# Page 59 Ex 4 by qfc10548

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```									 Method of elimination

Page 50 Ex 4

5x + 2y = 12

4x -y = 7

x    y
Eq 1                5     2   12      Same as 5x +
eq 2                4    -1    7      Same as 4x

Eq 1                5     2   12
Copy. Go to edit. Pas
(Eq 2 )*2           8    -2   14
special. Values only

x    y
Eq 1               5      2   12
Eq 2               8     -2   14
Eq1 +Eq2          13      0   26
1      0    2   This says x = 2

Sub x = 2 in any equation
to get y = 1
x    y
Eq 1              5     2   12
eq 2              4    -1    7

x    y
(Eq 1)/5          1   0.4   2.4
eq 2              4    -1     7

x    y
Eq 1              1   0.4   2.4
eq 2              4    -1     7

x    y
Eq 1              1 0.4 2.4
eq 2 + eq1*(-4)   0 -2.6 -2.6

x    y
Eq 1              1   0.4   2.4   This says 1x + .
eq 2 /-2.6        0     1     1   This says 1y =
5x + 2y = 12
me as 4x - y = 7

y. Go to edit. Paste
cial. Values only

ys x = 2

in any equation
x + .4y = 2.4 -----> x= 2
1y = 1 -----> y = 1
matrices. Section 2.2
Page 76. let's use the HW problem Page 74 # 33

6x-3y+3z = 6
x+5y+10z = 5            x

4x-5y-z = 4

eq 1                               6         -3    3
eq 2                               1          5   10
eq 3                               4         -5   -1
The nubmers in blue are the coefficients of the variables x,y
by 3 (colums) matrix. This is a square matrix. We simplify s
numbers 6,5,and 4 are in a colum. This is a 3 (rows) by
by 1 for short.

x     y         z
eq 1                         6    -3         3      6
eq 2                         1     5        10      1
eq 3                         4    -5        -1      4
The numbers in red are the same as
The numbers in red are the same as
However in this arrangement, they fo
AUGMENTED 3 by 4 Matrix

Interchange eqn. 1 and 2 to get a
position

x     y         z
eq 1                         1     5        10     1
eq 2                         6    -3         3     6
eq 3                         4    -5        -1     4

eq 1                         1     5        10     1
(eq 1)*(-6) + eq 2           0   -33       -57     0
(eq 1)*(-4) + eq 3           0   -25       -41     0

Eq 1                         1     5        10     1
Eq 2                         0   -33       -57     0
Eq 3                         0   -25       -41     0

Eq 1                         1     5        10     1
New Eq 2 = (Eq 2)/(-33)      0     1   1.72727     0
Eq 3                         0   -25       -41     0
Eq 1                1     5        10   1
Eq 2                0     1   1.72727   0
Eq 3                0   -25       -41   0

Eq 1                1    5         10   1
Eq 2                0    1    1.72727   0
(Eq2)*25) + Eq 3    0    0    2.18182   0

Eq 1                1    5         10   1
Eq 2                0    1    1.72727   0
Eq 3                0    0    2.18182   0

x    y          z
Eq 1                1    5         10   1
Eq 2                0    1    1.72727   0 y+
(Eq 3)/ 2.1818182   0    0          1   0
z=
6
5
4
he variables x,y,z and form a 3 (rows)
x. We simplify say a 3 by 3 matrix. The
(rows) by 1 (colum) matrix. We say 3

6x-3y+3z = 6
1x+5y+10z = 5
4x-5y-1=4
are the same as above.
are the same as above.
ngement, they form an
Matrix

2 to get a 1 in the (1,1)

Use paste special
under EDIT when
pasting cells without
formulae.

Lets make the blue #
1 by dividing by - 33
x+5y+10z= 1 ---> x = 1
y +1.72727z = 0 --> y = 0
z=0