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."