` `
log(`w`) + 3 = 0

`w` + 5 `x`^{2} - ` y` - 4 = 0

`x` + log(` y`) = 0

`w` + `x ` +` y` - ` z`^{0.5} + 10 = 0

Without performing any analysis of the equation, we might suppose that it would be necessary to guess the values of all 4 unknowns and iterate on all of them simultaneously, using techniques which we have not yet discussed.

However, experience with problems involving direct solution where we could reorder and rearrange sets of equations to solve without any iteration whatsoever, tells us that we should look more closely at the equations.

Here is an incidence matrix for the
equations as written above in the rows and the variables in order
`w, x, y, z` in the columns.

It is obvious that the first equation contains only the unknown
`w` and so can immediately be solved after rearranging to the
formula:

`w` = exp(-3) = 0.0498

`w` is now known and so is no longer a variable. We can
if we wish remove the `w` column from the incidence table
and look at the remaining 3 variable problem. In particular,
`w` is no longer an unknown in the second equation, which
now effectively contains only `x` and `y`.
However, all the other equations also contain at least both
`x` and `y` and so it is not possible to go any further with
`direct' solution.

In fact, it can be seen that the second and third equations now
contain both `x` and `y` and ** only**
these unknowns. They must be solved simultaneously, and for this purpose
could be rewritten with the value of `w` substituted in as:

0.0498 + 5 `x`^{2} - ` y` - 4 = 0

`x` + log(` y`) = 0

Although the above equations will have to be solved simultaneously somehow,
once we have done this and determined values for `x` and `y`
it can be seen that in the last equation there now remains only the single
unknown, `z`. This can then be determined by rearrangement:

`z` = (10 + `w` + `x ` +` y`)º

- The `head' of the equation set which is a subset of the equations which
can be solved directly by rearrangement. In this case the head contains
only one equation, which
may be solved for
`w`, but in some cases the whole set may be solved this way. - The `partition' which is a subset of the equations which
**must**be solved simultaneously. In this case there are 2 equations in the partition which require simultaneous solution for`x`and`y`. - A `tail' of the set which can be solved by simple rearrangement
once the partition has been solved. In this case there is only one equation
in the tail, to be solved for
`z`.

- Start to find the `head' by looking for any row (equation) with only one nonzero entry (variable). This can be solved on its own. Remove the variable (column) from the set and look for any other single variables. We can reorder the `head' subset of the table to be lower triangular as is the case for any set of equations which can be solved directly.
- Start to find the `tail' by looking for any variable (column) which only occurs in a single equation (row). This variable can be solved for once all others in the problem have been determined. Remove the equation (row) and look for any other variables which now occur only in one equation. The table for the tail may also be made lower triangular.
- What is left after the head and tail are removed must be the
partition. This has an incidence matrix which
**cannot**be made lower triangular no matter how rows and columns are interchanged.

`x` + log(` y`) = 0

Suppose we knew `x` and rearranged the first equation to
give ` y`:

` y` = 0.0498 + 5 `x`^{2} - 4

If we then substituted the values of `x` and `y`
into the LHS of the second equation then it should equal zero.

However, if we only ** guess ** the value of `x`,
but nonetheless calculate ` y` and substitute it into
the second equation, the latter will evaluate to zero if we
have chosen the correct value of ` x`.
In effect the two equations if rearranged appropriately
and treated in sequence:

` y` = 0.0498 + 5 `x`^{2} - 4

`f` = `x` + log(` y`)

effectively result in the evaluation of a single function of
`x`:

`f`(`x`) = `x` + log[0.0498 + 5 `x`^{2} - 4]

This is just a single equation in `x` and can be
solved using any suitable single equation solving method.
However the function evaluation is carried out in **two** steps
rather than in one.
** Note that we do not actually make the substitution shown above.
Doing the evaluation in two steps saves us the need to carry out
any algebra other than the rearrangement of the first equation.**

This method of solving sets of equations is called `tearing' and the
single variable `x` for which we solve is called the
`tear variable'. In a more complicated system of equations it
may be necessary to tear, i.e. guess and iterate
on, more than one tear variable, and so it will not be possible to use the simple solution schemes which have been described for single unknowns.
However, many practically useful problems may be reduced to
iteration in a single variable.

Inspection of the incidence matrix for the partition will show how many tear variables are needed for a given problem: this is equal to the number of variables (columns) which have nonzero table entries above the diagonal of the matrix. For this example it can be seen that there is only one.

Next - Section 3.5.2: Additional Notes

Return to Section 3 Index