1.
Unknowns: x1, x2, t, p
There are four unknowns and four equations, therefore the set may be solved.
Incidence Table
| Equation | x1 | x2 | t | p |
| (1) | 1 | 1 | ||
| (2) | 1 | 1 | ||
| (3) | 1 | |||
| (4) | 1 | 1 | 1 | 1 |
Rearranging and reordering gives the following incidence matrix:
| Equation | t | x1 | x2 | p |
| (3) | 1 | |||
| (2) | 1 | 1 | ||
| (1) | 1 | 1 | ||
| (4) | 1 | 1 | 1 | 1 |
This matrix is indeed lower triangular and so the equations may be solved in the following order:
(3) t = 4/a
(2) x1 = -3/(t + t2)
(1) x2 = 1 - x1
(4) p = (-t x2)/x1
Incidence Table:
| Equation | x1 | x2 | y1 | y2 |
| (1) | 1 | 1 | 1 | 1 |
| (2) | 1 | |||
| (3) | 1 | 1 | ||
| (4) | 1 | 1 | 1 |
Rearranging and reordering gives the following incidence matrix:
| Equation | x1 | y1 | x2 | y2 |
| (2) | 1 | |||
| (3) | 1 | 1 | ||
| (4) | 1 | 1 | 1 | |
| (1) | 1 | 1 | 1 | 1 |
This matrix is indeed lower triangular and so the equations may be solved in the following order:
(2) x1 = -3
(3) y1 = 1 - x1
(4) x2 = y1/x1
(1) y2 = x1 + x2 - y1
Unknowns: k, P, T, v,
There are five unknowns and five equations, therefore the set may be solved.
Incidence Table:
| Equation | k | P | T | v | ![]() |
| (1) | 1 | 1 | 1 | ||
| (2) | 1 | 1 | |||
| (3) | 1 | ||||
| (4) | 1 | 1 | 1 | ||
| (5) | 1 | 1 |
Rearranging and reordering gives the following incidence matrix:
| Equation | ![]() | v | T | P | k |
| (3) | 1 | ||||
| (5) | 1 | 1 | |||
| (2) | 1 | 1 | |||
| (4) | 1 | 1 | 1 | ||
| (1) | 1 | 1 | 1 |
This matrix is indeed lower triangular and so the equations may be solved in the following order:
(3)
= 0.76
(5) v = 1/
(2) T = a + bv + cv2
(4) P = RT/v
(1) k = (10A + B/T)/P
When i = 1, 2 the equations are:
(1) f1 = (1 +
) l1
(2) f2 = (1 +
) l2
(3)
= v1/l1
(4)
= v2/l2
(5)
/
=
(6) l1 = 10
unknowns:
, l1,
,
l2, v1, v2
There are six unknowns and six equations, therefore the set may be solved.
Incidence Table:
| Equation | ![]() | l1 | ![]() | l2 | v1 | v2 |
| (1) | 1 | 1 | ||||
| (2) | 1 | 1 | ||||
| (3) | 1 | 1 | 1 | |||
| (4) | 1 | 1 | 1 | |||
| (5) | 1 | 1 | ||||
| (6) | 1 |
Rearranging and reordering gives the following incidence matrix:
| Equation | l1 | ![]() |
v1 | ![]() |
l2 | v2 |
| (6) | 1 | |||||
| (1) | 1 | 1 | ||||
| (3) | 1 | 1 | 1 | |||
| (5) | 1 | 1 | ||||
| (2) | 1 | 1 | ||||
| (4) | 1 | 1 | 1 |
This matrix is indeed lower triangular and so the equations may be solved in the following order:
(6) l1 = 10
(1)
= (f1/l1) - 1
(3) v1 =
l1
(5)
=
/
(2) l2 = f2/(1 +
)
(4) v2 =
l2
Section 3.2.1: Ordering Equations Questions
Course Organiser Last Modified 2/9/00