The equation
yN2 == 1 - yCO - yCO2
adds no information relevant to estimating a
, b
, or c
so you really have two equations and three unknowns.
However, as an example you can solve for a
and b
in terms of c
, yCO
, and yCO2
:
Solve[yCO == (0.015 a + 0.05 b)/(a + b + c) && yCO2 == (0.07*a)/(a + b + c), {a, b}]
{{a -> -((1.42857 c yCO2)/(-0.1 + 2. yCO + 1. yCO2)),
b -> (0.428571 (-4.66667 c yCO + 1. c yCO2))/(-0.1 + 2. yCO + 1. yCO2)}}
A slightly cleaner solution would be to "rationalize" all of the coefficients:
Solve[yCO == ((15/1000) a +(5/100) b)/(a + b + c) && yCO2 == ((7/100)*a)/(a + b + c), {a, b}]
{{a -> -((100 c yCO2)/(7 (-1 + 20 yCO + 10 yCO2))),
b -> -((10 c (14 yCO - 3 yCO2))/(7 (-1 + 20 yCO + 10 yCO2)))}}