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Competency standard :
Solve problems related to the concept
of operating the real.
Hal.: 2
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Learning Materials :
System Real Numbers
On integer operations
Operations on Numbers fraction
Conversion Numbers
Comparison, scale and Percent
Implementation of Real Numbers In
completing the Program Expertise .
Hal.: 3
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basic competence 1 :
Applying the operation on the real .
Hal.: 4
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1. System real number
Learning Objectives
:
Students can:
Distinguish the various real
Hal.: 5
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Scheme real number :
Real number
Rational number
Fraction number
Integer
Positive integer
(natural number)
Prime number
Hal.: 6
Irrational number
0(zero)
1
Negative integer
Composite number
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Finish
Adaptif
Rational numbers
 Rational number is number that can be
expressed with the form a
b
where a and b integer .
example : 5/6, 7/8 ,-3/4, etc.
Hal.: 7
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Irrational number
 is number that can not be expressed with the
form a
b
example :
Hal.: 8
5,
7, ∏, etc.
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2. integer operations
Learning objectives :
 Students can :
Complete the operation Answer and
integer .
 1.
Complete the operation multiplication
and division of integer .
 2.
Hal.: 9
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a. Addition and subtraction
 addition integer can equipment with equipment or the
movement along the track represents the number lines.
First on the positive integer and zero, and then to develop a
rounded whole.
One model of learning to use the steps in the line is to
make agreements, for example:
Addition : Positive, advanced
Negative, back
Subtraction : Positive, advanced
Negative, back
Hal.: 10
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-2 + 3 = ....
step
3
-7
-6
-5 -4
-3
-2 -1
0
1
2
3
4
5
6
7
-2
So that the results of the additional
Hal.: 11
: -2 + 3 = 1
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5 + (-2) = ....
step
5
-7
-6
-5 -4
-3
-2 -1
0
1
2
3
4
5
6
7
2
So that the results of the additional
Hal.: 12
: 5 + (-2) = 3
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-2 - (-5) = ....
step
-7
-6
-2
-5
-4
-3
-2 -1
0
1
2
3
4
5
6
7
-5
So that the results of the additional
: -2 - (-5) = 3
etc
Hal.: 13
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b. Integer multiplication
Multiplication and division of integer is the development of multiplication and
division of the original. Which is still a problem is often a sign of problems in
the operation.
Can be explained with the pattern, for example, how to fill the box. . . in the
table below:

–3
–2
–1
0
1
2
3
1
...
...
...
0
1
2
3
0
...
...
...
0
0
0
0
–1
...
...
...
0
...
...
...
So that will be obtained:
Results marked with the same time the number of positive results and if
the sign is negative the result is different, and each number (rounded) 0
multiplied by the result 0 (zero).
Hal.: 14
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c. Integer division
With a distribution of b is declared with a: b or
a: b = c if and only if a = b x c.
0
Implications, among other :  0
b
0
 c b  0,
1). For each
0
0
0
b
0 : b = 0 because 0 x b = 0  0 = 0  b correct
2). c x 0  0 = 0  c. because 0  c =00 for each c, then c how can any value,
including negative numbers, the results
0 of c : 0 said not a .
a
c
0
3). For each a  0, a : 0 = c may not happen, because there is no value c that
a
0
meet the a = c  0, or frequently stated ∞ the result is not defined.
Hal.: 15
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3. Fraction number
Initial concept of a fraction is part of the whole (with the buckle geometry)
instance :
3
16
1
4
How the operation fraction number ?
instance :
that this result can equipment
6
12
Hal.: 16
3
4
3
4
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2
3
Adaptif
exercise
1. Ali has 2 meter wire, will be given to friends
to create interest. If each person gets the 2 /
5 meters, how many people get that part?
2. Of 2 ¾ kg sugar akan made bread recipe. If a
recipe requires sugar as much as ½ kg. How
many bread recipes that can be made?
3. If a job is done by 4 people will finish in 30
days. If done by the usual 6 will be
completed within 80 days. How much time
needed to complete the work if it is done
jointly by the 3 and 4 experienced people?
Hal.: 17
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4. The fractional conversion
Change the Ordinary to the fraction or
a decimal fraction .
Change the Ordinary to the form of a
fraction or vice versa Percent .
Fractional change to Decimal Percent
of, or otherwise .
Hal.: 18
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Change the fraction to fraction
Ordinary Decimal
 Change the Ordinary to fraction decimal
fraction, or vice versa.
Example :
3
= ….. Do with the usual .
4
0,7 = 7/10
0,08 = 8/100 = 2/25, etc.
Hal.: 19
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Change the fraction to the
percent of
To change the fraction to a normal
form of percent, we multiply the
fraction by 100% .
Meanwhile, to change the percent into
a fraction of normal, we must divide
the number by 100 percent .
Hal.: 20
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example
 4/5 = …… %
= 4/5 x 100 %
= 80 %
2/3 = …… %
= 2/3 x 100 %
= 66 2/3 %
15 % = ……
= 15/100
= 3/20
etc.
Hal.: 21
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Change the fraction to a
decimal form of a percent
To change a decimal to fraction of
percent, we multiply the fraction by
100%.
Meanwhile, to change the form of a
percent into a decimal fraction, we
must divide the number by 100
percent.
Hal.: 22
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example
 0,03 = ……%
= 0,03 x 100%
= 3%
0,056 = ……%
= 0,056 x 100%
= 5,6%
35% = ……. (decimal fraction )
= 35/100
= 0,35
425% = ………
= 425/100
= 4,25
etc.
Hal.: 23
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exercise
1. Indicate the following fraction into
the form of a decimal and percent :
a. 3/5
c. 2/3 (three decimal places )
b. 7/8
d. 5/6 (three decimal places )
2. Specify the form of a percent below the
fraction and decimal fraction normal :
a. 35%
c. 12 ½ %
b. 8 %
d. 16 2/3 %
Hal.: 24
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5. ratio
There are two kinds of comparisons, namely:
1. Comparison worth .
2. Comparison Value turn .
Try our eye on each comparison is .
Hal.: 25
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Kind of comparison
Worth
Turn value
Many
(Fruit)
Price
(Rupiah)
Speed.
(Km / hr)
Time
(Hours)
1
2
3
4
…
6
…
x
200
400
…
…
…
…
…
…
60
30
20
…
…
5
…
x
1
2
…
…
…
…
…
…
Hal.: 26
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Kind of comparison
Comparison worth
a = k  b, k  R
when b then a, proportionate increase in straight
when b then a , proportionate increase in straight
instance :
s=vt
Comparison Value turn
when b then a 
when b  then a 
instance : v = s/t
Hal.: 27
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Comparison of advanced
Learning Experience
With 24 workers, a job is complete bulk
sewing planned within 48 days. After
working for 12 days with 24 workers,
stopped work for 9 days because of it.
How many workers who should be
added so that work can be completed
on time?
Hal.: 28
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Concerning the settlement of the above :
Turn the value of comparison, so that:
The remaining work to 48–12 = 36 day which
should be completed by 24 people.
But the only remaining 48–12–9 = 27 day.
So obtained:
24 people

36 day
x people

27 day
24 27
864
then:
x
=
36/27
x
24
=
32

 27 x  24.36  27 x  864  x 
 x  32
x
36
27
So the additional 8 workers
Hal.: 29
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example
1. With fixed velocity, a car requires 5
liters of petrol to 60 km distance. How
many liters of gasoline needed to travel
150 km distance?
2. The distance between two cities can be
a vehicle with an average speed of 72
km / hr for 5 hours. What is the
average speed of vehicles to travel long
distances if the 8 hours?
Hal.: 30
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solution:
Because comparison worth then :
150/60 x 5 km = 12,5 liters.
Comparison value turn, then : 5/8 x
72 km/hr 72 8
= x45 km/hr.
5
Hal.: 31
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exercise
1. Making cake mix liquids consisting of
coconut oil and water with the
comparison 1: 18. How many liters of
coconut oil is needed to obtain 9.5
liters of fluid mixtures?
2. A map of the oblong drawn with
scale:
1: 120,000 and has a length:
width is 4:3. Meanwhile, around 112
cm map. actually a broad set
described by this map?
Hal.: 32
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KE MENU AWAL
Adaptif
Basic competence 2 :
Applying the operation have an
important position on the number.
Hal.: 33
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learning materials :
Concepts have an important position
and the nature-nature.
Numbers have an important position
in the operation.
The simplification have an important
position.
Hal.: 34
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exponent
Ex: IN PROCESS AMUBA cleavage SEL
Hal.: 35
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Learning objectives
1. Explain the definition of rank
2. Explain the nature have an important
position
3. Have an important position to operate the
appropriate concept unanimously positive.
Hal.: 36
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The designation ON REAL Numbers
If a number is real and n is a positive integer, then an is
defined as a multiple of n times, written
a =axaxax...xa
n
n factor
with,
a
is a form of public have an important position of the
n,
a called the principal or base , a R,
n called the rank of (exponent)
Hal.: 37
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start
Looking exponent =
looking logarithm
= 2048
2...
2log
?
2
3
4
5
6
2
4
8
16
32
64
?
2
22
2
Numbers have
an important
position notation
Hal.: 38
20
21
22
2
23
…
9
11
512
?
… 2048
2?9?
2211?
2
2222
1
1
222
after …
PERIOD
many amuba
2048 = ...
24
25
2...6
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definition :
a1 = a
an  a

a

a

a

...

n factor form a




Are all the results obtained form of rational number?
What rational number with the rank of integer numbers is always
rational?
Whether the division of two rational number (not a divider 0) always
produce rational number?
How does nature have an important position the operation?
Hal.: 39
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Consider the following examples :
7
. 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2,
7 factor
5
• (-3) = (-3) x (-3) x (-3) x (-3) x (-3)
5 factor
•  1 8 1 1 1 1 1 1 1 1
   x x x x x x x
2 2 2 2 2 2 2 2
2
8 factor
In the example above, 2, -3, and is a number (base), while 7, 5, and 8 is the rank of
(exponent).
Hal.: 40
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The attributes have an important position :
Attributes have an important position with the designation unanimously positive
is as follows.
Suppose a, b  R, and m, n is a positive integer, then
a m x a n  a m n
S-1
a m : a n  a mn , m  n
S-2
(a m ) n  a mn
S-3
S-4
S-5
( a x b) n  a n x b n
n
an
a
   n ,b  0
b
b
Hal.: 41
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 :
With experiment

2222
4 factor number
2
2 2  2  2  2  2
6 factor number 2
Means 24+6 = 210
(4 + 6) factor number 2
 kkkkkkkkk
kk  k
3 factor number k
9 factor number k
Means k3+9

(3 + 9) factor
number k
aaa…a
 a a  a  …  a
p factor number a
q factor number a
(p + q) factor number a
Contoh:
32  33 =
32+3
35
=
76  713= 76+13 = 319
Hal.: 42
= k12
Means
ap+q
 ap  aq = ap+q

x5  x 12= x5+12 = x17
4
3 5  3 45  3 9

3
 
 
   
 
4
4



4
4
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With experiment:
?
2  2 2  2  2  2 result ( = )
2=232 =
 227 - 4
27 : 24 = 2 __________
2222
Means 27 : 24 = 27 - 4
p factor number a
aax__________
aaxaaxa…
 ax a a …  a
=
= a  a  a...  a
a a  a … a
(p >q)
p – q factor
q factor number a
p : aq =
p-q
Means

a
=a
Example:
ap - q
8
5
85
3
6
4
2
2
2
2
3 :3 = :    3  =  
 3  3
 3  3
13-8
5
13
8
7
=
7
7 :7 =
ap :
aq
Hal.: 43
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Adaptif
With experiment:
b 
ap

  
p
p
p
p
a
a  a  a  
b
b
b
b
p factors
p number of factors a
b 
ap
ap
aaaaaa…a
_______________________
=
= ____
bbbbbb…b
bp
p number of factors b

ap
p
____
a
therefore :

b
bp
Hal.: 44
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exercise
Write operation results in the
following form of the most simple:
1. 94 = ……
36
2. 53 x 23 = ……
3. 103
23
Hal.: 45
= ……..
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Basic competence 3 :
Applying the operation on the
Irrasional.
Hal.: 46
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Learning materials :
The concept of the Irrasional
Production Numbers In the form of
roots.
Simplification of the root.
Hal.: 47
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Learning objectives
1. Explain the definition irrasional
2. Operate a number of roots.
3. Simplify the number irrasional.
Hal.: 48
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Root ( apersepsi )
 The Square Root
Extent of a garden 225 m2. If the garden is shaped
square, the length of the garden?
Square footage is 15 m, from
225  15
because 152 = 225.
Root of the withdrawal process
For each a  0 and b  0,
b a
if and only if
a2 = b
Why b can not be negative?
Hal.: 49
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Learning experience
mr Karyo have sebidang the extent
of land recorded in the Office land
is 1369 m2. In fact the form of a
square yard. How long is the size of
the ground yards?
How many meters of iron rod to
create the necessary framework of
iron below
12 m
6m
12 m
Hal.: 50
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Adaptif
how the value of :
4
(3)
2
Settlement:
4
4
(3) 
2
4
9
4
3 
2
3
2
4
(3)  (3)   3
2
Which is true!
Hal.: 51
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Adaptif
example: how the value of
1369 m2
30 m
hm
(1)
30 m
1369
berapa
meter?
(1)
(2)
(3) h m
(2  30 + h)m
30 m
900 m2
(2)
30 m
hm
(3)
hm
then:
2  30h + h2 = 1369 – 900
(2  30 + ….) …. = 469
Retrieved
Hal.: 52
:h=7 So the result 37
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Adaptif
method can be written :
13 6 9
302 =
900
469
(2  30 + …)  … = 469
0
Hal.: 53
= 30 + 7
(–)
(–)
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Cube root
Water tank inside the cuboid can accommodate as many as
8 m3 of water. Size of the bathtub?
8 m3
?=p
Cistern size is 2 m,
of p
3
3
Hal.: 54
pp=8p=2
8 2
a b
because 23 = 8.
If and only if b3 = a
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3
example, how the value of :
(p + s)3 = p3 + 3p2s + 3ps2 + s3
20 +......
6
3  202
3
17576 =?
203 = 8 0 0 0
3  20
17576 =?
= p3 + (3p2 + 3ps + s2 )s
9576
6
(1200 + 60 6... + ...6 2)  ...
0
so:
Hal.: 55
3
1 7 5 7 6 = 26
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Root operation
ab 
Basic Operation :
n
ab 
n
a b
For a  0 and b  0
n
a  b n  A, n  2,
n n
a = a, from n
a real.
Answer and Reduction
can be simplified if a similar radical
example : 75  147  48
=5
3 7 3 4 3
=(5–7+4) 3 =2 3
Multiple forms of Roots
with the nature
n a n b  n a.b
The form of the distribution of roots a
 a , for a  0 and b  0
b
b
Hal.: 56
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root
Rationalizing the denominator of a fraction
a
or
b
Forms:
6
2

6
x
2
a
b
2

multiplying
6
2
2
2
b
b
 3 2
forms:(ab + cd) where a, b, c, and d rasional, and at least
one of the b and d irrational
a b c d
........
........
x
=
a bc d
a bc d a b c d
1
1
Hal.: 57
2
1

1
2
1
2
x
2
1
1 2 2  2 3  2 2
2

 2 2 3

1

2

1
2
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Adaptif
Basic competence 4 :
Applying the concept of logarithm
Hal.: 58
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Adaptif
Learning materials :
The concept of logarithm
The logarithm operation.
Hal.: 59
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Adaptif
Learning objectives
1.
2.
3.
4.
Hal.: 60
Explain the concept of logarithm
Explain the nature logarithm.
Logarithm operation with nature logarithm.
Problem solving program expertise relating
to the logarithm.
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LOGARITHM
In the form of 2log 128
2 is the base logarithm
2log
128 = 7  27 = 128
In general it was stated that :
alog
b=c
In the form of alog b
b is the base logarithm
alog
b = c  ac = b
c replaced alog b
c = alog b
Appropriate number of discussion
c
= b have an important position, then a >
 a
alog b
0, b > 0 and a  1, because 1log b = c
1c = b, for b  1 never found
a
= b
value of c.
logarithm of b to the base a ditulis alog b is number if a
exponent number result of b
Hal.: 61
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LOGARITHM
Note : ab = c
ab = ….
c
alog
Hal.: 62
…b = c
a... = c
search results promotion
find square root of b
find that designation of a
result c
= find that designation of a
result c
= alog c = …
b = c  ac = b with a > 0 , a  1 and b > 0
Isi dengan Judul Halaman Terkait
Adaptif
The properties of logarithm :
if a > 0 , a  1 , m > 0 , n > 0 and x  R, then :
1.
alog
a
ax = x
logn
n
2.
a
3.
a q log a p  p
q
4.
alog
(m.n) = alog m + alog n
5.
alog
(m/n) = alog m - alog n
6.
alog
mx = x. alog m
g
7.
alog
Hal.: 63
m=
log m if g > 0 , g  1 etc.
g
log a
Isi dengan Judul Halaman Terkait
Adaptif
exercise:
1. Simple form of 2log 4 + 2log 12 – 2log 6 is …..
a. 2
2. If known
b. 3
3log
c. 4
d. 6
e. 8
2 = p, then 9log 8 is... .
3. if 5log 2 = p then log 2,5 = …
Hal.: 64
Isi dengan Judul Halaman Terkait
Adaptif
exercise:
An optical equipment needed to observe
the stars at the top to a height of six, the
limit of naked eye view. However, the tool
has limitations. L limit the height of a
telescope with optical inch in diamater D
formulated in L = 8.8 + 5.1 log D
a. Look for the height limit for the telescope
similar to the 6 inch diameter .
b. Find the diameter of a lens that has a
height limit of 20.6.
Hal.: 65
Isi dengan Judul Halaman Terkait
Adaptif
Competency standards :
Solve problems related to the
concept of matrix.
Hal.: 67
Isi dengan Judul Halaman Terkait
Adaptif
Learning materials:
Kind of Matrix.
Matrix operation.
Hal.: 68
Isi dengan Judul Halaman Terkait
Adaptif
Basic competence 1 :
Describe your kind of matrix.
Hal.: 69
Isi dengan Judul Halaman Terkait
Adaptif
Learning objectives
1.
2.
3.
4.
5.
Hal.: 70
Explaining the matrix
Explain the notation and matrix-element.
Distinguish the type of matrix.
Complete similarity matrix.
Explaining Transpose Matrix.
Isi dengan Judul Halaman Terkait
Adaptif
Matrix
 Matrix is the order of the numbers-which
consists of these lines and columns.
 Each number in the matrix is called the
entry or elements. Ordo (size) of the matrix
is the number of rows times the number of
columns.
Notation:
Matrix: A = [aij]
Element: (A)ij = aij
Ordo A: m x n
Hal.: 71
A=
a11
a21
:
ai1
:
am1
a12…….a1j ……a1n
a22 ……a2j…….a2n
:
:
:
ai2 ……aij…….. ain
:
:
:
am2……amj……. amn
row
column
Isi dengan Judul Halaman Terkait
Adaptif
General form Matrix
 a11 a12  a1n 
a a  a 
2n 
 21 22

A=   


am1 am 2  amn 
Hal.: 72
Isi dengan Judul Halaman Terkait
Adaptif
example:
P=
 2 1 8 
7
 Ordo
4
11


 6  5 1 
matrix P = 3 x 3
Entry that is located on line 1 column 2 is 1
Entry that is located on the first row third column is
Members of the matrix P in row 3 =
6
6
Members of the matrix P in column 1 =
Hal.: 73
Isi dengan Judul Halaman Terkait
 5 1
 2
7 
 
 6 
Adaptif
SQUARE MATRIKS
Matrix is a square matrix is the
number of rows and columns the
same number of.
Trace(A) = 1 + 2 + 3
1
2
4
2
2
2
3
3
3
Main diagonal
Trace from matriks is sum the diagonal elements of the main
Hal.: 74
Isi dengan Judul Halaman Terkait
Adaptif
Zero matrix and identity
zero matrix is matrix of all elements zero
0
0 0
0
0
0
0
identity matrix is a square matrix of elements of main diagonal
elements 1 and other 0
I3
I2
Hal.: 75
I4
1
0
1
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
1
0
0
1
0
0
0
0
1
Isi dengan Judul Halaman Terkait
Adaptif
Transpose
A=
4
2
6
7
5
3
-9
7
AT = A’ =
4
5
2
3
6
-9
7
7
Definition:
Transpose the matrix A is the matrix ATcolumns are rows of the A-line, lines is from the
columns of A .
[AT]ij = [A]ji
nxm
If A is m x n matrix, the matrix transpose AT size ………..
Hal.: 76
Isi dengan Judul Halaman Terkait
Adaptif
Properties matrix transpose
1. Transpose from matrix A transpose is A: (AT )T = A
A
(AT)T = A
AT
example:
Hal.: 77
4
5
2
3
4
2
6
6
-9
5
3
-9
7
7
4
5
7
2
3
7
6
-9
7
7
Isi dengan Judul Halaman Terkait
Adaptif
Properties matrix transpose
2. (A+B)T = AT + BT
T
T
A+B
=
A
(A+B)T = AT
Hal.: 78
T
+
B
+
T
B
Isi dengan Judul Halaman Terkait
Adaptif
Properties matrix transpose
3. (kA)T = k(A) T from scalar k
T
T
kA
k
T
(kA)
Hal.: 79
=
A
T
k(A)
Isi dengan Judul Halaman Terkait
Adaptif
Properties matrix transpose
4. (AB)T = BT AT
Hal.: 80
B
AB
=
T
(AB)
= BTAT
AB
T
T
T
Isi dengan Judul Halaman Terkait
A
Adaptif
matrix symmetry
Matrix A is symmetry if and
only if A = AT
A=
A =
Hal.: 81
1
2
3
4
4
2
2
3
2
5
7
0
A’ =
3
7
8
2
4
0
2
9
4
2
2
3
A symmetry
= AT
Isi dengan Judul Halaman Terkait
Adaptif
orthogonal matrix
Matrix A orthogonal if and only if AT = A –1
A=
B=
0
-1
1
0
0 1
AT=
½√2 -½√2
½√2 ½√2
BT=
-1
0
= A-1
½√2 ½√2
-½√2 ½√2
= B-1
(A
A-1-1)T = (A
ATT)-1
if A is orthogonal matrix, then (A-1)T = (AT)-1
Hal.: 82
Isi dengan Judul Halaman Terkait
Adaptif
similarity 2 matrix
&
summation 2 matrix
Hal.: 83
Isi dengan Judul Halaman Terkait
Adaptif
similarity 2 matrix
condition :
*
*
have the same ordo
The same entry and the same location for each i and j, aij = bij
Hal.: 84
Isi dengan Judul Halaman Terkait
Adaptif
Similarity 2 matrix
Two matrix with the same size and if
each entry is associated same .
A=
4
2
1
3
1
2 2
2
1
1
2
B=
1
2
4
2
1
3
A=B
1
2
3
2
1
3
2
4
x
2
4
2
2
2
2
2
2
2
2
?2
2?
2?
4
5
6
?4
5?
6?
9
0
7
?9
0?
7?
E=
Hal.: 85
2
2
C=
G=
1
D=
F=
H=
Isi dengan Judul Halaman Terkait
C≠D
E = F if x = 1
G=H
Adaptif
example:
find the values a, b , and
If given
matrix A = matrix B!
Hal.: 86
Isi dengan Judul Halaman Terkait
c
Adaptif
Example (next):
a  b 3 
A=  5 b  c 


 6
B= 
 c
Hal.: 87
3

4
a= - 5
b= -1
c5
=
Isi dengan Judul Halaman Terkait
Adaptif
Basic competence 2 :
Complete the operation matrix
Hal.: 88
Isi dengan Judul Halaman Terkait
Adaptif
Learning objectives
Explain the operation matrix.
Complete the operation summation
and reduction matrix.
Complete the matrix multiplication
operation with scalar.
Complete the two matrix
multiplication operations.
Determine the inverse matrix square
ordo two.
Complete SPL with matrix.
Hal.: 89
Isi dengan Judul Halaman Terkait
Adaptif
summation 2 matrix
Procedure:
Enumerate each entry / element
A matrix with each entry / element
matrix B which seletak
(rows and columns the same)
Hal.: 90
Isi dengan Judul Halaman Terkait
Adaptif
Summation and reduction two matrix
example
A=
A+B=
A-B=
10
22
1
-1
B=
10+2
10+2
22+6
22+6
1+7
1+7
-1+5
-1+5
10-2
22-6
1-7
-1-5
2
6
7
5
=
=
12
12
28
28
88
4
8
16
-6
-6
• What two conditions so that the matrix can summand?
•Answer: ordo two are the same matrix
A = [aij] and B = [bij] the same size,
A + B defined: (A + B)ij = (A)ij + (B)ij = aij + bij
Hal.: 91
Isi dengan Judul Halaman Terkait
Adaptif
example:
  4 2
A= 

 3 9
1
B= 
 3
A+B=
Hal.: 92
5

1
?
Isi dengan Judul Halaman Terkait
Adaptif
answer :
 4
A + B =
 3
2
+

9
1
 3

5

1
 4  1 2  5

=

3   3 9  1
 3 7 
=

 0 10
Hal.: 93
Isi dengan Judul Halaman Terkait
Adaptif
example:
A=
 11 3 4 
 2 9 6 


 7 0  5
B=
 4
 0


 15
8
3

2

Find the value A + B!
WHY ?????
Hal.: 94
Isi dengan Judul Halaman Terkait
Adaptif
Condition 2 matrix can summand :
*
Hal.: 95
Have the same ordo
Isi dengan Judul Halaman Terkait
Adaptif
Exercise 1
1. Find the value p, q, r, s,and t if given matrix
A = matrix B with:
30
q

9
23
A= 
16
p

 5 150  20
92 
68 
4 
40

51
B=
 30 8  5 81
3x3 23

t


 r
9 40
 10

s
51
 2
p, q , r, s, t whole number
Hal.: 96
Isi dengan Judul Halaman Terkait
Adaptif
Exercise 2
2.
 11
9

4
A= 
 2
3
7
13
8
0
1
6
6
5

7
9

4
 2 0
 7  11

B=  4
6

 4 12
9  1
8  9
3 9

2  8
Find the value A+B!
Find the value B+A!
Hal.: 97
Are the results the same?
Isi dengan Judul Halaman Terkait
Adaptif
Solution for number 2
A+B=
0
 11 3
 9  7 1

  4 13  6

8
6
 2
5
7 
9

4
+
 2 0
 7  11

4
6

12
4
9  1
8  9
3 9

2  8
9
4
 13 3
 2  18 7  2
=  0 19  3 18 


6
20
8

4


Hal.: 98
Isi dengan Judul Halaman Terkait
Adaptif
Solution for number 2
B+A=
 2 0
 7  11

4
6

12
4
9  1
8  9
3 9

2  8
+
0
 11 3
 9  7 1

  4 13  6

8
6
 2
5
7 
9

4
9
4
 13 3
 2  18 7  2
=  0 19  3 18 


6
20
8

4


Hal.: 99
Isi dengan Judul Halaman Terkait
Adaptif
exercise: summation two matrix
(next)
K=
C=
C+D
K+L
1
3
5
4 -9
7 0
9 -13
5
6
1
7
2
3
=
=
7 3 1
-2 4 -5
9 -4 3
L =
25
30
5
35
10 15
D =
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
D+C
=
L+K
=
What conclusions you? summation whether the matrix is commutative?
Hal.: 100
Isi dengan Judul Halaman Terkait
Adaptif
Quiz: summation two matrix
Quiz:
C=
A =
3
-8
0
4
7
2
-1
8
4
0
0
0
0
0
0
1.C + D =…
2.C + E = …
3.A + B = …
Hal.: 101
D =
B =
3
7
2
5
2
6
-1
8
4
0
0
0
0
0
0
E =
Feedback: C +D =
Isi dengan Judul Halaman Terkait
2
7
2
5
2
6
6
-1
2
9
9
8
-2
16
8
Adaptif
Results scalar time with the matrix
 example:
A=
5
6
1
7
2
3
5A =
5x5
5x6
5x1
5x7
5x2
5x3
=
25
30
5
35
10
15
What is the relationship between H with A?
H=
250 300
50
350 100
150
H = 50A
given matrix A = [aij] and scalar c, multiplication scalar
cA have entries as follows :
(cA)ij = c.(A)ij = caij
Note: On the set of Mmxn, matrix multiplication with skalar are closed
(the matrix with the same ordo)
Hal.: 102
Isi dengan Judul Halaman Terkait
Adaptif
Results scalar time with the matrix (next)
K
3x3
K=
4K =
5K =
Hal.: 103
1
3
5
4 -9
7 0
9 -13
4 16 -36
12 28 0
20 36 -52
5 20 -45
15 35 0
25 45 -65
Isi dengan Judul Halaman Terkait
Adaptif
Exercise: Results skalar times
with the matrix
given that cA is the matrix zero. What
is your conclusion about the A and c?
A =
example:
c=7
cA =
c=0
0*2 0*7 0*2
A =
0
0
0
0
0
0
2
7
2
5
2
6
0*5 0*2 0*6
conclusion
cA =
7*0 7*0 7*0
=
7*0 7*0 7*0
0
0
0
0
0
0
case 1: c = 0 and A random matrix.
case 2: A zero matrix and c how can it.
Hal.: 104
Isi dengan Judul Halaman Terkait
Adaptif
Multiplication matrix
A=
2
3
4
5
8
-7
9
-4
1
-5
7
-8
2.1 +3.7+4.4+5.11
AB =
Hal.: 105
1
B=
7 -6
4
-9
11
3
-35
-49
-35
-94
-55
2
94 -35
=
Isi dengan Judul Halaman Terkait
-49 -35
-94 -55
Adaptif
Multiplication matrix (next)
Definition:
If A = [aij] sized m x r , and B = [bij] sized r x n,
then matrix result multiply A and B, is C = AB have elemenelements which is defined as follows :
r
(C)ij = (AB)ij = ∑ aikbkj = ai1b1j +ai2b2j+………airbrj
k=1
• condition:
A=
Hal.: 106
A
mxr
B
rxn
2
3
4
5
8
-7
9
-4
1
-5
7
-8
AB
mxn
B=
1
2
7
-6
4
-9
Isi dengan Judul Halaman Terkait
Find the value AB and
BA
Adaptif
Multiplication matrix (next)
A=
2
3
4
5
8
-7
9
-4
1
-5
7
-8
2.1 +3.7+4.4+5.11
A B =
B=
1
2
7
-6
4
-9
11
3
-35
-49
-35
-94
-55
94 -35
=
-49 -35
-94 -55
BA not defined
Hal.: 107
Isi dengan Judul Halaman Terkait
Adaptif
Multiplication matrix (next)
1. Given A and B, defined AB and BA. What your conclusion?
A
B
B
A
mxn
nxk
nxk
mxn
m=k
ABmxm
AB and BA
square matrix
ABnxn
2. AB = O zero matrix, if one of A or B certainly zero matrix?
A=
2
2
3
3
B=
3 -3
-2 2
AB =
0
0
0
0
AB zero matrix, not A or B matrix is zero
Hal.: 108
Isi dengan Judul Halaman Terkait
Adaptif
Exercise: Multiplication matrix (next)
Find the value if the time defined
2
4
2
A=
•
•
•
•
•
Hal.: 109
3
7
3
4
9
5
AB = ??
AC = ??
BD = ??
CD = ??
DB = ??
5
0
6
B=
D=
1
-9
8
5
2
0
0
6
1
2
0
C=
7 -11 4
3 5 -6
8 9 5 6
5 6 -9 0
-4 7 8 9
Isi dengan Judul Halaman Terkait
Adaptif
The root of matrix
example:
A=
2
1
3
2
A2 =
2
1
3
2
2
1
3
2
2
1
3
2
2
1
3
2
A3 = A x A2 =
2
1
3
2
A0 = I
An = A A A …A
n factor
An+m = An Am
Hal.: 110
Isi dengan Judul Halaman Terkait
Adaptif
Presented with the SPL of the
matrix
 SPL in the form of:
a11x1 + a12x2 + a13x3 +….. ..a1nxn
= b1
a21x1 + a22x2 + a23x3 +…….a2nxn = b2
:
am1x1 + am2x2 + am3x3 + ……amnxn = bm
 can be presented in the form of a matrix equation :
a11 a12……...a1n
a21 a22 ……..a2n
:
:
:
am1 am2…… amn
x1
x2
:
xn
=
x
b1
b2
:
bn
b
A: matrix coefficients
Ax = b
Hal.: 111
Isi dengan Judul Halaman Terkait
Adaptif
example: Presented with the SPL
of the matrix
SPL
x1 + 2x2 + x3
= 6
-x2 + x3
= 1
4x1 + 2x2 + x3 = 4
1.x1 +2.x2 + 1.x3
0.x1 + -1.x2 + 1.x3
4.x1 +2.x2 + 1.x3
Hal.: 112
6
=
1
4
=
1
2
1
x1
0
-1
1
x2
4
2
1
x3
Isi dengan Judul Halaman Terkait
6
=
1
4
Adaptif
multiplication matrix and identity matrix
A=
A.I =
I.A =
Hal.: 113
1
2
3
7
5
6
-9
3
-7
1
2
3
1
0
0
7
5
6
0
1
0
-9
3
-7
0
0
1
0
0
1
0
1
0
0
0
1
1
2
3
7
5
6
1
-9
3
-7
2
3
1
2
3
7
5
6
7
5
6
-9
3
-7
-9
3
-7
=
=
Isi dengan Judul Halaman Terkait
Adaptif
multiplication matrix and identity matrix
AB = A and BA = A, what your conclusion?
1
3
5
4 -9
7 0
9 -13
1
0
0
0
1
0
1
0
0
0
1
0
1
3
5
4 -9
7 0
9 -13
A
0
0
1
I
=
0
0
1
I
=
=
A
1
3
5
4 -9
7 0
9 -13
1
3
5
4 -9
7 0
9 -13
=
A
AB = A and BA = A, then B = I
(I identity matrix)
Hal.: 114
Isi dengan Judul Halaman Terkait
Adaptif
Inverse matriks
B is inverse from matriks A, if AB = BA
= I identity matrix, wrote B = A-1
A-1
A
A-1
=
A
I
=
Example
4
2
2
½ -½
2
-½
=
1
2
-½
2
2
1
A-1
4
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Hal.: 115
4
A-1
A
B
½ -½
B-1
=
=
A
2
1
½
-½
1
2
2
1
-½
-½
1
3
3
1
0
3
-2
Isi dengan Judul Halaman Terkait
0
0
1
I
4
B-1
1
B
=
1
0
0
0
1
0
0
0
1
I
Adaptif
Inverse matriks 2x2
4
2
½ -½
2
2
-½
=
1
1
0
0
1
I
A-1
A
a
b
c
d
1
0
0
1
A-1
=
A-1
d
 ab-cd
=
 -c
 ab-cd

=
a 
ab-cd 
-b
ab-cd
1
d
-b
ad - bc
-c
a
If ad –bc = 0 then A haven’t inverse.
Hal.: 116
Isi dengan Judul Halaman Terkait
Adaptif
example: Inverse matrix 2x2
A=
A-1
Hal.: 117
3
2
4
1
1
 3.1-4.2
=
-4
 3.1-4.2
   15
= I 4
3
3.1-4.2 
 5
-2
3.1-4.2

 53 
2
5
Isi dengan Judul Halaman Terkait
Adaptif
Quiz: inverse matriks
1. When matrix
a
c

b  does not have any inverse?
d 
ad-bc = 0
2. Find the value inverse matriks in the below
a.
b.
c.
d.
Hal.: 118
5
1
1
2
0
1
0
2
0
0
4
1
1
0
0
1
a.
2/3
-1/5
-1/5
5/3
b. haven’t inverse
c. haven’t inverse
d.
1
0
0
1
Isi dengan Judul Halaman Terkait
Adaptif
Quiz:
Please fill in the dots below
1.A symmetry then A + AT= ……..
2.((AT)T)T = …….
3.(ABC)T = …….
4.((k+a)A)T = ….....
5.(A + B + C)T = ……….
Key:
1. 2A
2. AT
3. CTBTAT
4. (k+a)AT
5. AT + BT + CT
Hal.: 119
Isi dengan Judul Halaman Terkait
Adaptif
Properties Inverse Matrix
1.Inverse from matrix if there is one:
Jika B = A-1 dan C = A-1, maka B = C
2.
(A-1)-1
A=
=A
4
2
2
2
½ -½
A-1 =
-½
(A-1)-1
1
½ -½
-½
?
1
=
1
0
0
1
A-1
4
2
2A 2
Hal.: 120
Isi dengan Judul Halaman Terkait
Adaptif
Properties (next)
3. If A have inverse then An have inverse and
(An)-1 = (A-1)n, n = 0, 1, 2, 3,…
A=
A3
=
(A3)-1 =
½ -½
4
2
2
2
4
2
4
2
4
2
2
2
2
2
2
2
A-1 =
-½
0.625
-1
1
=
104
64
64
40
-1
1.625
same
(A-1)3 =
Hal.: 121
½ -½
½ -½
½ -½
-½
-½
-½
1
1
1
Isi dengan Judul Halaman Terkait
=
0.625
-1
-1
1.625
Adaptif
Properties (next)
4.If k scalar not zero, then
(kA)-1 = 1/k A-1
4
(5A) = 5
2
2
=
2
20
10
10
10
0.1
(5 A)-1
=
-0.1
-0.1 0.2
same
½ -½
1/5 (A)-1 = 1/5
Hal.: 122
-½
1
0.1
=
-0.1
-0.1 0.2
Isi dengan Judul Halaman Terkait
Adaptif
Properties (next)
5.(AB)-1 = B-1 A-1
A=
4
2
2
2
(AB)-1 =
B-1 A-1 =
B=
16
24
10
14
½
2
2
-1
½ -½
1
B-1 =
½
5/4
½
-¾
-0.875 1.5
0.625 -1
=
-¾
-½
Hal.: 123
5
5/4
½
A-1 B-1 =
3
½ -½
-½
1
½
5/4
½
-¾
=
=
Isi dengan Judul Halaman Terkait
-0.875 1.5
0.625 -1
-0.5
1
0.75 -1.375
Adaptif
Complete equation Linear System
With Two Variables and Determinan
Matrix Method
General form
General form from linear equation
system with two variables
a1 x  b1 y  c1

a2 x  b2 y  c2
Hal.: 125
Isi dengan Judul Halaman Terkait
Adaptif
Method of solution of SPL
There is method in solving linear
equation system with two variables?
Hal.: 126
Isi dengan Judul Halaman Terkait
Adaptif
Method of solution of SPL
Methods in solving linear equation system
with two variables :
 Graphic Method
 Substitution Method
 Elimination Method
 Combined methods (the
Elimination and Substitution)
 Determinan Matrix Method
Hal.: 127
Isi dengan Judul Halaman Terkait
Adaptif
Rules Cramer
Linear equation system
a1 x  b1 y  c1

a2 x  b2 y  c2
With the elimination method can be obtained from the value of X:
a1 x  b1 y  c1  b2 a1b2 x  b1b2 y  c1b2
a2 x  b2 y  c2  b1 a2b1 x  b1b2 y  c2b1 
a1b2  a2b1 x  c1b2  c2b1
c1b2  c2b1
x
a1b2  a2b1
Hal.: 128
Isi dengan Judul Halaman Terkait
c1 b1
c2 b2
X
a1 b1
a2 b2
Adaptif
Rules Cramer
Linear equation system
a1 x  b1 y  c1

a2 x  b2 y  c2
With the elimination method can be obtained from the value of Y:
a2 x  b2 y  c2  a1 a1a2 x  a1b2 y  a1c2
a1 x  b1 y  c1  a2 a1a2 x  a2b1 y  a2 c1 
a1b2  a2b1  y  a1c2  a2c1
a1c2  a2 c1
y
a1b2  a2b1
Hal.: 129
a1 c1
a 2 c2
Y
a1 b1
Isi dengan Judul Halaman Terkait
a2
b2
Adaptif
Method Matrix determinant
Linear equation system
a1 x  b1 y  c1

a2 x  b2 y  c2
Determinant D from coefficient matrix of linear equation system is :
D
Hal.: 130
a1 b1
a2 b2
 a1b2  a2b1
Isi dengan Judul Halaman Terkait
Adaptif
Method Matrix determinant
Explore the value X that meet
the equality :
c1 b1
c2 b2
DX
c1b2  c2b1
X


a1 b1
D
a1b2  a2b1
a2 b2
Hal.: 131
Isi dengan Judul Halaman Terkait
Adaptif
Method Matrix determinant
Explore the value Y that meet
the equality :
a1 c1
DY a2 c2 a1c2  a2 c1
Y


a1 b1
D
a1b2  a2b1
a2 b2
Hal.: 132
Isi dengan Judul Halaman Terkait
Adaptif
The value from X and Y exist, if:
a b
0
p q
D≠0
or
or
aq  bp  0
or
Hal.: 133
aq  bp
Isi dengan Judul Halaman Terkait
Adaptif
example
Define the set of the settlement
system of equality:
2 x  5 y  14

5 x  2 y  7
Hal.: 134
Isi dengan Judul Halaman Terkait
Adaptif
Solution from linear equation system
2 x  5 y  14

5 x  2 y  7
Determinant D
D
a1 b1
a2 b2

2 5
5 2
 2  2  5 5
 4  25
 21
Hal.: 135
Isi dengan Judul Halaman Terkait
Adaptif
The value X that meet the :
c1 b1
DX
c2 b2
X 

a1 b1
D
a2
b2
 14
7

2
5
5
2

 14 2  7  5

 21
5
2

 28  35

 21
 63

 21
3
Hal.: 136
Isi dengan Judul Halaman Terkait
Adaptif
The value Y that meet the :
a1 c1
a 2 c2
DY

Y
a1 b1
D
a2
b2
2  14
5
7
2  7  5   14


 21
2 5
5 2
14  70

 21
84

 21
4
Hal.: 137
Isi dengan Judul Halaman Terkait
Adaptif
HP
So the collective system of
settlement is
{(3,-4)}
Hal.: 138
Isi dengan Judul Halaman Terkait
Adaptif
Exercise
Define the set of the settlement system of the following equation:
3x  y  9
1. 
2 x  5 y  11
2 x  y  5
2. 
3x  2 y  11
Hal.: 139
Isi dengan Judul Halaman Terkait
Adaptif
Homework
2 x  5 y  16
1. 
3x  7 y  24
x  3 y  2
2. 
3x  9 y  6
1
3
x

y 1
 4
3
5. 
1 x  2 y  0
 2
3
3x  7 y  9
3. 
6 x  14 y  9
1
1
x

y5
 3
2
4. 
 x  2 y  1
 3
Hal.: 140
Isi dengan Judul Halaman Terkait
Adaptif
Competency standards :
Solve problems related to the system
of linear and inequality and quadratic.
Hal.: 142
Isi dengan Judul Halaman Terkait
Adaptif
Learning materials :
Equation and inequality linear.
Quadratic equation and inequality.
Roots of the quadratic equation and
properties.
Establishing new quardtic equality.
Implementation of quadratic equality
and inequality in problem solving
Skills Program .
Hal.: 143
Isi dengan Judul Halaman Terkait
Adaptif
Basic kompetence 1 :
Determine the collective settlement of
linear equality and inequality.
Hal.: 144
Isi dengan Judul Halaman Terkait
Adaptif
Learning objectives
1. Explain the linear equation and inequality.
2. Complete equality and inequality one linear
variable.
3. Complete linear equation system of two
variables.
4. Problem solving skills program using the
linear equation and inequality .
Hal.: 145
Isi dengan Judul Halaman Terkait
Adaptif
Linear equation example
Finished equation below :
1. 3x -5 = 4
2. 3(3x + 6) = 5x + 2
3. ½ x + 2/3 = 3/4
Hal.: 146
Isi dengan Judul Halaman Terkait
Adaptif
Answer no 1 and 2
1. 3x – 5 = 4
3x
=4+5
3x
=9
x
=3
2. 3(3x + 6) = 5x + 2
9x + 18 = 5x + 2
9x – 5x
= 2 – 18
4x
= -16
x
= -4
Hal.: 147
Isi dengan Judul Halaman Terkait
Adaptif
Answer no 3
½ x + 2/3 = ¾
x 12
6x+8
=9
6x = 9 – 8
6x = 1
x = 1/6
Hal.: 148
Isi dengan Judul Halaman Terkait
Adaptif
Linear inequality example
Finished inequality below :
1. 3x -5 > 4
2. 3(3x + 6) < 5x + 2
3. ½ x + 2/3 ≥ 3/4
Hal.: 149
Isi dengan Judul Halaman Terkait
Adaptif
Answer no 1 and 2
1. 3x – 5 > 4
3x
>4+5
3x
>9
x
>3
2. 3(3x + 6) < 5x + 2
9x + 18 < 5x + 2
9x – 5x
< 2 – 18
4x
< -16
x
< -4
Hal.: 150
Isi dengan Judul Halaman Terkait
Adaptif
Answer no 3
½ x + 2/3 ≥ ¾
6x+8
6x
6x
x
Hal.: 151
x 12
≥9
≥9–8
≥1
≥ 1/6
Isi dengan Judul Halaman Terkait
Adaptif
Two varibels SPL example
Find the value x and y satisfy the
following linear equation system with
a mixture of methods (elimination
and substitution):
2x + 3y = 18
5x - y = 11
Hal.: 152
Isi dengan Judul Halaman Terkait
Adaptif
Answer :
 2x + 3y = 18 |x 1| 2x + 3y = 18
5x - y = 11 |x 3|15x – 3y = 33 +
17x
= 51
x
=3
2x + 3y = 18
2.3 + 3y = 18
6 + 3y = 18
3y = 18 – 6
3y = 12
y=4
Hal.: 153
Isi dengan Judul Halaman Terkait
Adaptif
Exercise
1.finished equation and inequality
below :
a. 2x – 6 = 8
b. 3x + 4 = x – 8
c. 5(2x + 3) = 7x – 3
d. ½ x – 2/3 = 5/6
e. 5x + 6 > 11
f. 6x – 8 ≤ 2x + 4
g. 2( 3x – 4) ≥ 9x + 7
Hal.: 154
Isi dengan Judul Halaman Terkait
Adaptif
Exercise (next)
2. find the value x and y from
equation system below :
a. 3x + 4y = 13 dan 5x – 3y = 12.
b. 4x – 2y = 0 dan 6x + 7y = 20.
Hal.: 155
Isi dengan Judul Halaman Terkait
Adaptif
Quadratic equation and inequality
1. Quadratic equation definition
2. Find the root quadratic
equation
3. Kind of quadratic equation root
4. Formula sum & multiply
quadratic equation root
Hal.: 157
5. Quadratic inequality
Isi dengan Judul Halaman Terkait
Adaptif
Basic kompetence 2 :
Determine the settlement of the
collective quadratic equation and
inequality.
Hal.: 158
Isi dengan Judul Halaman Terkait
Adaptif
Learning objectives
1. Explain the quadratic equation and
inequality.
2. Explain the root and properties quadratic
equation.
3. Complete quadratic equality and inequality.
Hal.: 159
Isi dengan Judul Halaman Terkait
Adaptif
Quadratic equation :
`an equation where the highest rank
of variables, namely two`
General form of quadratic equation :
ax  bx  c  0
2
Hal.: 160
with
a  0, a, b, c  R
Isi dengan Judul Halaman Terkait
Adaptif
Quadratic equation
example
a = 2, b = 4, c = -1
2x 2  4x  1  0 
a = 1, b = 3, c = 0
x 2  3x  0 
a = 1, b = 0, c = -9
x2  9  0 
Determine the settlement of quadratic equation in x means that the search for
value so that if the value of x in the equation is disubsitusikan,
then the equation will be valued properly.
Settlement of square root is also called square-root of the equation.
Hal.: 161
Isi dengan Judul Halaman Terkait
Adaptif
There are three ways to determine the
square root of the equation, namely :
 Factorization
 Complete perfect square
 Quadratic formula (formula a b c)
Hal.: 162
Isi dengan Judul Halaman Terkait
Adaptif
Factorization
To complete the equation ax² + bx + c = 0 with factorization,
First time find two numbers that meet the following requirements as .
•Time results is same ac
• sum is same b
Suppose the two numbers that are eligible x1 and x 2 ,
and x1  x2  a  c then x1  x2  b
Basic principles that are used to complete the quadratic equation
Factorization is the nature of multiplication, namely :
If ab = 0, then a = 0 or b = 0 .
So, will change if the formula or standard form of equation
square ax² + bx + c = 0 .
• for a = 1
factorization form ax² + bx + c = 0 to be :
( x  x1 )( x  x2 )  0 or ( x  x2 )  0
• for a ≠ 1
factorization form ax² + bx + c = 0 to be :
( ax  x1 )( ax  x2 )
 0  ( ax  x1 )  0 or ( ax  x2  0)
a
Hal.: 163
Isi dengan Judul Halaman Terkait
Adaptif
example :
Complete the following quadratic
equation with factorization :
1. x2 + 7x + 12 =0
2. x2 - 4x – 21 = 0
3. 6x2 -7x – 20 = 0
Hal.: 164
Isi dengan Judul Halaman Terkait
Adaptif
answer no 1 and 2 .
1. X2 + 7x + 12 = 0
(x + 3)(x + 4) = 0
X1 = -3, x2 = -4
2. x2 – 4x – 21 = 0
(x + 3)(x – 7) = 0
x1 = -3, x2 = 7
Hal.: 165
Isi dengan Judul Halaman Terkait
Adaptif
answer no 3 .
3. 6x2 – 7x -20 = 0
(6x – 15)(6x + 8) = 0
6
(6x – 15)(6x + 8) = 0
3
.
2
(2x – 5)(3x + 4) = 0
x1 = 5, x2 = -4
Hal.: 166
Isi dengan Judul Halaman Terkait
Adaptif
complete perfect square
Quadratic equation ax² + bx + c = 0, change to be general
form perfect quadratic with below :
a. Ensure the coefficient of x² is 1, if not value 1
divide the number of such coefficients to the value 1.
b. Add segment with the left and right half
Coefficient of x and make to square .
c. Make a left segment of perfect square,
While the right segment simplified .
Hal.: 167
Isi dengan Judul Halaman Terkait
Adaptif
Example :
Complete the following quadratic
equation with complete square
(perfect square) :
2x2 - 8 x - 42 = 0
Hal.: 168
Isi dengan Judul Halaman Terkait
Adaptif

answer :
2x2 -8x -42 = 0
:2
X2 – 4x – 21 = 0
X2 – 4x
= 0 + 21
X2 – 4x
= 21
X2 – 4x + 22 = 21 + 22
( x – 2 )2
= 25
X–2
= +/- 5
X1 = 5 + 2 = 7
X2 = -5 + 2 = -3
Hal.: 169
Isi dengan Judul Halaman Terkait
Adaptif
quadratic formula (formula a b c)
By using the rules complete perfect square which has been
published previously, you can search in the formula for
complete quadratic equation .
if x1 and x 2 namely root quadratic equation
ax² + bx + c = 0, then :
 b  b 2  4ac
x1 
2a
Hal.: 170
and
 b  b 2  4ac
x2 
2a
Isi dengan Judul Halaman Terkait
Adaptif
Exercise
Complete the following quadratic
equation with factorization, perfect
square and with formulas :
1. x2 + 5x + 4 = 0
2. x2 + 6x = 40
3. 6x2 – 11x – 10 = 0
Hal.: 171
Isi dengan Judul Halaman Terkait
Adaptif
value from b² - 4ac namely diskriminant, that is D = b² - 4ac .
Some kind of square root of the equation based on the value of D .
a. If D > 0, then quadratic equation has two real roots
Different.
b. If D = 0, then quadratic equation has two real roots
the same or have the root is often called the twin (same) .
c. If D < 0, then square root of the equation that does not have any
Real (imaginary) .
Hal.: 172
Isi dengan Judul Halaman Terkait
Adaptif
Roots of quadratic equations :
or
 b  b  4ac
x1 
2a
2
 b  b 2  4ac
x2 
2a
If the note is the root of the second, then obtained:
b
x1  x 2  
a
If both roots are multiplied, then obtained :
x1  x 2 
c
a
Both of the above mentioned formula of the sum and the product
of roots of quadratic equations.
Hal.: 173
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Adaptif
1. The sum of roots of quadratic equations:
x1 + x2
Hal.: 174
 b  b 2  4ac
2a
=
 b  b 2  4ac
2a
=
 b  b 2  4ac  b  b 2  4ac
2a
=
 2b
2a
=

+
b
a
Isi dengan Judul Halaman Terkait
Adaptif
2. The product of roots of quadratic equations
x1 . x2 =
  b  b 2  4ac 




2a


 b
2
=


b
2
2a 2
 4ac
b 2  b 2  4ac
=
4a 2
.

  b  b 2  4ac 




2a


2

b 2  b 2  4ac
=
4a 2
4ac
=
4a 2
c
=
a
Hal.: 175
Isi dengan Judul Halaman Terkait
Adaptif
The sum and the product of roots of root quadratic equations
If x1 and x2 of the both are roots of quadratic equations ax2 + bx +c = 0
(a  0), the sum and the product of roots of quadratic equations
determined with the formula
b

x1 + x2 =
a
=
x1 . x2 c
a
Formula the sum and the product of roots can using for
distinguish the characteristics of the root-root of the
quadratic equation has two different real roots. Such as the
following:
Hal.: 176
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The sum and the product of roots of quadratic equations:
1. One is the root of the root of the opponent or the other are
often perceived to be radical opposite :
x1 = - x2
 x1 + x2 = 0



b
= 0
a
b = 0
2. One is the root of the inverse root of the other or are often
perceived to be radical the reverse :
1
X2
x1 =
 x1 . x2 = 1

Hal.: 177
c
=1 
a
a =c
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The sum and the product of roots of quadratic equations:
3. One root same with 0:
x1 = 0
c
 x1 . x 2 =
a
c
(0) x2 =
a
c

0 =
a

c = 0
Hal.: 178
b
 x1 + x 2 = 
a
b
 (0) + x2 = 
a

x2 = 
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b
a
Adaptif
The sum and the product of roots of quadratic equations:
4. The second root has the same sign or are often perceived
to be radical with the same :
x1  0 and x2  0 or x1  0 and x2  0
x1 . x2  0

c
a
> 0, a and c marked the same
5. Both root does not have any signs that are often perceived to
be the same or roots of the different signs :
x1  0 and x2 0 or x1  0 and x2  0
x1 . x2  0

Hal.: 179
c
 0, a and c marked the same
a
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Adaptif
quadratic inequality is inequality that have a variable with the highest
rank of the two .
a.
b.
c.
d.
Steps to seek settlement of the collective quadratic inequality :
Indicate quadratic inequality in the form of quadratic equation (made
with the right segment 0) .
Find the square root of the equation.
Create a line number which is the roots, set the sign (positive or
negative) on each interval .
Set of intervals obtained from the settlement that meets these
inequality.
Hal.: 180
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Adaptif
Exercise
Finished :
1. x2 + 5x + 4 > 0
2. x2 + 6x ≤ 40
3. 6x2 – 11x – 10 ≥ 0
Hal.: 181
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Adaptif
Basic competence 3 :
Applying quadratic equation and
inequality
Hal.: 182
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Learning objectives
1. Establishing quadratic equality based on
roots is known.
2. Establishing quadratic equality based on
roots of the quadratic equation others.
3. Problem solving skills program based on the
quadratic equation and inequality.
Hal.: 183
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Apersepsi:
general form quadratic equation, namely:
ax2 + bx +c = 0
to determine formula of the sum and the product
of roots of quadratic equations, we remember that
roots from equation ax2 + bx +c = 0 (a  0) can
determine with formula
x1,2 =  b  b 2  4ac
2a
 b  b 2  4ac
x1 =
2a
2

b

b
 4ac
x2 =
2a
Hal.: 184
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Adaptif
Problem 1
Find the roots of :
x2 - 5x + 6 = 0
Horeeee
I can………..!
Hal.: 185
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Adaptif
Problem 2
Find the quadratic equation, if roots is
x1 = 2 and x2 = 3
?????
Emmm what
equation???
Hal.: 186
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Problem solving
 Find the roots of
x2-5x+6=0
 Answer :
 x2-5x+6=0
 (x-2)(x-3)=0
 x-2=0 or x-3=0
 x=2
x=3
 Hp {2,3}
Hal.: 187
 Find the quadratic
equation, if roots is
x1=2 & x2=3
 then:
 x=2
x=3
 x-2=0 or x-3=0
 (x-2)(x-3)=0
 x2-2x-3x+6=0
 x2-5x+6=0
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Note
 Find the roots of
x2-3x+2=0
 answer:
 x2-3x+2=0
 (x-1)(x-2)=0
 x-1=0 or x-2=0
 x=1
x=2
 Hp {1,2}
Hal.: 188
 Find the quadratic
equation, if roots is
x1=1 & x2=2
 then:
 x=1
x=2
 x-1=0 or x-2=0
 (x-1)(x-2)=0
 x2-2x-x+2=0
 x2-3x+2=0
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Note
 Find the roots of
x2-2x-3=0
 Answer :
 x2-2x-3=0
 (x-3)(x+1)=0
 x-3=0 or x+1=0
 x=3
x=-1
 Hp {-1,3}
Hal.: 189
 Find the quadratic
equation, if roots
is x1=-1 & x2=3
 then:
 x=-1
x=3
 x-(-1)=0 or x-3=0
 (x+1)(x-3)=0
 x2+x-3x-3=0
 x2-2x-3=0
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What you get…?
If quadratic equation have roots of
x1=a dan x2=b then quadratic
equation is :
x=a
x=b
x-a=0 or x-b=0
(x-a)(x-b)=0
x2-ax-bx+ab=0
x2-(a+b)x+ab=0
Hal.: 190
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Adaptif
Conclusion
To arrange a quadratic equation has
been known radical with x1=a or
x2=b then equality is:
(x-a)(x-b)=0 or
x2-(a+b)x+a.b=0
Hal.: 191
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Adaptif
Note
If x1 and x2 is roots of quadratic equation
ax2 + bx +c = 0 (a  0),
the sum and the product of roots of quadratic equations is :
x1 + x2 =  b
a
c
x1 . x2 =
a
Hal.: 192
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Adaptif
If x1 and x2 roots of quadratic equation ax2 + bx + c = 0 (a ≠ 0).
1.
The opposite of roots (x1 = - x2 )  b = 0.
1
)  a = c.
x2
2.
The reversal of roots ( x1 =
3.
A root equal to zero ( x1 = 0)  c = 0 and x2 = 
4.
5.
A both of roots the same sign 
c
a
A both of roots the different sign 
Hal.: 193
b
a
> 0.
c
a
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< 0.
Adaptif
Note
Complete quadratic equations
Quadratic equation
Roots
ax2 + bx + c = 0
x1, x2
Prepare quadratic equations
Hal.: 194
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Adaptif
Composing quadratic equations if the roots are given
a. Product of factor
(x- x1) (x – x2) = 0
b. Using the sum and the product of roots
x2 – (x1 + x2) x + (x1 . x2) = 0
Hal.: 195
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Adaptif
Example 1:
Roots of quadratic equations 3x2 + 6x – 8 = 0 are p and q.
Compose of quadratic equations if roots are
1
1
and .
p
q
Answer :
remember!! Formula the sum and the product of roots: x2 – (x1 + x2)x + x1 . x2 = 0
Given quadratic equation 3x2 + 6x – 8 = 0
Of roots of p and q, then:
pq 
Hal.: 196
6
 2
3
and
p.q 
8
8

3
3
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Adaptif
Advanced answer
instance:
Quadratic equations have requested of roots of x1
and x2,
1
1
and x2 
then x1 
p
1 1

p q
pq
 x1  x 2 
p.q
2
 x1  x 2 
8
3
3
 x1  x 2 
4
x1  x 2 
Hal.: 197
q
1 1
.
p q
1
 x1 .x 2 
p.q
1
 x1 .x 2 
8
3
3
 x1 .x 2  
8
x1 .x 2 
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Adaptif
Advanced answer
Substitute x1  x 2 
3
3
and x1 .x 2  
8
4
to equation x2 – (x1 + x2)x + x1 . x2 = 0
to get:
x2 

3
 3
x      0,
4
 8
both segment to multiply 8
8x 2  6 x  3  0
So, quadratic equation is 8x2 – 6x – 3 = 0
Hal.: 198
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Adaptif
example 2:
The qudratic equation 3x2 + x – 2 = 0, prepare quadratic equation if roots is a
square from roots of equation to determine!
Answer :
remember!! Formula the sum and the product of roots: x2 – (x1 + x2)x + x1 . x2 = 0
Given quadratic equation 3x2 + x – 2 = 0, then:
1
pq 
3
Hal.: 199
and
2
2
p.q 

3
3
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Adaptif
Advanced answer
instance:
asked the equality y2 – (y1 + y2)y + y1 . y2 = 0
with roots y1 and y2
b
2
2
then,
  x1  x 2
a
b
2
2
y1  y 2    x1  x 2
a
c
2
2
y1 . y 2 
 x1 .x 2
a
Hal.: 200
 x1  2 x1 x 2  x 2  2 x1 x 2
2
2
  x1  x 2   2 x1 .x 2 
2
2
 1
 2
     2  
 3
 3
1 4
 
9 3
13

9
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c
2
2
 x1 .x 2
a
  x1 .x 2 
 2
  
 3
4
 
9
2
2
Adaptif
Advanced answer

b
2
2
 x1  x 2
a
 x1  2 x1 x 2  x 2  2 x1 x 2
2
2
  x1  x 2   2 x1 .x 2 
2
2
 1
 2
     2  
 3
 3
1 4
 
9 3
13

9
c
2
2
 x1 .x 2
a
  x1 .x 2 
 2
  
 3
4
 
9
2
2
Substitution:
b 13
 
a 9
c 4

a 9
and
to equation
y2 – (y1 + y2)y + y1 . y2 = 0
to get:
y2 

13
4
y   0,
9
9
both segment to multiply 9
9 y 2  13 y  4  0
So, quadratic equation is 9y2 – 13y + 4 = 0
Hal.: 201
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Adaptif
Example 3
Given α and β is the roots of equation 4x2 – 3x – 2 = 0. find the quadratic
equation if the roots of (α + 3) and (β + 3)!
answer:
Given quadratic equation 4x2 – 3x – 2 = 0 of the roots of α and β, then:
α+β= b = 3
and α . β = c =  2
a
4
a
4
Instance quadratic equation namely, x2 – (x1 + x2)x + x1.x2 = 0 have the roots
of x1 and x2, then x1 = (α + 3) and x2 = (β + 3), then we get:
x1 + x2 = (α + 3) + (β + 3)
= (α + β) + 6
= 3 + 6 = 27
4
4
Hal.: 202
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Adaptif
Advanced answer
x1 . x2 = (α + 3) . (β + 3)
= (α . β) + 3(α + β) + 9
2
3

= 4 + 3( 4 ) + 9
2
4
43
= 4
=  +
9
36
+
4
4
27
Substitution x1 + x2 =
4
and
x1 . x2
43
=
to
4
equation x2 – (x1 + x2)x + x1.x2 = 0, and get
x2 –
43
27
x +
= 0, both segment to multiply 4
4
4

4x2 – 27x + 43 = 0
So, equality that is required 4x2 – 27x + 43 = 0.
Hal.: 203
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Adaptif
Exercise
1.
Set equality x2 + bx + c = 0, a ≠ 0.
composing new quadratic equation the roots of:
a. opposite with the roots of given
b. reversal with the roots of given
c. larger n times from the roots of given
d. m more than from the roots of given
e. constitutes exponent 3 from the roots of given.
2.
Set equality x2 – 3x + 1 = 0.
Composing new quadratic equation the roots of:
a. opposite with the roots of given
b. reversal with the roots of given
c. larger 3 times from the roots of given
d. 3 more than from the roots of given
e. constitutes exponent 3 from the roots of given.
Hal.: 204
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Adaptif
Hal.: 205
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Adaptif
Ways to solve quadratic equations are :
Factoring
Completing the square
Formula
Hal.: 206
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Adaptif
Story problem related to the quadratic equation
Budi will answer the question from their
teacher about election with both the
number the following :
difference both the number is two and the
product both the number is 168.
how the number is ?
Hal.: 207
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Adaptif
Matter how complete the story
Reading problems
Translate in mathematical model
Finishing
Check to result
Hal.: 208
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Adaptif
Finishing
Budi will answer the
question from their
teacher about election
with both the number
the following :
Difference both the
number is two and the
product both the number
is 168.
how the number is ?
Hal.: 209
Given :
∙ Election both the
number
∙ Difference is 2
∙ The product is 168
Asked :
how the number is ?
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Adaptif
next
Budi will answer the
question from their
teacher about election
with both the number
the following :
Difference both the
number is two and the
product both the number
is 168.
How the number is ?
Hal.: 210
answer :
Instance number is a and b
a – b = 2 ………………1)
a . b =168 ………………2)
from equation 2)
a . b =168 => a = 168 / b
substitution to equation 1)
a-b = 2
(168 / b) – b = 2
b2 + 2b – 168 = 0
(b - 12)(b + 14) = 0
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Adaptif
Next
For b – 12 =0
b = 12
For b +14 = 0
b = - 14
Remember number
always positive
then value b to use is
12
a–b=2
 a – 12 = 2
 a = 14
check
∙a–b=2
14 – 12 = 2
∙ a . b = 168
14 . 12 = 168
So, number is 12 and 14
Hal.: 211
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Adaptif