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.
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)