This is documentation for Mathematica 3, which was
based on an earlier version of the Wolfram Language.
 3.4.7 Advanced Topic: Existence of Solutions Using Reduce, you can find out under exactly what conditions a particular set of equations has solutions. Solve tells you whether any generic solutions exist. There is no value of x which solves these simultaneous equations. Reduce thus simplifies the logical statement x==1&&x==2 to the explicit value False. In[1]:= Reduce[ x == 1 && x == 2 , x ] Out[1]= There is a solution to these equations, but only when a has the special value 1. In[2]:= Reduce[ x == 1 && x == a , x ] Out[2]= The solution is not generic, and is rejected by Solve. In[3]:= Solve[ x == 1 && x == a , x ] Out[3]= But if a is constrained to have value 1, then Solve again returns a solution. In[4]:= Solve[ x == 1 && x == a && a == 1, x ] Out[4]= This equation is true for any value of x. In[5]:= Reduce[ x == x , x ] Out[5]= This is the kind of result Solve returns when you give an equation that is always true. In[6]:= Solve[ x == x , x ] Out[6]= When you work with systems of linear equations, you can use Solve to get generic solutions, and Reduce to find out for what values of parameters solutions exist. Here is a matrix whose element is . In[7]:= m = Table[i + j, {i, 3}, {j, 3}] Out[7]= The matrix has determinant zero. In[8]:= Det[ m ] Out[8]= This makes a set of three simultaneous equations. In[9]:= eqn = m . {x, y, z} == {a, b, c} Out[9]= Solve reports that there are no generic solutions. In[10]:= Solve[eqn, {x, y, z}] Out[10]= Reduce, however, shows that there would be a solution if the parameters satisfied the special condition a==2b-c. In[11]:= Reduce[eqn, {x, y, z}] Out[11]= For nonlinear equations, the conditions for the existence of solutions may be very complicated. Here is a very simple pair of nonlinear equations. In[12]:= eqn = {x y == a, x^2 y^2 == b} Out[12]= Solve shows that the equations have no generic solutions. In[13]:= Solve[eqn, {x, y}] Out[13]= Reduce gives the complete conditions for a solution to exist. In[14]:= Reduce[eqn, {x, y}] Out[14]=