How to solve 5 simultaneous equations having 3 linear and 2 non linear equations

Question

How do i solve the below equations? $$ x+y+z-1=0 $$ $$ 2a+2b+4z-2.0272=0$$ $$ x+2y+b-1.5778=0$$ $$ 1.0115xb-ya=0$$ $$ 1.0207a^2-z(x+y+z+a+b)=0$$

Answer 1

I'd probably start eliminating variables one-by-one as long as the polynomial degrees stay low. Let's start with $x$. If you solve the first equation for $x$, you obtain $x=1-y-z$. Substitute that in wherever you see $x$, and your five equations become four:

$$2a+2b+4z-2.0272=0$$ $$y-z+b-0.5778=0$$ $$1.0115b-1.0115by-1.0115bz-ya=0$$ $$1.0207a^2-az-bz-z=0$$

Now, the second of these two equations is easy to solve for $y$, giving us $y=0.5778+z-b$. This replaces all remaining $y$s, turning four equations into three:

$$2a+2b+4z-2.0272=0$$ $$ab-az+1.0115b^2-2.023bz-0.5778a+1.0115(0.4222)b=0$$ $$1.0207a^2-az-bz-z=0$$

The top remaining equation can be solved for $a$ without too much trouble: $a=1.0136-b-2z$.

Plugging this into the other two, we're left with two equations in $b$ and $z$, and these equations are quadratic in both variables. You can use the quadratic formula to solve for $z$ in terms of $b$ in one of them, and substitute each of the two expressions you obtain into the other, each giving you one nasty equation in $b$ with no other variables. This one can be solved numerically using something like Newton's method, and then back-substitution will get you all the way home.

In this case, we never saw anything higher than degree $2$ until the very end, which was helpful. If it had gotten out of hand more quickly, we'd probably have to resort to something like Groebner basis calculations, which are prohibitive unless you've got software like Maple or Mathematica.