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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]= 
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