How to solve the equations system?

Question

I have a system of equations that I don't know how to solve.

1) $x = a - y$ ;

2)$ y = b \times sin(90 -z)$;

3) $z = \dfrac{(x - c )^2 }{b^2 \times e^2}$

$a, b, c, d$ and $e$ are known. How can I solve for $x, y$ and $z$?

Answer 1

As suggested by pbs, you eliminate $x$ for the first equation, replace it into the third equation to get $z$ and now you are left with $$y=\cos \left(\frac{(a-c-y)^2}{b^2 e^2}\right)$$ which not the most pleasant I know.

Let me use your numbers $a=200,b=50,c=85,e=2$. So, the equation is $$y=\cos \left(\frac{(115-y)^2}{10000}\right)$$ and so you want to solve for $y$ $$f(y)=y-\cos \left(\frac{(115-y)^2}{10000}\right)=0$$ What you can first notice is that $$f(0)=-\cos \left(\frac{529}{400}\right) \simeq -0.245753$$ Looking at the plot of f(y) as a function of $y$ reveals an almost linear behavior.

For solving nonlinear equations such as the present one, Newton method is one of the simplest tools. Starting from a "reasonable" guess $y_0$, the procedure will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ For the equation we look for $$f'(y)=\frac{(y-115) \sin \left(\frac{(115-y)^2}{10000}\right)}{5000}+1$$ So, let us apply Newton starting at $y=0$; this will generate the following iterates :$0.251357$, $0.251346$ which is the solution for six significant figures.

So, now, using $y=0.251346$, going backward, we find $x=199.749$ and $z=1.31673$.

What you also can do is to define $$y_{n+1}=\cos \left(\frac{(115-y_n)^2}{10000}\right)$$ and start with $y_0=0$. In this case, the following iterates will be obtained : $0.245753$, $0.251222$, $0.251344$, $0.251346$.