# Section 3.2.2: Ordering Equations Solutions

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

2.
Unknowns: x1, x2, y1, y2
There are four equations and four unknowns, therefore the set may be solved.

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

3.

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

4.

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

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