System of 3 simultaneous equation where one is non-linear Hi I have a problem where I need to solve the following set of equations:
$$
v = U u 
$$
$$
u = 1 -Uv
$$
$$
U^2 = u^2 + v^2
$$
I have tried subbing $u$ and $v$ into the expression for $U^2$ but it seems to get very messy very quickly.
Any help solving for $u$,$v$ and $U$ would be greatly appreciated.
 A: $v = Uu, u = 1-Uv, U^2 = u^2 + v^2$
I solve the first for $U = v/u$ and substitute into both of the others:
$$u = 1-\frac{v^2}{u}, \frac{v^2}{u^2} = u^2 + v^2$$
Then the first becomes
$$ u^2 = u - v^2$$ and the second becomes
$$v^2 = u^4 + u^2v^2$$
The former rearranges into $u = v^2 + u^2$ and putting that into the original third equation, I find
$U^2 = u$.  So
$$\frac{v^2}{u^2} = U^2 = u,$$ so $v^2 = u^3$ or $v = u^{3/2}$
Finally, $u = 1-Uv$ becomes $u = 1-u^2$, which quadratic equation solves to
$$u = {{-1\pm\sqrt{5}}\over {2}}$$
A: You have three equation
$$ v = U u , \tag{1} $$
$$ u = 1 -Uv , \tag{2}$$
$$ U^2 = u^2 + v^2. \tag{3}$$
Substitute $\,v\,$ in equation $(1)$ into equation $(2)$ to get
$$ u = 1 - U^2u. \tag{4} $$
Solve this linear equation for $\,u\,$ to get
$$ u = \frac1{1+U^2}. \tag{5} $$
Use equation $(1)$ to get
$$ v = \frac{U}{1+U^2}. \tag{6} $$
Substitute both these values in equation $(3)$ to get
$$ U^2 = \frac1{(1+U^2)^2} + \frac{U^2}{(1+U^2)^2} = 
\frac{1+U^2}{(1+U^2)^2} = \frac1{1+U^2}. \tag{7} $$
Let $\,x := U^2\,$ to get
$$ x = \frac1{1+x} \tag{8} $$
which implies that $\,x\,$ is a solution to the quadratic
$$ x^2+x-1 = 0. \tag{9} $$
This quadratic has two solutions for $\,x\,$ and since $\,x=U^2\,$
each value of $\,x\,$ has two square root values for $\,U.\,$ Thus,
there are four solutions for $\,U\,$ and equations $(5)$ and $(6)$
give the corresponding values for $\,u\,$ and $\,v.$
