 \\ \frac{a_3}{x} + \frac{b_3}{y} + \frac{c_3}{z} = d_3 $

$ a_i, b_i, c_i, d_i $: known constants
$ x, y, z $: unknown variables

First of all, is this equation set linear?

I want to learn method(s) of solving this kind of equation sets by using paper and pencil only. The point in which I'm stuck is that, when I equate the denominators, the equation turns into a non-linear form like below:

$ a_1yz + b_1xz + c_1xy = d_1xyz \\ a_2yz + b_2xz + c_2xy = d_2xyz \\ a_3yz + b_3xz + c_3xy = d_3xyz \\ $

What is the proper way(s) of solving this equation set?

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    $\begingroup$ Let $\xi = \frac 1x$, $\eta = \frac 1y$, $\zeta = \frac 1z$, then you have a linear equation for $\xi$, $\eta$, $\zeta$. $\endgroup$ – martini Sep 6 '12 at 8:19
  • $\begingroup$ @martini Wow, so simple? $\endgroup$ – hkBattousai Sep 6 '12 at 8:21
  • $\begingroup$ Why not? ;-) ${}{}$ $\endgroup$ – martini Sep 6 '12 at 8:23
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    $\begingroup$ You just have to remember that the new variables can't be zero (and that you have to re-substitute in the end.) $\endgroup$ – Simon Markett Sep 6 '12 at 8:29
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The equation is linear in $u:=1/x$, $v:=1/y$ and $w:=1/z$. Do that substitution and your problem is easily solved.

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