title: Solving a system of equation with Sympy, python2.7
body: <p>I want to solve a system of equations. But I want to be able to precise the value to "get", and as a function of "what".</p>
  
  <p>To better understand, I take an exemple from <a href="https://stackoverflow.com/questions/22156709/solving-system-of-nonlinear-equations-with-python#_=_">here</a>, wich I modfified:</p>
  
  <pre><code>import sympy as sp
  x, y, z = sp.symbols('x, y, z')
  rho, sigma, beta = sp.symbols('rho, sigma, beta')
  f1 = sigma * (y - x)
  f2 = x * (rho - z) - y
  f3 = x * y - beta * z
  print sp.solvers.solve((f1, f2, f3), (x, y, z))
  </code></pre>
  
  <p>in </p>
  
  <pre><code>import sympy as sp
  x, y, z, w = sp.symbols('x, y, z, w')
  rho, sigma, beta = sp.symbols('rho, sigma, beta')
  f1 = sigma * (y - x)
  f2 = x * (rho - z) - y
  f3 = x * y - beta * w
  f4 = z - w
  print sp.solvers.solve((f1, f2, f3, f4), (x, y, z))
  </code></pre>
  
  <p>So, as you can see, I replace <strong>z</strong> by <strong>w</strong> in the last equation and I add a new to precise <strong>z = w</strong>. 
  But, <strong>sympy (on python 2.7) is unable to solve this new system of equation!!</strong></p>
  
  <p><em>So my question:</em> How to get the result for x, y, z as a function of rho, sigma, beta. And more generally, how do we precise the variable "response variable".</p>
  
  <p><em>I think that could be very helpfull because, often, you don't want to developp your system of equation before asking python to solve it.</em></p>
  
  <p>In the same way, if I take a more complex example:</p>
  
  <pre><code>import sympy as sp
  x, y, z, w, u = sp.symbols('x, y, z, w, u')
  rho, sigma, beta = sp.symbols('rho, sigma, beta')
  f1 = sigma * (y - x)
  f2 = x * (rho - u) - y
  f3 = x * y - beta * w
  f4 = z - w
  f5 = w - u
  print sp.solvers.solve((f1, f2, f3, f4, f5), (x, y, z))
  </code></pre>
  
  <p>The response I get is: </p>
  
  <pre><code>[]
  </code></pre>
  
  <p>But as you see I have z = w = u
  Son I should get the same answer!</p>