Algebra and Application 11 (Repaired)

8/10/2019 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. ( )
http://www.crayonpedia.org/mw/Berkas:Jawab_aljabar_13.jpg


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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


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


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.

Algebra and Application 11 (Repaired)

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