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Algebra and Application 11 (Repaired)
Transcript of Algebra and Application 11 (Repaired)
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ENGLISH FOR MATH
ALGEBRA AND APPLICATION OF JUNIOR HIGH SCHOOL
Oleh : 2ndGROUP
Atikah (E1R011007)
Bq Sri Rahayu Kartini (E1R011009)
Luh Putu Asri Parwati (E1R011023)
Satria Irawansyah (E1R0110..)
Pendidikan Matematika
Pendidikan Matematika dan Ilmu Pengetahuan Alam
Fakultas Keguruan dan Ilmu Pendidikan
Universitas Mataram
2012
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ALGEBRA AND APPLICATION
A. Algebraic Form
1.
Define Algebraic Form
An algebraic form is an expressions showing combination between coefficient and
variable which is represented by algebraic equations. For example, 3a5+12x6 is called
algebraic form.
3a5+ 12x
6
2.Definition of Like and Unlike Terms
Two or more terms that have exactly the same variables are called like terms, and if
the variables are different, its called unlike terms. Look at the example below.
terms
variable
coefficient
Algebraic form
terms
Example :
1. Simplify the following algebraic forms
a. 7 a b. yz yz yz c. (b b b b) + (c c c c)
Solution :
a. 7 a = 7ab. yz yz yz = (yz)3
c. (b b b b) + (c c c c) = b4 +c4
2. Find the terms, the variable, and the coefficient of 12x3+ 6a32b.
Solution :
~ The terms are 12x3, 6a3 and2b.
~ The variable arex, a, and bbecause the values of x, a and b can vary.
~ The coefficient are 12, 6, and -2. Its value does not change.
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3.Addition and Subtraction of Like and Unlike Terms
If an algebraic forms has like terms, it can be simplified by adding of subtracting.
Unlike terms, cannot be simplified. In essence, the nature of the occurring sum and
reduction in the number of real, happens also to summing and reduction in algebraic
forms, as follows,
Commutative nature ofa + b = b + a, a and b are real number
Associative nature(a + b) + c = a + (b + c), where a, b, and c real number
Distributive nature ofa (b + c) = ab + ac, where a, b, and c the number of real
4. Multiplication and Division of Like and Unlike Terms
Example :
Find out the like terms of 2pq + 7p2q8pq + 5p2q.
Solution:
Look at the algebraic forms of 2pq + 7p2q8pq + 5p2q.
There are some like terms, 2pq and8pq also 7p2q and 5p2q.
Like terms
Like terms
2pq + 7p2q8pq + 5p
2q
Example :
Simplify the following algebraic forms.
1. 2pq + 3p2q5pq + 3p2q
2. 4b5ca6ca2+ 6b5c5b5ca + 7ca2
Solution :1. 2pq + 3p2q5pq + 3p2q = 2pq5pq + 3p2q + 3p2q
=3pq + 6p2q
2. 4b5ca6ca2+ 6b5c5b5ca + 7ca2= 4b5ca5b5ca6ca2 + 7ca2+ 6b5c
=b5ca + ca2+ 6b5c
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You can multiply and divide the algebraic forms as they were done in integers. For
example, 3 a = a + a + a = 3a 15a : 3 = 5a
The following properties can be used to simplify the algebraic forms.
The multiplication distributive law across addition The property of The property of
5. Special Product of Algebraic Forms
The special products of algebra forms, a(b + c + d), (a + b) (c + b), and (a +d)2. To
find out the result of these multiplication, we use the distributive law of multiplication
across addition or subtraction. See the following steps :
Example :
1. Find out the results of the following operations.
a. 4a 9b b. 12a2: 3b
Solution :
a. 4a 9b = 4 a 9 b = 4 9 a b = 36ab
b. 12a2: 3b = (12 a a) : (3 b)
=
=
2. Look at the picture. Find out the area of the rectangle.
Solution :
The length of the figure is 2y units and its widht is 5x units.
Thus, the area of the ficture is 2y 5x = 2 y 5 x = 2 5 y x =
10xy units.
2y
5x
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With the same method, the result of a(b + c +d) and (a + b)2 are (a + b)2 ab +ac + ad
and a2 + 2ab + b2.
6. Exponent of Like and Unlike Terms
The properties of the exponent of integers can be used to solve the operation of
exponential algebraic forms.
The properties of exponent on integers are:
See the example below.
Example :
Simplify the following algebraic forms
1. 2. 4: 2ySolution :
1. = (3 4ab) + (3 2bc)= (3 4 a b) + (3 2 b c)
= 12ab + 6bc
2. 4: 2y = : 2 = 2xy
Example :
Simply the following exponential algebraic forms.
1. ((x)2)5 2. (3x5)3 3. (a5b3)7
Solution :
1. ((x)2)5= x2 5 = x10
2. (3x5)3 = 33x5 3= 27x15
3. (a5b3)7= a5 7b 3 7= a35b21
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B. Algebraic Fractions
Algebraic terms can be written as fractions, for example,
A fractionis a number which can be written in the form of , where a and b are integers
and b 0, and b is not a factor of a.1. Addition and Subtraction of Algebraic Fractions
There are two operations of addition or subtraction on algebraic fractions, as
describle below:
a. to add or subtrac two fractions with the same denominatir, writw the sum or
difference or given numerators over the given denominator.
1.
2.
b. fractions whose denominators are different, cannot be combined until they have been
changed to equivalent fractions with these denomonator. To get an equivalent
fractions we use the Least Common Multiple (LCM) the denominator.
LCM of algebraic fractions are product of multiplication of prime factors to the
highest power to which it is raised in any one of the given fractions.
Example :
Determinane the LCM of 6b and 4a. Find out the result of
Solution :
Yoy should find out the prime factors of 6b and 4a first.
The prime factor of 6b and 4a can be written as follows.
6b = 2 . 3 . b
4a = 22. a
The prime factors with the highest power of 6b and 4a are 22and 3.
Thus, the LCM of 6b and 4a is 22. 3 . a. B = 4 . 3 . a . b = 12ab.
You can find out the result of
from the LCM of the denominators as
follows.
Prime number
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2. Multiplication and Division of Algebraic Fractions
a. Multiplication of algebraic fractions
The multification of algebraic fractions, ike all other kinds of multiplication, that
is the product of the numerators divided by the product of their denominators.
Ifand
are fractions, then
where b and d .
b. Division of algebraic fractions
The division of algebraic fractions is the reverse of the multiplication proces. It
is like all kinds of division fractions.
Ifand
are fractions, then
where b , c and d .
Example :
Simplify the algebraic form of
Solution :
Example :
Simplify the algebraic fraction division of
Solution :
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3. Exponents of Algebraic Fractions
The operations of exponential algebraic fractions have its similartities to that of
whole number exponents in the simple operations.
The properties of exponent on algebraic fractions are:
Example :
Simplify the following power of algebraic fractions.
1.
2. ( )
Solution :
1.
2. ( )
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E. Application of Algebra
In this part, we will discuss some application of algebraic forms in daily life, for
example to determine discount.
1. Total Value and Value per UnitTo find out the value, we must know the value per unit first.
Total Value = Number of units x Value per unit
To find out the value per unit, we must know the total value first.
Value per unit =
2. Profit and Loss
If selling price is higher than buying price its make profit.
Profit = selling pricebuying price
If selling price is lower than buying price its make loss.
Loss = buying price- selling price
There are two ways to determine profit and loss by amount and percentage.
Profit percentages =
Loss percentages =
Example :
A book coasts Rp3,000.00. Ningsihs money is worth only to buy 10 books. If the
book prices decrease into Rp2,500.00, how many books she can buy?
Solution:
If one book prices is h, and Ningsihs money only enough to bought 10 books, we
can write that Ningsihs money is 10 x h = 10h.
If the book prices decrease into Rp2,500.00, the number of books that Ningsih
can buy is,
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Example :
1. Mr. Randy buy cow for Rp8,000,000.00. After one month the weight of the
cow is 250 kg. Then, he sell it for Rp40,000.00 per kg. What does he
make?Profit or loss? How much does he make?
Solution :
Buying price = Rp8,000,000.00
If selling price is x, and it formula is
Selling price = weight of cow h price per kg, we can write
h = 250 x Rp40,000.00 = Rp10,000,000.00
Thous, selling price of the cow is Rp10,000,000.00, and its higher than buying
price. So, Mr. Randy make a profit. His profit is,
Profit = Rp10,000,000.00Rp8,000,000.00 = Rp2,000,000.00
2. Anton bought a bike for Rp210,000.00. but this bike needs repairing before
being sold. The cost for its repair is Rp50,000.00. Impact, he had a loss of
Rp12,500.00. What percent did he make a loss?
Solution :
You must find out the total coast before the bike is sold.
If total coast is x, we can write
x = buying price + repair coast = Rp210,000.00 + Rp50,000.00 = Rp260,000.00
Thous, total coast that Antons paid is Rp260,000.00
Anton had a loss of Rp12,500.00. If percentage of Anton loss is y,
Y =
=
= 4,81 %
We can conclude that Anton had a loss of 4,81 % from buying price.
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3. Discount and Rebate
A price reduction is called discout, while rebate mean cutting price for things you
bought in a large amount, for example if you bought one dozen of books. Discount and
rebate has a same calculation.
4. Saving Interest and Cooperation Interest
Have you save your money in the bank? If you save your money in the bank you
will receive the mounthly interest. The amount of imnteres you get is depens on the
amount of your money in the bank. Interest usually represent in percent. How to
calculate saving interst?
Example :
Fandy bought USB flashdisk for Rp370,000.00. If he got 15% discount, how
much was a discount and how much is a sell price of a USB afer getting a
discount?
Solution :
Fandy got 15% discount for Rp370,000.00.
Thous, Therefor, the sell price of the USB after getting the discount is
R 370,000.00R 55,500.00 = R 314,500.00
Example :
1. Mr. Burhan save Rp7,000,000.00 in bank with the monthly interest rate of
1,2%. Determine the amount of Mr Burhans money after three years.
Solution :
Mr. Burhan capital Rp7,000,000.00.
Because the monthly interest rate of 1,2%. So the money after one year is 12
x 1,2% = 14,4%.
So, the interst rate after three years 3 x 14,4% = 43,2%
So, Burhans money after three years is
Capital + Interest = Rp7,000,000.00 + (
= Rp7,000,000.00 + Rp3,024,000.00
= R 10,024,000.00
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2. Mr. Dody borrowed Rp20,000,000.00 from cooperation with the monthly
interest rate of 1,5%. Mr. Dody wants to pay back each month in four years.
Determine how much he has to pay each month.
Solution :
The borrowing capital is Rp20,000,000.00.
The monthly interst rate of 1,5%.
So, the interest rate in four years is 4 x 12 x 1,5% = 72%
Hence, the total sum of interest should be paid in four years is
72% x Rp20,000,000,00 =
= Rp14,400,000.00
The total sum of money should be repaid by Mr Dody isRp20,000,0000.00 + Rp14,400,000.00 = Rp34,400.000.00
If the total sum of money will be installed for 4 years (48 months) then the
monthly installment will be
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C. Factoring Algebra
1. Factoring with distributive properties
In essence, factoring out a number of means in the form of specifying a number of
the factors multiplication. In this part, will learn the ways factoring an algebraic form
by using the distributive nature. By its nature, the form ax + ay algebra can be factored
into a(x + y), where a is a federation factor of ax and ay. For that, learn the following
example question.
Examples
Factorise following algebraic forms.
a. 5ab + 10b c. -15p2q2+ 10pq
b. 2x - 8x
2
y d.
1
/2a
3
b
2
+
1
/4a
2
b
3
Answer:
a. 5ab + 10b
For 5ab + 10b factoring, determine the federal factor of 5 and 10, then from ab and
b. Federal factor of 5 and 10 is 5. Federal factor from ab and b is b. So, 5ab + 10b
factored into 5b (a + 2).
b.
2x - 8x2y
Federal factor of 2 and -8 is 2. Federal factor of x and x2y is x.
So, 2x - 8x 2y = 2x (1 - 4xy).
c. -15p2q2+ 10pq
Federal factor of -15 and 10 is 5. Federal factor of p2q2and pq is pq.
So, -15p2q2+ 10pq = 5pq (-3pq + 2).
d.1/2a
3b2 + 1/4a2b3
Federal factor of 1/2and1/4is
1/4. Federal factor of a3b
2and a
2b
3is a
2b
2.
So,1/2a
3b
2+
1/4a
2b
3=
1/4a
2b
2(2a + b)
2. Two Squared Difference
Observe the shape of multiplication (a + b)(a - b). This form can be written
(a + b) (a - b) = a2- ab + ab - b2
= a2- b2
So, form a 2- b 2can be expressed in the form of multiplication (a + b) (a - b).
Form a 2- b 2called difference of two squares.
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Examples
Factorise following forms.
a. p2- 4 c. 16m29n2
b. 25x2- y2 d. 20p25q2
Answer:
a. p 2- 4 = (p + 2)(p - 2)
b. 25x2- y2= (5x + y)(5x - y)
c. 16m29n2= (4m + 3n)(4m - 3n)
d. 20p25q2= 5(4p2- q2) = 5(2p + q)(2p - q)
3. Factoring Quadratic Forms
a. Factoring the form ax2+ bx + c with a = 1
Note the two terms following multiplication.
(x + p)(x + q) = x2+ px + qx + pq = x2+ (p + q)x + pq
So, the form x2+ (p + q)x + pq can be factored into (x + p)(x + q).
For example, x2+ (p + q)x + pq = ax2+ bx + c so a = 1, b = p + q, and c = pq.
From that can be seen that p and q is a factor of c. If p and q are summed, the result
is b. Hence the need to factor in the form ax2+ bx + c with a = 1, determine the two
numbers is a factor of c and when the both of number is summed, the result is equal
to b.
Examples
Factorise following forms.
a. x2+ 5x + 6 b. x2+ 2x8
Answer:
a. x2+ 5x + 6 = (x + ...)(x + ...)
For example, x2 + 5x + 6 = ax2 + bx + c, obtained a = 1, b = 5, and c = 6.
To fill points, determine the two numbers is a factor of 6
, and when the second number is summed, the result is equal to 5. Factor of 6 is 6
and 1 or 2 and 3, that meets the requirements is 2 and 3.
So, x2+ 5x + 6 = (x + 2)(x + 3)
b. x2+ 2x - 8 = (x + ...)(x + ...)
By way as in (a), obtained a = 1, b = 2, and c = -8. Factors of 8 are 1, 2, 4, and 8.
As for c = -8, one of two numbers that look certainly has negative value. Thus,
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two qualified number is -2 and 4, because -2 4 = -8 and
-2 + 4 = 2. So, x2+ 2x - 8 = (x + (-2))(x + 4) = (x - 2)(x + 4).
b. Factoring form ax2+ bx + c with a 1
Previously, you were factoring the form ax 2+ bx + c with a = 1. Now you will
learn how to factor in the form ax 2+ bx + c with a 1.Attention the two terms
following multiplication.
(x + 3)(2x + 1) = 2x2+ x + 6x + 3 = 2x2+ 7x + 3
In other words, the form of 2x2+ 7x + 3 factored into (x + 3)(2x + 1). As for how to
factorise 2x2 + 7x + 3 is by turning the binomial multiplication stages above.
2x2+ 7x + 3 = 2x2+ (x + 6 x) +3 (describe 7x be the sum of the two terms is select
(x + 6x )
= (2x2+ x) + (6x + 3)
= x(2x + 1) + 3(2x + 1) (Factorise use distributive properties)
= (x + 3)(2x +1)
From the description you would have know how to factor in the form ax 2+ bx + c
with a 1 as follows.
1.Describe bx be the sum of two terms when the both of terms results are the same
multiplied by (ax2)(c).
2.
Factorise form obtained using the distributive nature
Examples
Factorise following forms.
a. 2x2+ 11x + 12 b. 6x 2+ 16x + 18
Answer:
a.
2x2+ 11x + 12 = 2x2+ 3x + 8x + 12
= (2x2+ 3x) + (8x + 12)
= x (2x + 3) + 4 (2x + 3)
= (x + 4) (2x + 3)
So, 2x2+ 11x + 12 = (x + 4)(2x + 3).
b. 6x2+ 16x + 8 = 6x2+ 4x + 12x + 8
= (6x2+ 4x) + (12x + 8)
= 2x(3x + 2) + 4(3x + 2)
= (2x + 4) (3x + 2)
So, 6x2
+ 16x + 8 = (2x + 4) (3x +2).
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D. Linear Systems with Two Variables
A linear system of two equations with two variables is any system that can be written
in the form.
ax + by = p
cx + dy = q
Where any of the constants can be zero with the exception that each equation must
have at least one variable in it.
Also, the system is called linear if the variables are only to the first power, are only in
the numerator and there are no products of variables in any of the equations.
Here is an example of a system with numbers.
3xy = 72x + 3y = 1
Before we discuss how to solve systems we should first talk about just what a solution
to a system of equations is. A solution to a system of equations is a value ofxand a value
ofythat, when substituted into the equations, satisfies both equations at the same time.
For the example above x = 2 and y = 1 is a solution to the system. This is easy
enoughto ckeck.
3(2)(1) = 72(2) + 3 (1) = 1
So, sure enough that pair of numbers is a solution to the system. Do not worry about
how we got these values. This will be the very first system that we solve when we get into
examples.
Note that it is important that the pair of numbers satisfy both equations. For instance
x = 1 and y =4 will satisfy the first equation, but not the second and so isnt a solution to
the system. Likewise, x = 1 and y = 1 will satisfy the second equation but not the firstand so cant be a solution to the system.
Two methods for solving systems
1.
Method of subtitusion
The first method is called the method of substitution. In this method we will
solve one of the equations for one of the variables and substitute this into the other
equation. This will yield one equation with one variable that we can solve. Once
this is solved we substitute this value back into one of the equations to find the
value of the remaining variable.
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In words this method is not always very clear. Lets work a couple of examples
to see how this method works.
Example 1: solve each of the following systems.
a. 3xy = 7 b. 5x + 4y = 1
2x + 3y = 1 3x6y = 2
Solution :
(a) . 3xy = 7
2x + 3y = 1
So, this was the first system that we looked at above. We already know
the solution, but this will give us a chance to verify the values that we wrote
down for the solution.
Now, the method says that we need to solve one of the equations for one
of the variables. Which equation we choose and which variable that we
choose is up to you, but its usually best to pick an equation and variable that
will be easy to deal with. This means we should try to avoid fractions if at all
possible.
In this case it looks like it will be really easy to solve the first equation for
yso lets do that.
3xy = 1
Now, substitute this into the second equation.
This is an equation inxthat we can solve so lets do that.
2x + 9x21 = 1
11x = 22
So, there is thexportion of the solution.
Finally, do NOT forget to go back and find the yportion of the solution.
This is one of the more common mistakes students make in solving systems.
To so this we can either plug thexvalue into one of the original equations and
solve foryor we can just plug it into our substitution that we found in the first
step. That will be easier so lets do that.
y = 3x7 = 3(2)7 = -1
So, the solution is x = 2 and y = -1 as we noted above.
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(b) 5x + 4y = 1
3x6y = 2
With this system we arent going to be able to completely avoid fractions. However, it looks like if we solve the second equation forxwe can minimize
them. Here is that work.
3x = 6y + 2
x = 2y +
Now, substitute this into the first equation and solve the resulting equation
for y.
Finally, substitute this into the original substitution to findx.
( )
So, the solution to this system is x = and y = -
.
As with single equations we could always go back and check this solution by
plugging it into both equations and making sure that it does satisfy both equations.
Note as well that we really would need to plug into both equations. It is quite
possible that a mistake could result in a pair of numbers that would satisfy one of
the equations but not the other one.
Lets now move into the next method for solving systems of equations. As we
saw in the last part of the previous example the method of substitution will often
force us to deal with fractions, which adds to the likelihood of mistakes. This
second method will not have this problem. Well, thats not completely true. If
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fractions are going to show up they will only show up in the final step and they will
only show up if the solution contains fractions.
b. This second method is called the method of elimination. In this method we
multiply one or both of the equations by appropriate numbers (i.e.multiply every
term in the equation by the number) so that one of the variables will have the same
coefficient with opposite signs. Then next step is to add the two equations
together. Because one of the variables had the same coefficient with opposite signs
it will be eliminated when we add the two equations. The result will be a single
equation that we can solve for one of the variables. Once this is done substitute
this answer back into one of the original equations.
As with the first method its much easier to see whats going on here with a couple
of examples.
Example :
a. 5x + 4y = 1 b. 2x + 4y = 1
3x6y = 2 6x + 3y = 6
Solution :
a. 5x + 4y = 1
3x6y = 2
This is the system in the previous set of examples that made us work with
fractions. Working it here will show the differences between the two methods and it
will also show that either method can be used to get the solution to a system.
So, we need to multiply one or both equations by constants so that one of the
variables has the same coefficient with opposite signs. So, since theyterms already
have opposite signs lets work with these terms. It looks like if we multiply the first
equation by 3 and the second equation by 2 theyterms will have coefficients of 12
and -12 which is what we need for this method.
Here is the work for this step.
So, as the description of the method promised we have an equation that can be
solved forx. Doing this gives, which is exactly what we found in the previousexample. Notice however, that the only fraction that we had to deal with to this point
is the answer itself which is different from the method of substitution.
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Now, again dont forget to findy. In this case it will be a little more work than
the method of substitution. To findywe need to substitute the value ofxinto either
of the original equations and solve fory. Sincexis a fraction lets notice that, in this
case, if we plug this value into the second equation we will lose the fractions at least
temporarily. Note that often this wont happen and well be forced to deal with
fractions whether we want to or not.
Again, this is the same value we found in the previous example.
b. 2x + 4y = 1
6x + 3y = 6
In this part all the variables are positive so were going to have to force an
opposite sign by multiplying by a negative number somewhere. Lets also notice
that in this case if we just multiply the first equation by -3 then the coefficients of
thexwill be -6 and 6.
Sometimes we only need to multiply one of the equations and can leave the
other one alone. Here is this work for this part.
Finally, plug this into either of the equations and solve forx. We will use the
first equation this time.
So, the solution to this system is x = 3 and y = 4.