a) In determining leading variables, you must look at the whole system, not just individual equations. You should not use the term "leading variable" unless the system has been reduced and is consistent. In the first system x, z are leading, and y is free. ("5y" is not one of the variables.) There are solutions (infinitely many: but you did not need to say that, and you did not need to give the solution set). If you do give a solution set, say explicitly what it is. For full credit on the first system you needed to say explicitly that there are solutions.

b) The second system is reduced, not consistent and therefore has no solutions. You did not need to say anything else.

c) The third system is NOT reduced. This was an error that several people made.

The coefficient of y in the second equation is not 1, and the column of y has not been cleared. Likewise, the column of z has not been cleared. Once the system is reduced, each leading variable occurs in only one equation, and with coefficient one. If the system is consistent, then there is exactly one leading variable for each nonzero equation (i.e., each equation that is not 0 = 0).

2.

In the second problem you were asked to carry through the
reduction of the system. You should subtract 3 times the
first equation from the second and then write BOTH. The
first step in reduction produces

x+y+z = a

0 = 6-3a

Unless you wrote down both of the above equations as a system you did not receive full credit.

The second equation shows that the system is inconsistent if a is different from 2 (no solutions in that case). If a = 2, there are infinitely many solutions (the second equation becomes 0 = 0, while the first now has one leading and two free variables).

It is best to answer all parts of the question explcitly, so you should also say that there is no value of a for which the system has a unique solution.

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