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.