Hello: I am having a hard time with this problem: 
For which values of "a" does the following system have zero solutions? One
solution? Infinitely many solutions? 
x1 + x2 + x3 = 4
x3 = 2
(a^2 - 4)x3 = a - 2 

I wasnt sure if I should substitute x3 into the third equation, then solve for
a. I re-read the chapter section and there are not any examples. 

x1 + x2 +        x3 = 4
                 x3 = 2
           (a²-4)x3 = a-2

Since x3 = 2, if we substitute 2 for x3 in the third equation we get

            (a²-4)(2) = a-2

              2a² - 8 = a - 2 
   
          2a² - a - 6 = 0

      (2a + 3)(a - 2) = 0

 2a + 3 = 0.    |     a - 2 = 0
                |         a = 2
     2a = -3    |
      a = -3/2  |

In order to have a solution at all, a has to be one of these.
So if a is neither of these, there are no solutions.

If a = -3/2 the bottome equation comes out x3 = 2, identical with the
second equation.

And if we substitute 2 for x3 in the first equation, we get

x1 + x2 + 2 = 4 or

x1 + x2 = 2 or

x1 = 2 - x2.

So there are infinitely many solutions in this case, because there are
infinitely many choices for x2 and each one gives a different value for x1.

If a = 2 , the bottome equation becomes 0x3 = 0, so any value of x3 would
satisfy that, but the middle equation tells us that x3 = 2.  So that still
gives us an infinite number of solutions.

So a = -3/2 or a = 2 gives infinitely many solutions. Any other value of a
gives no solutions.  There are no values of a which give just one solution.

Edwin