Try to solve: Seeing "x" begining each of the equations, subtract the first from the second to get y+ 2z= 4 and the first from the third to get

.

Mr F says: Careless here. It's
Now subtract the first of those two from the second to cancel the "y"s:

.

Mr F says: Wrong. A ripple effect from the above mistake. The correct result (found in a simpler way) is given in pst #6.
If [tex]\lambda- 5

[/itex] is NOT 0, we can divide both sides by that to get a specific value for z and then it would be easy to solve for x and y.

Mr F says: Mixing your latex tags. Tsk tsk.
As long as

is not 0, we have a unique solution.

Mr F says: Darn missing b ..... "As long as "
If

, in other words if

we cannot divide to find z but if it also happens that

that equations just says "0z= 0" which is true for all z.

If [tex]

\lamba= 5

[/itex] and

, there are an infinite number of solutions.

Mr F says: Darn missing d ..... mixing latex tags ..... darn \m "If and , there are an infinite number of solutions."
If

and

is NOT 14, then 0z would have to be equal to a non-zero number and that is impossible.

If

and

is not 14, there is no solution.

Mr F says: Darn missing d again ...... "If and is not 14, there is no solution."