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by **Luke53** on Mon May 16, 2011 2:57 pm

How to solve?
x/10 = y/14 = z/35 = 1/(10x + 14y + 35z)
Have no clue how to start solving this system.
Luke.

**Sponsor**

by **stapel_eliz** on Mon May 16, 2011 6:19 pm

Luke53 wrote:How to solve?
x/10 = y/14 = z/35 = 1/(10x + 14y + 35z)

One method might be to solve x/10 = y/14 and x/10 = z/35 for y and z in terms of x. Then plug this into x/10 = 1/(10x + 14y + 35z), and solve for the value(s) of x.

Whatever method you use, it's likely gonna be messy, since this isn't linear. :wink:

by **Luke53** on Tue May 17, 2011 4:36 pm

One method might be to solve x/10 = y/14 and x/10 = z/35 for y and z in terms of x. Then plug this into x/10 = 1/(10x + 14y + 35z), and solve for the value(s) of x.

Whatever method you use, it's likely gonna be messy, since this isn't linear.

stapel_eliz

This method works out very well, tried it out and works fine, (and not too messy):
y = 14x / 10 ; z = 35x / 10 susbt. this in the eqn. x/10 = 1/(10x + 14y +35 z) gives: x/10 = 1/(10x + 14*(14x/10) + 35*(35x/10)
x/10 = 1/(10x +19.6x + 122.5x) => x/10 = 1/152.1x and so 152.1x² = 10 => 152.1x² - 10 = 0 (quadratic) with solutions: x = 0.25641 or - 0.25641
It's easy now to find the values for y and z. (y = 0.358974 and z = 0.897435)

Thanks .

Luke.

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