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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 $z$?

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You could try substituting (1) into (3) and then (3) into (2). Then solve for $y$. Then use your result for $y$ to obtain the remaining unknowns. –  pbs Jul 21 at 8:37
    
pbs gave you the way. The problem will be solve the equation for $y$ and there will not be analytical solution. So, numerical methods, such as Newton, would probably be required. Also notice that $\sin(90-z)=\cos(z)$. Give me some numbers to play with. –  Claude Leibovici Jul 21 at 8:43
    
@pbs But I don't know how to get y square out of sin. –  Mike Shaw Jul 21 at 8:46
    
This is exactly the reason of my comment. –  Claude Leibovici Jul 21 at 8:46
    
@ClaudeLeibovici For instance, a = 200, b = 50, c = 85 , no d , e = 2 –  Mike Shaw Jul 21 at 8:53

1 Answer 1

up vote 0 down vote accepted

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 : $0.245753$, $0.251222$, $0.251344$, $0.251346$.

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My question is more like how to express x y z with a b c and e. Btw, I'm so sorry for not telling you z is in degree so is 90. But I don't know how to edit the question to add degree. –  Mike Shaw Jul 21 at 9:24
    
What you are being told, Mike, is that you can't express $x,y,z$ in terms of $a,b,c,e$. All you can do is find a numerical solution when given particular values of $a,b,c,e$. –  Gerry Myerson Jul 21 at 10:04

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