linear equation by Gaussian

As follows:

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2010-04-10 14:58:20 補充：
http://i707.photobucket.com/albums/ww74/stevieg90/11-14.jpg

gaussian elimination?

Stevie-G 死圖

(a)
x+y=8 => x=8-y
Sub this into the first to equation
p(8-y)+qy=4 and q(8-y)+py=4
sum up the two equations above, LHS+LHS, RHS+RHS
p(8-y)+qy+q(8-y)+py=4+4
8p-py+qy+8q-qy+py=8
ie 8p+8q=8
ie p+q=1

2010-04-10 12:21:43 補充：
first to equation->first two equations (typo)

2010-04-10 12:25:29 補充：
(b) for 3p=5q, need to solve p,q first with p+q=1
from 3p=5q, q=0.6p, put back into p+q=1
ie1.6p=1
p=1/1.6=5/8
then q=3/8

2010-04-10 12:30:37 補充：
sub back p=5/8, q=3/8 into the simutaneous eqn
multiply the first two by 8, the third by 4
5x+3y=32 (eqn 1) and 5y+3x=32 (eqn 2) and 4x+4y=32 (eqn 3)
either do eqn1 - eqn2, eqn1 - eqn3, eqn2 - eqn3, LHS-LHS, RHS-RHS
you will get x=y
thus put back to x+y=8
x=y=4

2010-04-10 12:35:39 補充：
Gaussian elimination (if needed)
sub back p=5/8, q=3/8 into the simutaneous eqn
5x/8 + 3y/8 = 4 (eqn 1) and 5y/8 + 3x/8 = 4 (eqn 2) and x + y = 8 (eqn 3)
(5/8...3/8...4)
(3/8...5/8...4)
(1......1......4)

2010-04-10 12:45:28 補充：
should be
(5/8...3/8...4)
(3/8...5/8...4)
(1......1......8)
then (row 2 times 8)minus(row 1 times 8), leave it in row 2
(5/8...3/8...4)
(-2.....2.....0)
(1......1......8)
then row 2 plus (row 3 times 2), leave it in row 2
(5/8...3/8...4)
(0.....4.....16)
(1......1......8)

2010-04-10 12:49:33 補充：
then (row 3 times 5/8) minus row 1, leave it in row 3
(5/8...3/8...4)
(0.....4.....16)
(0.....1/4.....1)
then (row 3 times 16) minus row 2, leave it in row 3
and divide row 2 by 4
(5/8...3/8...4)
(0.....1.....4)
(0.....0.....0) (row echelon form)

2010-04-10 12:50:48 補充：
then (row 1 times 8) minus (row 2 times 3), leave it in row 1
(5....0....20)
(0.....1.....4)
(0.....0.....0)
then divide row 1 by 5
(1.....0.....4)
(0.....1.....4)
(0.....0.....0) (reduced row echelon form)
from this, we know, x=4, y=4

2010-04-10 12:59:48 補充：
Gaussian elimination for (a)
(p.....q.....4)
(q.....p.....4)
(1.....1.....8)
then (row 2 times p/q) minus row 1, leave it in row 2, then times row 2 by q
(row 3 times p) minus row 1, leave it in row 3
(p.........q.................4)
(0.....p^2-q^2.....4(p - q))
(0........p-q............8p-4)

2010-04-10 13:07:57 補充：
then (row 3 times (p+q)) minus row 2, leave it in row 3
p(8p+8q-8)=0
to have consistence, p+q=1

2010-04-10 13:08:09 補充：
p=0 is rejected