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Getting Full SolutionsEliminating Variables

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 ]


There is a solution to these equations, but only when a has the special value 1.

In[2]:= Reduce[ x == 1 && x == a , x ]


The solution is not generic, and is rejected by Solve.

In[3]:= Solve[ x == 1 && x == a , x ]


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 ]


This equation is true for any value of x.

In[5]:= Reduce[ x == x , x ]


This is the kind of result Solve returns when you give an equation that is always true.

In[6]:= Solve[ x == x , x ]


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}]


The matrix has determinant zero.

In[8]:= Det[ m ]


This makes a set of three simultaneous equations.

In[9]:= eqn = m . {x, y, z} == {a, b, c}


Solve reports that there are no generic solutions.

In[10]:= Solve[eqn, {x, y, z}]


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}]


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}


Solve shows that the equations have no generic solutions.

In[13]:= Solve[eqn, {x, y}]


Reduce gives the complete conditions for a solution to exist.

In[14]:= Reduce[eqn, {x, y}]


Getting Full SolutionsEliminating Variables