We have two equations and two unknowns, x 1 and x 2.
.0202x 1 + .0062x 2 – .0113 = 0
.0062x 1 + .0270x 2 – .0078 = 0
.0202x 1 = –.0062x 2 + .0113
x 1 = –.3069x 2 + .5594
Having solved for x 1 , we can now substitute x 1 into the following x 2
equation:
.0062 (–.3069x 2 + .5594) + .027x 2 = .0078
–.0019x 2 + .0035 + .027x 2 = .0078
.0251x 2 = .0043
x 2 = .1713
x 1 = (–.3069)(.1713) + .5594 = .5068
x 3 = 1 – .5068 – .1713 = .3219
∂σ
∂
p
x
xx
2
2
=+−=... 027 210062 0078 0
∂σ
∂
p
x
xx
xx
2
2
2
22
1
2
22
2 1107 0811 2 2988 1107 0811 2 0811
2 0184 0710 1107 2 1646 0710 0811 2 2988 1107 0811
2 0811 2 0811 2 2988 1107 0811
2210123 0066 0054 0132 0003
=+−+
+−−
−++
=+−++−
[(. ) (. ) (. )(. )(. )] [ (. )
(. )(. )(. ) (. )(. )(. ) (. )(. )(. )]
(. ) (. ) (. )(. )(. )
(... ) (...
00190019 0054 0132 0054
2210135 0062 0078 0
−−+
=+−=
.)..
xx(. ) (. ).
∂σ
∂
p
x
xx
2
1
=+−=... 0202 120062 0113 0
∂σ
∂
p
x
xx
xx
2
1
1
22
2
2
2
12
2 0710 0811 2 1646 0710 0811 2 0811
2 0184 0710 1107 2 1646 0710 0811
2 2988 1107 0811 2 0811 2 1646 0710 0811 0
2 0037 0066 0002 0132 0003 0019
=+−+
+−
−−+=
=+=++−−
[(. ) (. ) (. )(. )(. )] [ (. )
(. )(. )(. ) (. )(. )(. )
(. )(. )(. )] (. ) (. )(. )(. )
(... ) (...
..)
..
(. ) (. ).
0054
0132 0019
2120101 0062 0113
−+
=+−xx