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Chapter I Review topics in Algebra 1

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1.Set of real numbers The word algebra originated from the Arabic word “al-jabr” which means

the science of reduction and cancellation. The algebraic symbolism used to generalize the operatiob of arithmetic was formulate in the sixteenth and seventeenth centuries.

Real number set of rational numbers and the set of irrational numbers make up. It

consists of the set of real numbers and two operations called addition and multiplication. Addiotion is denoted by the symbol “+” and multiplication is denoted by the symbol “x” or “”. If a nd b are real numbers, a+b denotes the sum of a and b, and ab or (ab) denotes their products.

If the numbers are repeating or terminating decimal they called rational number. The square roots of perfect squares also name rational number.

Examples:

1)

√0.16

2) 0.666

3)13

4)109

5)96

If the numbers are not repeating or terminating decimals. They called irrational number. For examples:

1) π

3

2) √23) 0.613514) √85) √11

Properties of real numbersLet us denote the set of real numbers by R. These properties are statement derived from the basic axioms of the real numbers system. Axioms are assumptions on operation with numbers.Axioms of EqualityLet a, b, c, d ∈ R

1. Reflexive LawIf a=a

2. Symmetric LawIf b=c then c=b

3. Transitive LawIf b=c and c=d then b=d

4. Additional Law of EqualityIf a=b then a+c=b+c

5. Multiplication Law of Equality If a=b then a.c=b.c

Axioms for Addition and MultiplicationLet a, b, c, d, ∈R

1) A. Closure property for addition a+b ∈ R

Examples:1) 3+3=62) 7+(-4)=33) -8+4=-4

B. Closure property for multiplication a.b ∈ R Examples:

1) 3(7)=212) -8(3)=-243) 0.11=0

2) A. Commutative prroperty for addition a+b=b+a Examples:

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1)12+7=7+ 1

22) 0.3+(−56 )=−5

6+0.3

3) 13+21=21+ 1

3B. Commutative prroperty for multiplication a.b=b.a Examples:

1)45

(22 )=22( 45 )

2) 6.3=3.6

3)109

(−25 )=−25( 109 )

3) A. Associative property for addition (a+b)+c=a+(b+c) Examples:

1) (3+7)+0.4=3+(7+0.4)

2) (0.36+89)+12=0.36+(89+ 1

2 )3) ( 3

5+0.8)+ 3

8=3

5+(0.8+ 3

8 )B. Associative property for multiplication (a.b).c=a.(b.c) Examples:

1) (3.x).y=3.(x.y)

2) [5 (7 ) ] 14=5 [7 ( 1

4 )]3) [3x (6 x )]¿5=3 x [6 x (5)]

4) Identity property for multiplication a.1=a Examples:

1) 1.a3=a3

2)37

(1 )=37

3) 3.1=35) A. Inverse property for addition

a+(-a)=0 Examples:

1) 6+(-6)=0

5

2) 10+(-10)=03) -3+3=0

B. Inverse property for multiplication

a .1a=1

Examples:1) -2¿)=1

2) 8(18

)=1

3) -6(-16

)=1

6) Distributive property of multiplication over addition a(b+c)=ab+ac Examples:

1) 3(4+6)=3(4)+3(6)2) -6(7+1)=-6(7)+[-6(1)]3) a(7+5)=7a+5

1.2 Exponents and Radicals In the expression α n, α is the base and n is the exponent. The expression α n means

that the value α is multiplied n times by itself.Examples:1) 63= 6.6.6 =2162) 56= 5.5.5.5.5 =156253) 42= 4.4 =16

Integral and zero exponentsLaws of Integral and Zero ExponentsTheorem 1:

For any real number α, (α≠ 0)

a0=1Examples:

1) (6a0+3)0=1

2) 6α0+70=6(1)+1=73) 2α0+70=2(1)+1=3Theorem 2:For any real numbers α,

αm. α𝘯= αm+n

where m and n are integers.

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Examples:1) α5.α4=a5+4=a9

2) 4xy2 ( 2 x2 y2 )=8 x1+2 y2+2=8 x3 y4

3) xa+3 . xa+4=x2 a+7

Theorem 3:For any real numbers a+b,

(ab)n=anbn,where n is any integer.Examples:1) (5x)2=55x2=25x2

2) (-2x)3=-23x3=-8x3

3) [x(x-3)]2=x2(x-3)2

=x2(x2-6x+9) =x4-6x3+9x2

Theorem 4:For any real numbers a(am)n=amn

where m and n are integers.Examples:1) (-x2)3=-x2(3)=-x6

2) [(3x+4)2]3=(3x+4)6

3) (-x2y3z)4=-x8y12z4

Theorem 5:For any real numbers a and b (b≠0),

( ab)

n

=an

bn

where n is any integer.Examples:

1) (a2

b3 ¿¿2=a4

b6

2) (34

¿¿3=33

43 =2764

3) (x

y+2¿¿2

=x2

( y+2)2 =x2

y2+4 y+4Theorem 6:For any real numbers a(a≠0),

am

an =am−n

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where m and n are integers. Examples:

1)a7

a5 =a7−5=a2

2) x3 y4 z5

xyz=x3−1 y4−1 z5−1=x2 y3 z4

3)x4 y4

x4 y4 =x4−4 y4−4=x0 y0=1 (1 )=1

Theorem 7:For any real numbers a(a≠0),

a−n= 1

an

Where n is any positive integer.Examples:

1) 3 x3 y−2=3 x3

y2

2) ( 4 x2 y )−2= 1

( 4 x2 y )2= 1

8 x4 y2

3) ¿

Fractional Exponents: Radicals Since not all numbers are integers, we can’t expect exponents to always whole number or zero. Exponents can be form fractional. Fractional exponents may seem unfamilliar for they are usually expressed as radicals.

For expression x12 is the same as √2 (read as square root of 2), and x

23 is

the same as 3√ x2 (read as cube root of x squared). The expression n√am is called a

radical. The symbol √ is called a radical sign, where n is the index, a is the radicand and m is the power of the radicand.

amn=

n√am

Laws of RadicalsTheorem 1:For any real numbers a,

n√an=aExamples:

1) √42=4

2) 3√¿¿¿=x2 y

3) 3√33=3

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Theorem 2:For any real numbers a,and b.

n√a . n√b=n√abExamples:

1) √3 .√3=√3.3=√9=32) √4 .√3=√4.3=√123) √a .√b=√a . b

Theorem 3:For any real numbers a,and b, (b≠0)

n√an√b

=n√ ab

Examples:

1)3√a3√b

=3√ ab

2)√4√5

=√ 45

3)4√ x4√ y

=4√ xy

Theorem 4:For any real numbers a ,

mn√a=m√ n√a=

n√m√aExamples:

1) 6√64=3√√64= 3√8=2

2) 4√16= 2√√16=2√4=¿2

3) 3√100= 3√100=√100=10Theorem 5:For any real numbers a

kn√akm=n√am

Examples:

1) 6√24=2.3√22.2=3√22=3√4

2) 6√93=3.2√93.1=2√9=3

3)Addition and Sutraction of Radicals

To add and subtract radicals, first we need to combine the like terms with similar radicals.Examples:

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1) √2+3√2−2√2=2√22) √8+√18+√32=√4.2+√9.2+√16.2=2√2+3√2+4√2=9√23)

y √x3 y−√ x3 y3+x √xy3= y √x2 . xy−√x2 . x . y2 . y+ x√ x . y2 . y=xy√ xy−xy √ xy+ xy √xy=xy √ xy

Multiplication and Division of RadicalsTo multiply and divide radicals with the same index, multiply, or divide

the radicals and copy the common index.Examples:

1) √3.√3=√32=3

2) 3√ xy.3√ x2 y . 3√xz=3√ xy . x2 y . xz=3√ x4 y2 z=x3√xy2 z

3) 3√16÷3√−2=3√16 ÷ ¿¿)=3√−8=−2

1.3 Polynomials

Polynomials was used to describe any algebraic expression. The algebraic expression, 5x+4 and x3+x2+1 are examples of polynomials in variable x. A polynomial with just one term 2x is called a monomial. If the polynomial is the sum or difference of two terms as in -9x+7, then it is called a binomial. If it has three terms like x 2+2x+1, then it is called a trinomial. In general a polynomial consisting of a sum of any numbers of terms is called a multinomial.

In the binomial, 5x+4 the number 5 is called the numerical coefficient of x while x is the literal coefficient and the numbers 4 is the constant term.

Addition and Sutraction of Polynomials

To determined the sums and differences of polynomials, only the coefficients are combined. By similar terms are refer to the terms with the same coefficients. Those with different literal coefficient are called dissimilar or unlike terms.

Examples:

1. Find the sum of 2x-3y+5 and x+2y-1,=(2x-3y+5)+( x+2y-1)=2x+x-3y+2y+5-1=3x-y++4

2. Find the differences between 2x-3y+5 and x+2y-1=(2x-3y+5)-( x+2y-1)

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=2x-3y+5+(-x-2y+1)=2x-x-3y-2y+5+1=x-5y+6

3. Subtract 2(4x+2y+3) from 5(2x-3y+1)=5(2x-3y+1)- 2(4x+2y+3)=10x-15y+5-8x+4y+6=2x-11y+11

Multiplication of PolynomialsExamples:

1) xm.xn=xm+n

2) x−2.x2=x0=13) Multiply a+2b+3c by 5m.

= a+2b+3c(5m) in multiplication, we apply the =5am+10bm+15cm distributive property

Division of PolynomialsTo divide a polynomial by a monomial, divide each term of the polynomial by the

monomial.

xm

xn =xm−n∧x−n= 1xn

Examples:

1)x5

x2 =x3

2) x−5=1

x5

3) Divide 7 x2−5 xby x x is the divisor and 7 x2−5 x as the dividend, we have

7 x2−5 xx

=7 x2

x-5 xx

=7 x−5

1.4 FactoringFactors and Greatest Common Denominator

If the two of more numbers are multiplied, each number is a factor of the product. In the example above, 18 is expressed as the product of different pair of whole numbers.

18=2.918=3.618=18.1A prime number is a whole number, greater than1, whose only factors are 1 and

itself. A composite number is a whole number greater than 1, that is not prime.Examples:

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1) Find the prime factorization of 84.84=2.42 the least prime factor of 84 is 2 =2.2.24 the least prime factor of 42 is 2 =2.2.3.7 the least prime factor of 21 is 3All of the factors in tha last row are prime. Thus, the prime factorization of 84 is 2.2.3.7 or 22.3.7.

2) Factor 20a2b20a2b=2.10.a.a.b

=2.2.5.a.a.bThe greatest common factor of two or more integer is the product of the prime

factors common to the integers.Examples:1) Find the GCF of 54, 63, and 180.

54=2.③.③.3 factor each number

63③.③7

180=2.2.③.③.5 then circle the common factorsThe GCF of 54, 63, and 180 is 3.3 or 9.

2) 8 a2b∧18 a2 b2 c

8 a2b= .2② .2 . a⃝� . a⃝� . b ⃝�

18 a2 b2 c=②.3.3.a ⃝� . a ⃝� . b ⃝� . b . c

¿2 a2 bThe GCF of 8 a2b∧18 a2 b2 c is 2 a2 b.

Factoring Using the Distributive PropertyTo multiplied a polynomial by a monomial by using the distributive property.

Multiplying Polynomials Factoring Polynomials3(a+b)=3a+3b 3a+3b=3(a+b)x(y-z)=xy-xz xy-xz= x(y-z)3y(4x+2)=3y(4x)+3y(2) 12xy+6y=3y(4x)+3y(2)

=12xy+6y =3y(4x+2)Examples:

1) Use the distributive property to factor 10 y2+15 y

10 y2=2.⑤. y ⃝� . y

15 y=¿3.⑤.y

The GCF is 5y

10 y2+15 y=5y(2y)+5y(3)

=5y(2y+3) distributive property2) Factor 21 ab2−33 a2 bc

21 ab2=③.7. a ⃝� . b ⃝� . b

33 a2 bc=③.11. a ⃝� . a . b ⃝�.c the GCF is 3ab

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21 ab2−33 a2 bc= 3ab(7b)-3ab(11ac)

=3ab(7b-11ac) distributive propertyFactoring by Grouping

Polynomial with four or more terms, like 3xy-21y+5x-35, can sometimes be factored by grouping terms of the polynomials. One method is to group the terms into binomials that can each be factored using the distributive property. Then use the the distributive property again with a binomial as the common factor.

Examples:1) Factor 3xy-21y+5x-35

3xy-21y+5x-35= (3xy-21y)+(5x-35) =3y(x-7)+5(x-7)

=3y+5(x-7)Check by using FOIL ;(3y+5)(x-7)=3y(x)+3y(-7)+5(x)-5(7)

=3xy-21y+5x-352) Factor 8 m2 n−5 m−24 mn+15

8 m2 n−5 m−24 mn+15=(8 m2 n−5 m ¿+¿¿m (8 mn−5 )+(−3 ) (8 mn−5 )

=m−3 (8mn−5)Check:m−3 (8mn−5 )=m (8mn)+m (−5 )+(−3 ) (8mn )+¿−3)(-5)

¿8 m2 n−5 m−24 mn+15Factoring Trinomials

When two numbers are multiplied each number is a factor of the product. Similarly if two binomials are multiplied, each binomials is factor of the product.

Consider the binomials 5x+2 and 3x+7. You can use the FOIL method to find their product.(5x+2)( 3x+7)=(5x)(3x)+(5x)(7)+(2)(3x)+(2)(7)

=15x2+35x+6x+14 =15x2+(35+6)x+14 =15x2+41x+14

You can be use this pattern to factor quadratic trinomials, such as 2 y2+7 y+6Factors of 12 Sum of Factors

1.12 1+12=13 no2.6 2+6=8 no3.4 3+4=7 yes

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2 y2+ (3+4 ) y+6 Select the factors 3 and 4.

2 y2+3 y+4 y+6

(2 y2+3 y )+(4 y+6) Group terms that have a

y (2 y+3 )+2(2 y+3) common monomials factor

( y+2)(2 y+3) Factor (use the distirbutive property)

Therefore 2 y2+7 y+6= ( y+2)(2 y+3)

Example:

Factor 5 x−6+x2

The trinomials 5 x−6+x2 can be written as x2+5 x−6. For this trinomials, the constant terms is -6 and the coefficient of x is 5. Thus, we need ti find two factors two factors of -6 whose sum is 5.Factors of -6 Sum of factors 1, -6 1+(-6)=-5 no-1, 6 -1+6=5 yes

Select the factors -1 and 6

Therefore, x2+5 x−6=(x−1)(x+6)

Factoring Differences of SquareThe product of the sum and ifference of two expressions is called the differences

of squares. The process for finding this product can be reversed in order to factor the differenceof squres. Factoring the difference of square can also be modeled geometrically.

a2−b2= (a−b ) (a+b )Examples:

1) factor a2−64

a2−64=¿ ¿ (a−8 ) (a+8 )

a .a=a2∧8.8=64 use the difference of square

2) factor ax2−100 y2

ax2−100 y2=¿ ¿ (3 x−10 y )(3 x+10 y)

3 x .3 x=9 x2∧10 y .10 y=100 y2

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Perfect Square and FactoringNumbers such as 1,4,9 and 16 are called perfect squares. Since they can be expressed as the square of an integer. Products of the form ¿ are called perfect squares and the expansions of these products are called perfect square trinomials.

(a+b)2=a2+2ab+b2

(a−b)2=a2−2 ab+b2

Finding a Product Factoring

( y+8)2= y2+2 ( y ) (8 )+82 y2+16 y+64=( y)2+2 ( y ) ( 8 )+¿

¿ y2+16 y+64 ¿( y+8)2

Examples: Determine whether 16 a2+81−72 ais a perfect square trinomial.

1) 16 a2+81−72 a=16 a2−72 a+81

¿ (4 a )2−2 ( 4 a ) (a )+ (a )2

¿(4 a−9)2

2) x2+22 x+121=¿ ¿(x+11)2

Solving Equations by Factoring Zero Product PropertyFor all numbers a and b, if ab=0, then a=0, b=0 or both a and b equal 0.Example:

1) Solve 16t(9-t)=016t(9-t)=0, then 16t=0 or 9-t zero product property16t=0 or 9-t=0 solve each equationt=0 9=tcheck: Substitute 0and 9 for t in the original.

16t(9-t)=016(0)(9-0)=0 or 16(9)(9-9)=0

0(9)=0 144(0)=0 0=0 0=0

SOLUTION SET: (0,9)2) (y+2)(3y+5)=0

If (y+2)(3y+5)=0, then y+2=0 or 3y+5=0y+2=0 or 3y+5=0y=-2 3y=-5

y=−53

Check: (y+2)(3y+5)=0

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(-2+2)[(3)(-2)+5]=0 or (−53

+2) [(3 )(−53 )+5]=0

0(-1)=0 13

(0 )=0

0=0 0=0

SOLUTION SET: (-2, −53

)

1.5 Rational ExpressionsA fraction where the numerator and denominator are polynomials, and is defined

for all values of the variable that do not make the denominator zero.

Reducing Rational Expression to Lowest TermsWe need to lowest term the fraction, if the numerator and denominator have no

common factor.

Examples:

1)4 a2 bc3

6 ab3 c4 =2.2. a . a . b . c . c . c

2.3 . a . b .b . b . c . c . c . c= 2 a

3 b2 c

2)x2+2 xy+ y2

x2− y2 =(x+ y )(x+ y)

( x+ y )−(x− y )=

x+ yx− y

3) x3+8 y3

4 x+8 y=x+2 y¿¿= x2−2 xy+4 y2

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Multiplying and Dividing Rational Expressions

In multiplication if pq∧r

s are rational expressions and q and s are real numbers

not equal to 0, then pq

.rs= pr

qs.

Examples:

1)43

.15= 4

15

2)c

a2−b2.(a+2 b)(a−b)

¿ c(a+b ) ( a−b )

.(a+2 b)(a−b)

=c ( a+2 b )

a+bIn dividing algebraic fractions, multiply the dividend by the reciprocal of the divisor. The reciprocal of a fraction is its multiplicative inverse.

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Examples:

1)43

÷65=4

3.56=20

18∨10

9

2)87

÷ 3=87

.12= 8

14∨4

7

3) y2−16y−5

÷2 y−8xy−5 x

=( y−4)( y+4)

y−5.

x( y−5)2( y−4)

= xy+4 x2

Adding and Subtracting Rational Expressions.To add and subtract rational expressions, it is the important that the least common

denominator is accurately determined.Examples:

1)56−2

3+ 1

8=20−16+3

24= 7

24

2)45+ 3

5+ 2

5=4+3+2

5=9

5

3) 3 x−2 y+2 x2− y2

x+ y=

3 x ( x+ y )−2 y (x+ y )+2 x2− y2

x+ y=3 x2+3 xy−2 xy+2 y2+2 x2− y2

x+ y=

5 x2+xy−3 y2

x+ y

Simplifying Complex Rational ExpressionsA factor which contains one or more fractions either in the numerator or

denominator or in both.Examples:

1)

4313

=43

.31=

123

∨¿4

2)3

2+13

= 36+1

3

= 373

=3.37=9

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1.6 Rational ExponentsWe defined an if n is any integer (positive, negative or zero). To define a

power of a where the exponent is any rational number, not specifically an integer.

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That is, we wish to attach a meaning to a1n∧a

mn , where the exponents are fractions.

Before discussing fractional exponents, we give the following definition.Definition

Examples 1:1) 2 is a square root of 4 because 22=42) 3 is a fourth root of 81 because 34=813) 4 is a cube root of 64 because 43=64

Definition

.

The symbol √ is called a radical sign. The entire expression n√a is called a radical,

where the number a is the radicand and the number n is the index that indicates the order of the radical.

Examples 2:

1) √4=22) 4√81=3

3) 3√64=4Definition

The nth root of a real number

If n is a positive integer greater than 1∧a and b are real number such that bn=a,

then b is an nth root of a.

The principal nth root of a real number. If n is a positive integer greater than 1, a is a real

number, and n√a denotes the princial nth root of a, then

If a>0, n√a is the positive nth root of a.

If a<0, and n is odd, n√a is the negative nth root of a.

n√0=0

If n is a positive integer greater than 1, and a is a real number, then if n√a is a real number

a1n=n√a

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Examples 3:

1) 2512=√25=5

2) −81

3=¿3√−8=−2¿

3) ( 181

)1/4=4√ 181

=13

Definition

Examples 4:

1) 932=(√9¿3=33=27

2) 823= ( 3√8 )2=22=4

3) −2743=¿)4=(-3)4=81

It can be shown that the commutative law holds for rational exponents, and therefore

(a¿¿m)¿1/n=¿¿)m

From which it follows that n√am=( n√a)m

The next theorem follows from this equality and the definition of amn

Theorem 1

Examples 5:Theorem 1 is applied in the following:

1) 932=√93=729 =27

If m and n are positive integers that are relatively prime, and a is a real number,

then if n√a is a real number

amn =( n√a)

m ⇔ amn =¿¿)m

If m and n are positive integrers that are relatively prime, and a is a real number,

then if n√a is a real number

amn =

n√am ⇔ amn =(a¿¿m)¿

1/n

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2) 823= 3√8

2=3√64=4

3) −2743=¿)4=3√531441=81

Observe that amn can be evaluated by finding either ( n√a)m or n√am. Compare

example 4 and 5 and you will see the computation of ( n√a)m in example 4 is simpler than

that for n√am in example 5.

The laws of positive-integer exponents are satisfied by positive-rational exponents with one exception: For certain values of p and q, (ap)q≠apq for a<0. This situation arises in the following example.

Examples 6:1) [(-9)2]1/2=811/2=9 and (-9)2(1/2)=(-9)1=-9

Therefore [(-9)2]1/2≠(-9)2(1/2).2) [(-9)2]1/4=811/4=3 and (-9)2(1/4)=(-9)1/2 (not a real number)

Therefore [(-9)2]1/4≠(-9)2(1/4). The problems that arise in example 6 are avoided by adopting the following rule: If m and n are positive even integers and a is a real number, then (a¿¿m)¿1/n=│a│m/n

A particular case of this equality occurs when m=n. We then have (a¿¿n)¿1/n=│a│ (if n is a positive even integer) or, equivalently, n√an=│a │ (if n is even)

If n is 2, we have √a2=│a │Examples 7:

1) [(-9)2]1/2=│-9│=92) [(-9)2]1/4=│-9│2/4=91/2=3

Definition

If m and n positiv e integer that are relatively prime and a is a real number and

a≠0, then if n√a is a real number.

a−m

n = 1

amn

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Example:8

1) 8−23 = 1

823= 1¿¿

=1

22=14

2) 8−23 =(8

−13 )2=( 1

813 )2=( 1

2 )2=14

3)x

13

x14

=x13.

1

x14=x

13 . x

−14 =x

( 13 )−1

4=x112

ASSESSMENTTEST IName the property that justifies each step.

1. Simple 6a+(8b+2a)

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a. 6a+(8b+2a)=6a+(2a+8b)b. =(6a+2a)+8bc. =(6+2)a+8bd. =8a+8b

2. Simplify 6 a2+( 6a+a2 )+9 a

a. 6 a2+( 6a+a2 )+9 a=6 a2+¿)+9a

b. ¿(6a¿¿2+a2)+¿¿+9a)

c. ¿(6a¿¿2+1 a2)+¿¿+9a)

d. =6+1(a¿¿2)+(6+9)a¿e. ¿7 a2+15 a

Simplify and express the following.

1.am+5

am−2

2. [( x+ y )0 +a0+b0

a+b+c]-2

3. (a¿¿−2+ y)¿-2

4. (3¿¿7 x+5)(3¿¿ 4 x−4 )¿¿5. (9 xy¿¿2)(4 x¿¿3 y )¿¿

Rational Expression (simplify)

1.9 n

63 n÷ 9 n

2.−15 m3 n2 p2

−35 m2n5 p

3.x+ y

x2− y2

4.3m−1

9 (m−1 )2−4

5.4 mn+6

10 m+8 n

6. 2 x2+3 x−510 x+25

7.x2−5 x−24

4 x2−27 x−40

8.25 a2+70 a+49

25 a2−49Factor each polynomial into two binomials

1. a2+ 12a+ 272. y2+ 21y+ 1103. n2-4n+ 4

22

4.x2-12x + 205. x2+ 11x -12

Answer the following word problemsand multiple choice questions1. The area of a rectangle is (x2–12x + 35). If the length is (x-5),find the width.

(hint:“x5” times“something” will give you “x2–12x + 35.”).2. The area of a rectangle is 3a2+ 5a–28. If the length is (a+ 4), find the width.3. A rectangle has an area of 3x2+ 5x –12. What factors are the length and width of

the rectangle?a. (3x + 4)(x –3)b. (3x –4)(x + 3)c. (3x + 3)(x –4)d. (3x –3)(x + 4)

4. The area of a certain rectangle is 5n2–6n–27. Which factors are the width and length of the rectangle?a. (5n + 3)(n –9)b. (5n –3)(n + 9)c. (5n + 9)(n –3)d. (5n –9)(n + 3)

5. If the area of a certain rectangle is 6m2–2a –28, and the length is (2m + 4), what is the width?a. (3m + 7)b. (4m –7)c. (4m + 7)d. (3m –7)

TEST IIA. Determine whether each statement is true or false.

1. Every integer is also a real number.2. Every irrational number is also an irrational number.3. Every natural number is also a whole number.4. Every real number is also a rational number.

B. State whether each decimal represents a rational o irrational number.5. √46. √57. 08. 39. 0.6358635810. √866

C.Determine which real number property is shown by each of the following

1.−14

+ 14=0

2. 2(1)=2

3.14

(4)=1

23

4. -7+(-4)=-4+(-7)5. 0.3(0)=0.36. 5[3+(-1)]=5(3-1)

7. (8+98

)+0.45=8+(98

+0.45)

8. 5(8+8)=5(8)+5(8)9. 6x+(8x+10)=(6x+8x)+1010. 5a+2b=2b+5a

D.Simplify and express the following expressions with positive and negative integrals only.

1. 50

2. 10 m4

30 m

3.16 b4 c−4 bc3

4. y3 . y 4

5. (5 xy )6

6. (ab)3

7. (x¿¿3 y2)3 ¿8. [(−5)¿¿2]2 ¿

9. x5 y6

xy=

10.a7

a3

E.Simplify the following radicals1. (5√2)¿2. (3 a

3√4 x2 ¿(4 3√3 xy )

3. 4 √ 916

4. √2(3+√3¿5. 5√2+3√26. √18−2√27+3√3−6 √87. √16 b+√4 b

8.−12√24

3√29. √8+√50

10. 4 √x7 y10

F.Simplify the following polynomials.

24

1. (5 x−1 )+(10 x2+7 x )2. (20 x¿¿2+2)+(15 x2−8 )+(3 x2−4)¿3. ( x2+ y2+8 )+( 4 x2−2 y2−9 )4. (−3 x2+5 y−4 xy+ y2 ¿¿(2 x2−4 y+7 xy−6 y2)5. 2 x2+6 x+5∧3 x2−2 x−1

6. ( x+2 ) ( x2−2x+3 )7. ab (2 a+1 )

8. x2−3 x−10x+2

9.x6+2 x4+6 x−9

x3+3

10. (3 x3−11 x2 y+11 x y2−2 y3 )÷ ( x−2 y )G.Find the factors and its GCF

1. 212. 4, 123. 634. 3045. 18, 356. 12 an2 , 40 a4

7. 7 y2 ,14 y2

8. 15 , 10

9. 6 a2, 18 b2 , 9 b3

10. 18 x2 y2 , 6 y2 ,42 x2 y3

H.Factor each trinomials. If possible. If the trinomial cannot be factored using integers. Write prime.

1. 6 mx−4 m+3 x−2r2. 3 my−ab+am−30 y

3. a2−2 ab+a−204. 3 m2−5 m2 p+3 p2−5 p3

5. 4 ax−14 bx+35 by−10 ay

6. y2+12 y+277. c2+2 c−38. x2−5 x−249. 7 a2+22 a+310. 6 y2−11 y+4

I.Factor out each polynomials1. x2−492. x2−36 y2

25

3. 16 a2−9b2

4. 2 a2−255. 2 z2−986. n2−8 n+167. 4 k2−4 k+18. x2+6 x−99. 1−10 z+25 z2

10. 50 x2 40 x+8J. Solve and simplify each polynomials

1.a+1

a3−a+2

a2+ a+3

a

2.5 x3

7 y4 .21 y2

10 x2

3.9 x5

36 x2

4.5−a

a2−25

5.10 a2−29 a+106a2−29 a+10

÷10 a2−19 a+6

12 a2−28 a+15

6.1

x+h−1

xh

7.x6−7 x3−84 x2−4 x−8 ÷ ¿¿)

8.

ab−b

aab+

ba

9.t2−2t−15

t 2−9.t 2−6 t+912−4 t

10.a−1+b−1

a−2−b−2

26

Chapter II Equations and Inequalities

27

2.Equations

To sove an equation means to isolate the variable having a coefficient of 1 on one side of the equation. By using Addition Property of Equality.Examples:

1) solve r+16=-7 r+16=-7r+16+(-16)=-7+(-16) add -16 to each side r=-23 the sum of -16 and 16 is 0check: r+16=-7 -23+16=-7 -7=-7

2) x+(3.28)=-17.56 x+(3.28)=-17.56x+(3.28)+(3.28)=-17.56+3.28

x=-14.28check: x+(3.28)=-17.56

-14.28+(-3.28)=-17.56-17.56=-17.56

3) y+21=-7y+21+(-21)=-7+(-21)

y=-28check: y+21=-7

28

-28+21=-7 -7=-7

Equations by Using SubtractionThe property that used to subtract the same number from each side of an

equation is called the subtraction property of equality.Examples:

1) x+15=-6x+15-15=-6-15x=-21check: x+15=-6-21+15=-6 -6=-6

2) b-(-8)=23 b+8=23b+8-8=23-8 b=15check: b-(-8)=2315-(-8)=23 23=23

Equations by Using Multiplication and DivisionTo solve the equation by using multiplication, we use the multiplication property of equality.For any numbers a,b, and c, if a=b, then ac=bcEamples:

1)512

= r24

24 ( 512

)=( r24

)24 multiply each side by 24

10=rCheck:512

= r24

replace r with 10

512

=1024

512

= 512

29

2) 24=-2a24=-2a−12

(24 )=−12

(2 a )

−¿12=a:̌24=¿−2a24=¿−2a(-12)24=24

To solve the equation by using division, we use the division property of equality.For any numbers a,b,c with c≠ 0 ,

If a=b, then ac=b

c.

Examples:1. -6x=11

−6 x−6

= 11−6

divide each side by -6

x=−116

Check: -6x=11−6¿)=11 11=11

2. 4x=244 x4

=244

X=6Check: 4x=244(6)=24 24=24

2.2 Appplication of Linear EquationsIn many applications of algebra, the problems are stated in words. They are called

word problems, and they give relatiomships between known numbers and unknown numbers to be determined. In this section we solve word problems by using linear equations. There is no specific method to use. However, here are some steps that give a possible procedurefor you to follow. As you read through the examples, refer to these steps to see how they are applied.

1. Read the problem carefully so that you understand it. To gain understanding, it is often helpful to make a specific axample that involves a similar situation in which all the quatities are known.

2. Determine the quantities that are known and those that are unknown. Use a variable to represent one of the unknown quantities inthe equation you

30

will obtain. When employing only one equation, as we are in this section, any other unknown quantities should be expressed in terms of this one variable. Because the variable is a number, its definition should indicate this fact. For instance, if the unknown quantity is a length and lengths are mesured in feet, then if x is a variable, x should be defined as the number of feet in the length or, equivalently, x feet is the length. If the unknown quuantity is time, and time is measured in seconds, then if t is the variable, t should be defined as the number of seconds in the time or, equivalently, t seconds is the time.

3. Write down any numerical facts known about the variable. 4. From the information in step 3, determined two algebraic expressions for

the same number and form an equation from them. The use of a table as suggested in step 3 will help you to discover equal algebraic expressions.

5. Solve the equation you obtained in step 4. From the solution set, write a conclusion that answers the questions of the problem.

6. It is important to keep in mind that the variable represents a number and the equation involves numbers. The units of measurement do not appear in the equation or its solution set.

7. Check your results by determining whether the condition of the word problem are satisfied. This check is to verify the accuracy of the equation obtained in step 4 as well as the accuracy of its solution set.

Example 1If a rectangle has a length that is 3cm less than four times its width and its perimeter is 9cm, what arethe dimension?Solution w: the number of centimeters in the width of the rectangle4w-3: the number of centimeters in the length of the rectangle

(4w-3)cm

w cm w cm

(4w-3)cmw+(4w-3)+ w+(4w-3)=19

10w-6=19 10w=25

w=52

4w-3=4¿)-3

=7Example 2

31

A man invested part of $15,000 at 12 percent and the remainder at 8 percent. If his annual income from the two investments is $1456, how much does he have invested at each rate?

Solutionx: the number of dollars invested at 12 percent

15,000-x: the number of dollars invested at 8 percent

12 percent investment x 0.12 0.12x8 percent investment 15,000-x 0.08 0.08(15,000-x)0.12x+0.08(15,000-x)=1456 0.12x+1200-0.08x=1456

0.04x=256 x=6400 15,000-x=15,000-6400

=8600Thus the man has $6400 invested at 12 percent and $8600 at 8 percent.Example 3.A father and daughter leave home at the same time in separate automobiles. The father drives to his office, a distance of 24 km, and the daughter drives to school, a distance of 28 km. They arrive at their destinations at the same time. What are their average rates, if the father’s average rate is 12km/hr less than his daughter’s?

Solution: r: the number of kilometers per hour in the daughter’s average rater-12: the number of kilometers per hour in the father’s average rate

Number of Kilometers ÷ Number of Kilometers = number of hours In Distance per hour in rate in time

Number of Dollar × Rate = Number of Dollars invested in Interest

32

Daughter 28 r 28r

Father 24 r-12 24

r−12Equation:28r

= 24r−12

Solve the equation by first multiplying on both sides by the LCD:

r (r−12 ) 28r

=r (r−12 ) 24r−12

(r-12)28=r(24) (r-12)7=r(6) 7r-84=6r 7r-6r=84 r=84 r-12=84-12

=72Therefore, the daughter’s average rate is 84km/hr and the father’s average rate is 72km/hr. 2.3 Quadratic Equation in One Variableax2 + bx + c =0, a≠0where a, b, and c are real number constants and a≠0, is called a second degree polynomial equation, or quadratic equation , in the variable x. the word “ quadratic “ comes from quadrate, meaning square or rectangular when a quadratic equation is written in the above manner, it is said to be in standard form.

The following are examples of quadratic equations in x, written in the form, with the indicated values of a, b and c.8x2 + 16x -5 = 0 ( a = 8, b = 16, c = -5 ) 2x 2 - 10 = 0 ( a =2, b = 0, c = -10 ) 5x2 – 7x = 0 ( a = 5, b = -7, c = 0 )Theorem 1

If r and s are real numbers, then rs = 0 if and only if r =0 or s=0This theorem can be extended to a product of more than two factors. For

instances, if r, s , t, u єR, then rstu = 0 if and only if at least one of the numbers r, s, t or u is 0.To find the solution set of the equation X2 +3x – 10 = 0We factor the left side and obtain ( x+5 )( x-2 )By applying Theorem 1, it follows that the equations gives a true statement if and only ifX+5 = 0 or x-2 = 0

33

The solution of the first of these equation is -5 and the solution of the second is 2. Therefore the solution set of the given equation.Example 1:1+5x/6 = 2x2 /3(6) (1) + (6) 5x/6 = (6) 2x2 /36+ 5x =4x2

-4x2 +5x +6 = 04x 2-5x -6 = 0 ( 4x+3 ) (x-2 ) = 04x=3 = 0 x-2 =0 4x =-3 x =2 X =-3/4 The solution set is {-3/4, 2 } Suppose we have a quadratic equation of the formX 2 = dThat is, there is no first degree term. Then an equivalent equation is

X2 –d = 0

And, factoring the left member, we obtain( x- d )( x + d )We set each factor equal to zero and solve the equations X - √ d = 0 x + √d = 0x = √d x = -√dtherefore the solution set of the equation x2 = d is { √d , -√d }. We can abbreviate this solution set as {± √d }. Thus x2 = d if and only if x= ±√d

Example 2:a. X2 = 25 b. x2 =13 c. x2 =16

X = ±√ 25 x =√13 x = ± √-16X = ± √5 x = ± 4ἰ

The solution set of the equation x2 + 6x-1 =0

we first add 1 to each side and obtain x2+ 6x =1

we now add to each side the square of one half of the coefficient of x, or 32. We obtain x2 +6x +9 = 1+9

The left side is now the square of x + 3. Thus we have(x +3 )2 =10

Taking the square root of both sides of the equation, we have X = 3 = ±√10 X = -3 ±√10

34

This method is called completing the square To complete the square of x2 + kx, add the square of one half the coefficient of x; that is, add (k/2)2.Quadratic Formula

If a≠0, the solutions of the equation ax2+ bx + c =0 are given by

x=−b±√b2−4 ac2a

Example 3: Use quadratic formula to find the solution set of the equation

6x2= 10 + 11x

×=−b ±√b2−4 ac2a

¿−(−11)±√¿¿¿

¿ 11±√121+24012

¿ 11±√36112

¿ 11±1912

¿ 11+1912

¿ 11−1912

¿ 3012

¿−812

¿52

¿−23

The solution set is {−23

,52 }.

2.4 INEQUALITIES Trichotomy Property of Order

If a and b are real numbers, exactly one of the following three statements is true.A<B B<A A = B

Transitive Property or orderIf a, b and c are real numbers, and if a<b and b<c then a<c.

The domain of a variable in an inequality is the set of real numbers for which the members of the inequalities are defined.Examples:

4x – 9 < 10 x – 9/10 ≤ x 3 < 5x + 7 ≤ 15

35

An example of a quadratic inequality having domain R isX2 + 5x > 13

The inequality4x/x + 3 < 6 is rational.

Any number in the domain for which the inequality is true is a solution of the inequality, and the set of all solution is called the solution set. An absolute inequality is one that is true for every number in the domain. For instance, if x is a real numbers x + 1 < x + 2 and x2 ≥ 0 are absolute inequalities. A conditional inequality is one for which there is at least one number in the domain that is not in the solution set. To find the solution set of a conditional inequality we proceed in a manner similar to that used to solve an equation that is we obtain equivalent inequalities until we have one whose solution set is apparent.PROPERTIES OF INEQUALITIES

If a, b and c are real numbersa. Addition Property

If a < b, then a + c < b + cb. Subtraction Property

If a < b, then a – c < b – cc. Multiplication Property

If a < b and c > 0, then ac < bcIf a < b and c < 0, then ac > bc

Example 1Find and show on the real number line the solution set of the inequality.

3x – 8 < 73x – 8 + 8 < 7 + 83x < 151/3 (3x) < 1/3 (15)X < 5

Example 2Find and show on the real number line the solution set of the inequality

X – 7/4 ≤ x(4)x – 7/4 ≤ (4)xX – 7 ≤ 4xX – 7 – 4x + 7 ≤ 4x – 4x + 7-3x ≤ 7-1/3 (-3x) ≥ (-1/3)7X ≥ -7/3

2.5 POLYNOMIAL AND RATIONAL INEQUALITIESA quadratic and inequalities is one of the form ax2 + bx + c < 0

36

A critical number of the inequality above is a real root of the quadratic equation ax2 + bx + c = 0.To solve the inequality

X2 – 8 < 2xWe first write an equivalent inequality having all the non-zero terms on one side

of inequality sign. Thus we have x2 – 2x – 8 < 0` (x + 2) (x – 4) < 0Example I Find and show on the real number line the solution set of the inequality. x 2 + 5x ≥ - 6 x 2 +5x + 6 ≥ 0 (x+3) (x+2) ≥ 0Figure 1

-4 -3 -2 -1 0 1 2 The critical numbers are -3 and -2. The points corresponding to these numbers are plotted in figure 1 and the following intervals are determined.(-∞ , -3) (-3 , -2) (-2 , +∞)

Figure 2 ] [ -4 -3 -2 -1 0 1 2

Thus, the solution set of the given inequality is (-∞ , -3] U [-2 , +∞) , appearing in Figure 2.Example 2 Find the solution set of each of the following inequalities

x2- 6x + 9 < 0 (x-3) 2< 0 Because there is no value of x for which (x-3) 2 is negative, there is no solution. Therefore, the solution set is

9x2 + 12x + 4 ≤ 0 (3x + 2)2 ≥ 0 Because (3x + 2)2 is non-negative for all values of x , the solution set is the set of all real numbers.2.7 Equations and Inequalities Involving Absolute Value The absolute value of areal number a denoted by /a/ , is given by / a / = { a, if a ≥ 0 }

/a / = { -a, if a < 0 }Example 1

37

Find the solution set of the equation

/ 3x-2 / = / 8-4x / 3x – 2 = 8 – 4x 3x + 4x = 8 + 2 7x = 10 x = 10/7

3x – 2 = - ( 8 – 4x ) 3x – 2 = - 8 + 4x 3x – 4x = - 8 + 2 -x = -6

x=6

The solution set is {10/7 , 6}Example 2 Find the solution set of the equation / 2x – 7 / <9 -9 <2x – 7 <9 -9 + 7 <2x <9 + 7 -2 <2x <16 -1 <x <8 Therefore the solution set is the open interval (-1, 8)

Theorem 1 If a and b are real numbers, then / ab / = / a / / b /Example 3 / ( 3x + 2 – 8 / <1 / 3x -6 / <1 / 3 (x-2) / <1 / 3 / / x-2 / <1 3 / x-2 / <1 / x-2 / <1/3The Triangle Inequality If a and b are real numbers, then / a + b / ≤ / a / + / b /If a = 2 and b = 7 , then / a + b / = / 2 + 7 / / a / + / b / = / 2 / + / 7 / = / 9 / = 2 + 7 = 9 = 9 If a = -2 and b = -7 , then / a + b / = / -2 + -7 / / a / + / b / = / 2 / + / -7 / = / -9 / = 2 +7

38

= 9 = 9

ASSESSMENTTEST IA. Find the solution set of the equation.

1. 3 (6x – 5) = 11 – ( 4 + 8x )2. 49x2 – 64 = 0 3. 2 + 3 = 0 6 – y y – 24. (x – 4) ( x + 2) = 75. 2p2 – 4p – 5 = 06. √2x + 5 + x = 57. √y + 2 + √y + 5 - √8 – y = 08. / r + 1 / = 6 r – 19. / 8x - 2 / = x + 710. / 9x + 7 / = 11

B.Find the solution set of the inequality and write it with interval notation. Show the solution set on the real number line.

11. 3x – 2 ≤ 2012. x2 < 6413. 2x + 1 > 1 x – 514. 3x – 4 ≤ 12

39

2x – 3 15. 2x2 – 3 x ≥ 516. 5x + 8 ≥ 2x – 217. -7 ≤ 8 – 4x ≤ 10 -518. / x – 10 / ≤ 1719. x2 + 3x – 10 ≥ 1 20. / 4x – 5 / < 15

C.Solve for x in terms of the other symbols

21. d/10x – d/5 = 1/x 22. rsx2 + s2 x + rtx + st = 0 23. x2 + b2 = 2 bx + a2 24. 5x2 – 4xy – x + y -1 = 0 25. x2 + xy + 4x - 1 = 0 Show that the two inequalities are equivalent.

26. A woman leaves home at 10 A.M and walks to her office at he rate of 7 km/hr. At 10. 15 A.M. the womans daughter leaves home and rides her bicycles at a rate 15 km/hr along the same route to school. At what time does she overtake her mother? 27. How many liters of a solution that is 85 percent glycerine should be added to 30 liters of a solution that is 40 percent glycerine to give a solution that is 60 percent glycerine. 28. In a long distance race around a 600 m track the winner finished one lap ahead of the loser. If the average rate of the winner was 8 m/sec and the average rate of the loser was 7. 25 m/s how soon after the start did the winner complete the race? 29. An automobile radiator contains 9 liters of a solution that is 10 percent antifreeze and 95 percent water. How much of the solution should be drained and replaced with pure antifreeze to obtain a solution that is 45 percent antifreeze? 30. The perimeter of a rectangle must not be greater than 50 cm and the length must be 10 cm. What is the range of values for the width?

TEST IIA. Find the solution set of the equation.

1. X2 = 642. 6t2 – 11 = 03. 9x2 = x4. 3t/3t + 4 + 2/5 – 1/3t – 45. 64y2 – 80y + 25 = 0

B. Find the solution set of the equation by completing the square.6. X2 + 6x + 8 = 07. 4x2 + 4x – 3 = 08. 3y2 + 4y + 2 = 0

C. Find the solution set of the equation by using the quadratic form.

40

9. X2 + 1 = 6x10. 5y2 – 4y – 2 = 0

D.Find the solution of the inequality and write it with interval notation.1. 2x – 1 < 62. -3 > 4x + ¾3. 11 ≥ 5x – 4 > 14. 2x – 4 < 65. 1 < 4x – 1/3 < 56. 5x – 2 < 5x – 2/47. 10 – 3x > 4x – 5/-38. 5 ≤ 3x – 4 < 149. -1 < 7 – 2x/5 ≤ 510. 6 ≤ 2 – x ≤ 8

E.Find the solution set of the equation.11. / x – 8 / = 9 12. / 4y – 10 / = 5 13. / 3t -2 / = t214. / x – 1 / = x215. / 5t – 15 / = 20

CHAPTER III Points and

Equations

41

3.1 Points in a Plane Ordered pairs of real numbers are important in our discussions. Any two real numbers form a pair. When the order of appearance of the numbers is a significant, we call it an ordered pair. If x is the first real number and y is the second, this ordered pair is denoted by writing them in parenthesis with a comma separating them as (x,y).

The set of all ordered pairs of real numbers is called the number plane, denoted by R2 ,And each ordered pair (x,y) is a point in the number plane. The intersection of these two perpendicular line segments is the point P, associate with the ordered pair (x, y). Refer to figure 1, the first number x of the pair is called the abscissa or x-coordinate of P, and the second number y is called the ordinate or the y-coordinate of P.

The x and y axes are called the coordinate axes. They divide the plane into four parts called quadrants. The first quadrant is the one in which the abscissa and the ordinate are both positive, that is , the upper right quadrant. The other quadrant are numbered in the counterclockwise direction, with the fourth being the lowered right quadrant.

Pythagorean TheoremIn a right triangle, if a and b are the lengths of the perpendicular sides and c is the

length of the hypotenuse then, a2 + b2 = c2.Distance Formula The distance between two points P1 (x1 , y1) and P2 ( x2 , y2) is given by P1P2 = √(x2 – x1)2 + (y2 –y1)2

Converse of Pythagorean Theorem If a ,b and c are the length of the side of a triangle and a2 + b2 = c2, then the

triangle is a right triangle, and c is the length of the hypotenuse.Midpoint Formulas

42

If M (x,y) is the midpoint of the line segment from P1 (x1 , y1) to P2 ( x2 , y2), then x = x1 + x2 y = y1 + y2

2 2Solution

From the midpoint formulas, if M is the point (x,.y), then

x = 5 – 1 y = -3 + 6 2 2 = 2 = 3/2

3.2 Graphs of EquationsThe graph of an equation in R2 is the set of all points in R2 whose coordinate are

numbers satisfying the equation.Example 1 Draw a sketch of the graph of the equation y2 – 4x = 0Solution

y2 = 4x y = ± 2√x

y = 2 √x and y = -2 √xSymmetry of two points

Two points P and Q are said to be a symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ. Two points P and Q are said to be symmetric with respect to a third point if and only if the third point is the midpoint of the line segment PQ. Symmetry of a graph

The graph of an equation is symmetric with respect to a line l if and only if for every point P on the graph there is a point Q, also on the graph , such that P and K are symmetric with respect to l. The graph of an equation is symmetric with respect to a point R if and only if for every point P on the graph there is a point S, also on the graph such that P and S are symmetric with respect to R.Example 2

Draw a sketch of the graph of the equation. 2y = x3

2(-y) = (-x)3 -2y = -x3

Circle A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is called the center of the circle and the constant equal distance is called the radius of a circle. Equation of a Circle

An equation of the circle with center at the point (h,k) and radius r is (x – h)2 + (y – k)2 = r2

Example 3 Find an equation of the circle having the diameter with endpoints at A (-2,3) and

B (4,5).Solution

43

h=−2+42

¿1

k=3+52

=4

3.3 Equation of a line Slope

If P1 (x1 , y1) and P2 (x2 , y2) are any two distinct points on line l, denoted by m, is given by

m = y2 – y1

x2 – x1

If the slope of a line is positive, then as the abscissa of a point on the line increases, the ordinate increases. A line whose slope is negative, for this line as the abscissa of the point on the line increases, the ordinate decreases. If a line is parallel to the x axis, then y2 = y1 ; so the slope of the line is zero. If a line is parallel to the y axis, x2 = x1 ; thus the fraction y2 – y1 / x2 – x1is meaningless because we cannot divide by zero. Thus the slope of a vertical line is not defined.

Example 1 Determine the slope of a line.A (3,7) and B (-2,-4)m = -4 – 7 /-2 -3 = -11 / -5

= 11 / 5A (-3,4) and B (5,4)

m = 4 – 4 / 5 – (-3) = 0/8 = 0

The point slope form we choose the particular point (0,b) (that is , the point where the line intersects the y axis ) for the point ( x1,y1) , we have

y - b = m ( x – 0 ) y = mx + b

The number b, the ordinate of the point where the line intersects the y axis, is the y intercept of the line. Consequently, the preceding equation is called the slope-intercept form. Theorem 1

The graph of the equation Ax + By + C = 0

Where A, B, and C are constants and where not both A and B are zero is a line.Theorem 2

If l1 and l2 are two distinct nonvertical lines having slopes m1 and m2, respectively, then l1 and l2 are parallel if and only if m1 = m2 Theorem 3

Two nonvertical lines l1 and l2, having slopes m1 and m2 , respectively, are perpendicular if and only if m1m2 = - 1

44

ASSESSMENTTEST IA.Find the center and radius of the circle.

1. x2 + y2 + 4x – 6y -3 = 02. 3x2 + 3y2 + 4x -4 = 0

B. Determine the slope of the line and find an equation of the line.3. (1,-3) and (4,5) 4. (2,5) and (6,9)5. (9,8) and (-2,4)6. (6,2) and (-9,7)7. (4,2) and (-7,8)8. (-1,3) and (6,1)9. (4-6) and (5,6)10. (9,7) and (6,-4)

C.Find the slope and y-intercept of the line having the given equation .11. 2x – 5y - 10 = 012. 2x + 3y + 12 = 013. Prove that the quadrilateral having vertices at (2,1) , (6, -2) , (9,6) and

(7, 10) is a rectangle.14. Find an equation of the line through the point ( -1,6) and perpendicular

to the line whose equation is 4x + 2y – 5 = 0.15. Prove that the lines having the equations 2x + 5y + 20 = 0 and 5x – 2y -

10 = 0 are perpendicular.D. For the given points A and B , find the directed distances : (a) AB : (b) BA

16. A (2,4) and B (7,8)17. A (4, - 8) and B (5, - 10)18. A (7, - 10) and B (- 8, 7 )19. A (- 4, 5 ) and B (10,3)

45

20. Given that A is the point (-2, 3) and B is the point ( -x , 3) , find x such that (a) AB = -8 ; (b) BA = -8

21. Given that A is the point (-4, y) and B is the point (-4, 3) , find y such that (a) AB = -3 ; (b) BA = -3

22. A (10,-5) , B (-4, 6) , C (-2, 9)23. A (4,10) , B (7, -3) , C (-1, -1)

E. Find the center and radius of the circle.1. 2x2 + 2y2 – 2x + 2y + 7 = 02. x2 + y2 – 10x – 10y + 25 = 03. 3x2 + 3y2 + 4y – 7 = 04. x2 + y2 - 6x - 8y + 9 = 05. x2 + y2 + 2x + 10y + 18 = 0

F. Find an equation of the line satisfying the given condition. 6.(a) the slope is 4 and through the point ( -3 , 2)

(b) through the two points ( -2, -6) and ( 4,5 ) 7 (a) the slope is -7 and through the point (-5,4) (b) through two points (5,7) and (-6,9) 8. (a) through the point ( 2,-8) and parallel to the x-axis (b) through the point (3,5) and parallel to the y-axisG. Find the slope and y-intercept of the line having the given equation. 9. (a) x + 3y -6 = 0 ; (b) 4y – 9 = 0 10. (a) x – 4y -2 ; (b) 4x = 3y

46

Chapter IV Functions and their Plane

47

4.1 Functions A function can be thought of as a correspondence from a set x of real numbers x to a set y of real numbers y, where the number y is unique for a specific value of x. A function is asset of ordered pairs of real numbers (x , y) in which no two distinct ordered pairs have the same first number. The set of all admissible values of x is called the domain of the function , and the set of all resulting values of y is called the range of the function.Graph of a function If f is a function , then the graph of f is the set of all points (x , y) in R 2 for which (x , y) is an ordered pair in f. The graph of a function can be intersected in a vertical line at most one point.Example 1 The function h is defined by h{ (x , y) y = / x/ } The required formula of h, x can be any real number. Therefore the domain is ( - ∞ , + ∞ ). Because we observe from figure 1 that y can be any non negative number the range is [ 0 , + ∞ ]Quadratic Function

The general quadratic function is defined by F ( x ) = ax2 + bx + c

Where a, b and c are constants representing real numbers and a≠0. The graph of f is the same as the graph of an equation

Y = ax2 + bx + cIf the function f is the defined by

F ( x ) = -2 x2 + 8x – 5The graph of f is the same as the graph of an equation

Y = 2x2 + 8x – 5This equation is equivalent to

2(x2 – 4x) = -y – 5

48

To complete the square of the binomial with the parentheses, we add 2(4) to both sides of the equation and we have

2(x2 – 4x +4) = -y – 5 + 8(x – 2 )2 = 4p (y – k)

Where (h,k) is (2,3) and = -1/8. Therefore the vertex of the parabola is at 92,3) and the axis is the line x = 2. Because p<0 , the parabola opens downward we find a few points on the parabola. The zeros of the function are the values of x for which f(x) = 0.When the graph of a quadratic function opens upward the function has a minimum value, which occurs at the vertex of the parabola. There is no maximum value for such a function. When the parabola opens downward. The function has a maximum value occurring at the vertex ;it has no minimum value We now apply the method used in the solution to the general quadratic function defined by

f (x) = ax2 + bx + c -1 = ax2 + bx + c ax2 + bx = y – c a (x2 +b/a x) = y – ca (x2 + b/a x + b2/4a2 0 = y – c +b2/4a (x = b/2a )2 = 1/a ( y+b2 – 4ac/4a)The graph of this equation is a parabola having its vertex at the point where x = -b/2a. if a>0, the parabola opens upward and so f has a minimum value at the point where x + -b/2a Theorem 1 The quadratic function defined by f (x) = ax2 +bx +c, where x = -b/2a. if a>o, the extreme value is a minimum value, and if a,a0, the extreme value is a maximum valueExample use theorem 1 to find either a maximum or minimum value of the function g if g (x) = -5/2 x2 + 8x -10For the given quadratic function a = -5/2 and b =8. Because a<0, g has a maximumu value at the point x = -b/2a = -8/2(-5/2) = 8/5 The maximum value is G (8/5) = -5/2(8/5)2 + 8(8/5) – 10 =-5/2 (64/25) + 64/5 – 10 = -18/5Rational Functions

49

A rational function is of the form f ( x )= p(x )q (x) , where p(x) and q(x)are polynomial

functions and q ( x ) ≠ 0. A graphing calculator is a good tool for exploring graphs of rational functions.

Graphs of rational functions may have breaks in continuity. This means that, unlike polynomials functions which can be traced with a pencil that never leaves the paper, a rational function may not be traceable. Breaks in continuity can occur where there is a vertical asymptote or point discontinuity. Point of discontinuity is like a hole in the graph. Vertical asymptote and point of discontinuity occur for the values of x that make the denominator of the rational function zero.Graphing Rational Functions

Connection:Mathematical HistoryMathematician Maria Gaetana Agnesi was one of the greatest women scholars of

all time. In the analytic geometry section of her book Analytical Institutions, Agnesi

discussed the characteristics of the equation x2 y=a2 (a− y ) , called the “curve of Agnesi”.

The equation can be expressed as y= a2

x2+a2 .

Because the function described above is the ratio of two polynomial expression a3 and

x2+a2 is called a rational function. A rational function is function of the form

f ( x )= p(x )q (x)

, where p(x) and q(x) are polynomial functions and q(x)≠ 0

Examples of Rational Function:

f ( x )= xx−1

g ( x )= 3x−3

h ( x )= x+1( x+2 )(x−5)

The lines that graph of the rational function approaches is called Asymptote. If the function is not define whenx=a, then either there is a “hole” in the graph x=a.

50

POLYNOMIALS The expression x2+2xy+y2 is called a polynomial. A polynomial is a monomial or

a sum of monomials. The monomials that make up the polynomial are called the terms of the polynomial. The two monomials xy and xy ca be combined because they are like terms. Like terms are two monomials that are the same, or differ only by their numerical coefficient. An expression like m2+7mb+12cd with three unliked terms is called trinomial. An expression like xy+b3 with two unliked terms is called binomials. The degree of a polynomial is the degree of the monomial with the greatest degree. Thus, the degree of x2+2xy+y2 is 2.

Remember:

If a polynomial contains only one term, it called monomial; if two terms, it is called binomial; if it is contains three terms it is called trinomial. If a polynomial has more than three terms, it is called multinomial.

The following table shows examples of polynomials.

Polynomial No. of terms Class by terms

12 x y2

5 xy+3 z2 v

4 x2 y−3 x y2−xy3 x−5 y+a+10 b

5 x4+2 x3−x2−6 x+8

12345

Monomial BinomialTrinomial

MultinomialMultinomial

Example:

Determine whether or not each expression is a polynomial. Then state the degree of each polynomial.

a.27

x4y3 – x3

This expression is a polynomial. The degree of the first term is 4 + 3 or 7, and the degree of the second term is 3. The degree of the polynomial is 7.

b. 9 + √ x−3This expression is not polynomials because √ x is not a monomial.

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The FOIL Method is an application of the distributive property that make the multiplication easier.

FOIL Method of Multiplying Polynomial

The product of two binomial is he sum of the products of: F the first terms

O the outer terms I the inner terms L the last terms

Example:Find (k2 +3k +9) (k +3)(k2 +3k +9) (k +3)

¿k 2 (k+3) + 3k (k +3) +9(k +3) distributive property¿k2∙k+¿k2∙ 3+3 k ∙+9 ∙ k+3 ∙9 distributive property¿k 2∙ k+k 2∙ 3k2+9 k+9 k+27¿k2+6 k2+18 k+27 combined like terms

Dividing PolynomialsYou can use a process similar to long division of a whole numbers to divide a

polynomial by a polynomial when doing the division, remember that you can only add ad subtract like terms. Example: Simplify: c 2 –c –30 c –6

c

c−6√c2−c−30

c2−65 c−30

−c− (−6 c )=−c+6c∨5 c c+5

c−6√c2−c−30

c2−6

5 c−30

5 c−30

0Polynomial functions

A polynomial function is a function that can be defined by evaluating a polynomial. A function f of one argument is called a polynomial function if it satisfies

52

f ( x )=an xn+an−1 xn−1+…+a2 x2+a1 x+a0

For all arguments x, where n is a non-negative integer and a0, a1, a2, ..., an are constant coefficients.For example, the function f, taking real numbers to real numbers, defined by

f ( x )=x3

is a polynomial function of one variable. Polynomial functions of multiple variables can also be defined, using polynomials in multiple indeterminates, as in

f ( x , y )=2 x3+4 x2 y+x y5+ y2−7

An example is also the function f ( x )=cos (2arcos ( x ) )  which, although it doesn't look like

a polynomial, is a polynomial function on [−1,1]  since for every  from [−1,1] it is true

that   f ( x )=2 x2−¿1

Polynomial functions are a class of functions having many important properties. They are all continuous, smooth, entire, computable, etc

Graphs of Polynomial FunctionA polynomial function in one real variable can be represented by a graph.

The graph of the zero polynomialf(x) = 0is the x-axis.

The graph of a degree 0 polynomialf(x) = a0, where a0 ≠ 0,is a horizontal line with y-intercept a0

The graph of a degree 1 polynomial (or linear function)f(x) = a0 + a1x , where a1 ≠ 0,is an oblique line with y-intercept a0 and slope a1.

The graph of a degree 2 polynomialf(x) = a0 + a1x + a2x2, where a2 ≠ 0is a parabola.

The graph of a degree 3 polynomialf(x) = a0 + a1x + a2x2, + a3x3, where a3 ≠ 0is a cubic curve.

The graph of any polynomial with degree 2 or greaterf(x) = ao + a1x + a2x2 + ... + anxn , where an ≠ 0 and n ≥ 2is a continuous non-linear curve.

The graph of a non-constant (univariate) polynomial always tends to infinity when the variable increases indefinitely (in absolute value)

Polynomial graphs are analyzed in calculus using intercepts, slopes, concavity, and end behavior.

53

Polynomial of degree 2:f(x) = x2 − x − 2= (x + 1)(x − 2)

 

Polynomial of degree 3:f(x) = x3/4 + 3x2/4 − 3x/2 − 2= 1/4 (x + 4)(x + 1)(x − 2)

 

Polynomial of degree 4:f(x) = 1/14 (x + 4)(x + 1)(x − 1)(x − 3) + 0.5

 

Polynomial of degree 5:f(x) = 1/20 (x + 4)(x + 2)(x + 1 )(x − 1)(x − 3)+ 2

54

Polynomial of degree 2:f(x) = x2 − x − 2= (x + 1)(x − 2)

 

Polynomial of degree 3:f(x) = x3/4 + 3x2/4 − 3x/2 − 2= 1/4 (x + 4)(x + 1)(x − 2)

 

Polynomial of degree 4:f(x) = 1/14 (x + 4)(x + 1)(x − 1)(x − 3) + 0.5

 

Polynomial of degree 5:f(x) = 1/20 (x + 4)(x + 2)(x + 1 )(x − 1)(x − 3)+ 2

Inverse function

Definition of Inverse Function

Two function f and g are inverse function if and only if both of their compositions are the identity function. That is,

( f ∘ g )=x and ( g∘ f ) (x )=x

55

An inverse function is a function that "reverses" another function: if the

function f applied to an input x gives a result of y, then applying its inverse

function g to y gives the result x, and vice versa. i.e., f(x) = y if and only if g(y) = x.

A function f that has an inverse is said to be invertible. When it exists, the inverse

function is uniquely determined by f and is denoted by f −1, read f

inverse. Superscripted "−1" does not, in general, refer to numerical exponentiation.

In some situations, for instance when f is an invertible real-valued function of a real

variable, the relationship between f andf−1 can be written more compactly, in this

case, f−1(f(x)) = x = f(f−1(x)), meaning f−1 composed with f, in either order, is the identity

function on R.

Property of Inverse Function

Suppose f andf−1 are inverse function. Then f ( a )=band only if f−1 (b )=a

Definition of inverse Relationship

Two relationships are inverse relationship if and only if whenever one relation contains the element ( a, b ), the other relation contains the element ( b, a )

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ASSESSMENTTEST I.A. Draw a sketch of the graph of the function and determine its domain and range.

1. f = {(x,y) / y = 3x – 1 }2. F = { ( x, y / y = 2x2 }3. F = { ( x, y) / y = / 3x + 2 / }4. G = { ( x, y} = x 2 – 4 }

x-25. g = { ( x, y /y = (x 2 – 4 ) ( x – 3 ) }

x2 – x – 6 6. f : y = { -2 if x ≤ 3 }

{ 2 if 3 < x }7. g : y = { -4 if x< -2 }

{-1 if -2 ≤ x ≤ 2 } { 3 if 2 < x }

8. f : y = { 3x +2 if x ≠ 1 } { 8 if x = 1 }

9. G : y { 9 – x2 if x ≠ -3 } { 4 if x = -3}

10. G : y = { x2 – 4 if x < 3 } {2x – 1 if 3 ≤ x }

B. Find the zeros of the function.1. f(x) = x2 – 2x – 32. f(x) = 2x2 – 2x – 1

57

3. f(x) = x2 – 3x + 14. f(x) = 6x2 – 7x – 5

C. Use theorem 1 to find either maximum or minimum value of the function.5. f(x) = -1/2 ( x2 + 8x + 8)6. G (x = 1/8 ( 4x2 + 12x -9 ) 7. g (x) = 3x2 + 6x + 9 8. f(x) = 2 + 4x - 3x2

9. Find two numbers whose sum difference is 10 and whose product is a minimum.

10. Find two numbers whose sum is 20 and whose product is a maximum.

Chapter V

Exponential and Logarithmic

58

Functionc

EXPONENTIAL FUNCTIONAn equation of the form y=a ∙bx , w h ere a≠ 0 , b>0 ,and b ≠ 1 , ia called

exponential function with base b.The logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. Logarithms are exponents. They were once used t simplifies calculations, but the advent of calculators and computers caused calculation with logarithms to be used less and less.

Definition of Logarithm Suppose b>0 andb ≠ 1. forn>0 , there is a number p such that Lo gb n=pif and only if b p=n .

The chart below shows some equivalent exponential and logarithmic equations.

Exponential Equation Logarithmic Equation

52=25105=100,000

80=1

2−4= 116

912=3

log5 25=2log10 100,000=5

log 81=0

log 21

16=−4

log 93=12

  Integral ExponentsBasic Laws of Exponents

1. bx ∙ b y=bx+ y

59

2. bx

by =bx− y (b ≠ 0)

3. if b ≠ 0 , 1 ,−1 t henbx=b y∧only if x= y4. (ab )x cx by

5. ( ab )

x

=ax

bx(b ≠ 0 )

6. if x ≠0 ,a>0 , b>0 , then ax=bx if ∧only if a=b7. (bx )y

=B x y

8. b0=1 (b≠ 0 )

9. b−x= 1

bx

Each of the above rules should be familiar to you from algebra I. Here are some sample problems with their solutions.   1)  Watch the difference between these two:                      a)  (-3)-2                                 b)  -(3)-2      The first one is squaring a negative number and the second is squaring a positive number and then making the whole result negative.                      a)  = 1/(-3)2 = 1/9                   b)  1/-(3)2 = -1/9

c)  7 . 2-3 =                              d)  (7 . 2)-3 =      The first one raises the power then multiplies, while the second one multiplies first then raises the power.                      c)  = 7/8                                    d)  = 14-3 = 1/143 = 1/2744e) (3−2+3−3 )−1

=¿

¿( 19+ 1

27 )−1

¿( 327

+ 127 )

−1

¿( 427 )

1

=274

In the above example our first step is to work inside the grouping symbols and get a common denominator.  Then add the two fractions.  Only when you have a single fraction, is it permitted to invert the fraction.

f) ¿¿

¿( 1a−2 −

1b2 )

−1

¿( b2

a2b2−a2

a2b2 )−1

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¿( b2−a2

a2b2 )−1

¿ a2 b2

(b−a ) (b+a )

In the above example we again simplify inside the grouping symbols and get a common denominator.  Once we have a single fraction in step 3 we can invert the fraction.  Notice the factoring in the last step!

g) (3 a−1 )2

(3a−1 )−2

¿ (3 a−1 )4

¿81 a−4=81

a4

In this example, the numerator and denominator have the same base.  We can apply the division rule by subtracting the exponents.  Then simplify.  Remember, no negative exponents should be left in the answer.

An important type of rule can now be stated using exponents.  It is a growth or decay problem.  We can mathematically model this function by using the following:A(t) = Ao(1 + r)t

Where Ao is the initial amount at time t = 0r is the rate (as a decimal)t is the timeA(t) is the amount after the time t.If r > 0, then it is an exponential growth.If -1 < r < 0, then it decays exponentially.

1)  Suppose a bike costs $100 now and it increases at a rate of 5% per year.  What will be the cost of the bike in 4 years?

Solution:    Ao = 100, r = 5% = .05, t = 4                                         A(4) = 100(1 + .05)4 = 100(1.05)4 = 121.55 The bike will cost $121.55 in 4 years.  (Rounded to the nearest penny)

2)  Suppose a car is worth $15,000 new.  What will it be worth in 3 years if it decreases at a rate of 12% per year?

Solution:  This is a decrease problem with                                          Ao = 15000, r = -.12, t = 3                                          A(3) = 15000(1-.12)3 = 15000(.88)3 = 10222.08 The car will be worth $10222.08 in 3 years.

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1). 4−4

4−2 +4−3

¿ 4−4

4−2+4−3 ∙44

44

¿ 1

42+41

¿ 116+4

= 120

One of the easier ways to do this problem is to multiply both numerator and denominator by the positive power of the biggest negative exponent.  In this case we multiplied by 44.  This greatly simplifies the problem.Rational Exponents

All rules presented in the previous section were defined for integers only.  All of the properties in the last section can also be extended to include rational exponents according to the following definitions:

1) b1n=n√b

2)bmn =( n√b )=n√bm

Example:

1) 813= 3√8=2

2)  2)  8−13 = 1

3√8=1

2

3) 163 /4=( 4√16 )3=23+8

4) 16−34 = 1

( 4√16 )3= 1

23 =18

5) ( 827 )

13 =3√ 8

27=2

3

6) ( 827 )

−23 =( 27

8 )23=( 3√ 27

8 )2

=( 32 )

2

= 94

7)  (1001/2 - 361/2)2 = (10 - 6)2 = 42 = 16   

8)  x1/2(x3/2 + 2x1/2) = x2 + 2x   9)  ¿¿

We can use these rules to solve for x when x is the exponent.  This method will only work if the bases are the same.  Check back in section 5-1 for the appropriate rule!!Example: 

62

1)  16x = 25   We can write both sides in base two.      24x = 25   Now use the fact that the bases are the same, the exponents are =       4x = 5       And solve for x!!         x = 5/4     To check it, take the 5/4 root of 16 = 25!! 2)  271-x = (1/9)3-x   You need to make both bases the same.  How about 3!!       33(1-x) = 3-2(3-x)      Notice the power on the right side is negative.        3(1 - x) = -2(3 - x)    Because the bases are =, the exponents must be =         3 - 3x = -6 + 2x        Solve for x.         3 = -6 + 5x         9 = 5x         9/5 =

 The growth and decay formula can also be used with rational numbers.  Consider the following: 1)  The cost of a computer has been increasing at 7% per year.  If it costs $1500 now, find the cost:                      a)  2 years and 6 months from now                      b)  3 years and 3 months ago.              Solutions:                      a)  Ao = 1500, r = .07 and t = 2.5                            A(2.5) = 1500(1 + .07)2.5 = 1500(1.07)2.5 = 1776.44                        b)  Ao = 1500, r = .07, and t = -3.25                              A(-3.25) = 1500(1.07)-3.25 = 1203.91

Exponential FunctionsAny function in the form f(x) = abx, where a > 0, b > 0 and b not equal to 1 is called an exponential function with base b. Let's take a look at a couple of simple exponential graphs.f(x) = 2x

X f(x)3 82 41 20 1-1 1/2-2 1/4

63

-3 1/8 

Notice the domain is all real numbers and the range is y > 0.  As x gets larger (right), y gets very large.  As x gets smaller(left), y approaches zero asymptotically.  Notice also that the graph crosses the y-axis at (0, 1).  The above is the general shape of an exponential with b > 1.  This is an example of exponential growth.

 Now let's look at the graph off(x) = (1/2)x

X f(x)3 1/82 1/41 1/20 1-1 2-2 4-3 8

 

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 Observe that this graph is the reflection about the y-axis of the first graph.  The domain is still all real numbers and the range is y > 0.  The y-intercept is (0, 1).  This is the general form of an exponential graph if 0 < b < 1.  It is an example of anexponential decay.

 Look at the following graphs that illustrate the general properties of exponentials.

Do you see the similarities of each graph?How about this one?

65

Many of the functions associated with exponential growth or decay are functions of time.  We have already had one form:A(t) = Ao(1 + r)t

A second form looks like:A(t) = Aobt/k

where k = time needed to multiply Ao by b

 Rule of 72If a quantity is growing at r% per year then the doubling time is approximately 72/r  years.

For example, if a quantity grows at 10% per year, then it will take 72/10 or 7.2 years to double in value.  In other words, it will take you 7.2 years to double your money if you put it into an account that pays 10% interest.  At the current bank rate or 2%, it will take you 72/2 or 36 years to double your money!!  Boy, jump all over that investment!!

Sample Problems  1)  Suppose you invest money so that it grows at A(t) = 1000(2)t/8                   a)  How much money did you invest?                  b)  How long will it take to double your money?                        Solutions:                          a)  The original amount in the formula is $1000.                          b)  This means what time will it take to get $2000.

2000 = 1000(2)t/8

2 = 2t/8

1 = t/8 8 = t

It will take 8 years to double your money!!  2)  Suppose that t hours from now the population of a bacteria colony is given by:  P(t) =

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150(100)t/10                      a)  What is the initial population?                      b)  How long does it take for the population to be multiplied by 100)20/10 = 150(100)2 = 1,500,000                        c)  What is the population at t = 20?                     Solutions:                          a)  It is 150 from the original equation.                          b)  It takes 10 hours.  That's the definition of the exponential                               function.                          c)  P(20) = 150()  The half life of a substance is 5 days.  We have 4 kg present now.                      a)  Write a formula for this decay problem.                      b)  How much is left after 10 days?  15 days?  20 days?                          Solutions:                              a)  A(t) = 4(1/2)t/5                              b)  A(10) = 4(1/2)10/5 = 4(1/2)2 = 1 kg.                                    A(15) = 4(1/2)15/5 = 4(1/2)3 = 1/2 kg.                                    A(20) = 4(1/2)20/5 = 4(1/2)4 = 1/4 kg. 

3)  The value of a car is given by the equation V(t) = 6000(.82)t                      a)  What is the annual rate of depreciation?                      b)  What is the current value?                      c)  What will be the value in three years?                      Solutions:                          a)  It is 1 - .82 or .18 = 18%                          b)  The current value is given in the formula, $6000.                          c)  V(3) = 6000(.82)3 = 3308.21 Which is $3308.21 The number e and the function ex

Definition of the irrational number e    

 

Without getting into a discussion of limits right now, we can get an idea of what's happening by taking increasingly larger values of n.  We will talk about limits later on in

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the year.  Study the following table of values and use your calculator to double check the results:

N (1 + 1/n)n 

10 2.593742

100 2.704814

1000 2.716924

10,000 2.718146

100,000 2.718268

1,000,000

2.718280

 If you study this chart, you see that the number e approaches a value of 2.718 . . .  The function ex is called the natural exponential function.  The graph of  ex and e-x are graphed below:

Notice, they fit the pattern of the previous section.  The number e appears in many applications of physics and statistics.  We will take a close look at the number e and how it relates to compound interest.

Compound Interest Formula 

A ( t )=A0(1+ rn )

xt

Where:A(t) = amount after time t.Ao = Initial amountr = rate in decimal

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n = number of times compounded in a year.t = time in yearsThus, if the interest was paid semiannually, n = 2.  Paid quarterly would make n = 4,  Paid monthly, n = 12, etc.Sample Problems1)  Find the value of a $1 if it is invested for 1 year at 10% interest compounded quarterly.                  Solution:  Initial amount is $1 with r = .10, n = 4 and t = 1.                          A(1) = 1(1 + .10/4)4 = 1.1038  This means that at the end of a year, each dollar invested in worth 1.1038 or slightly more than $1.10.  The effect of compounding adds another .0038 % to the interest rate.  Thus the effective annual yield is 10.38%.   2)  You invest $5000 in an account paying 6% compounded quarterly for three years.  How much will be in the account at the end of the time period?                Solution:  Initial amount is $5000, with r = .06, n = 4 and t = 3                          A(3) = 5000(1 + .06/4)12 = $5978.09.  This account pays $978.09 in interest over the three years.   3)  What is the effective annual yield on $1 invested for one year at 15% interest compounded monthly?                  Solution:  Initial amount $1, with r = .15, n = 12 and t = 1                           A(1) = 1(1 + .15/12)12 = 1.1608.  The effective annual yield is 16.08%.  This is a relatively big increase because of the number of times compounded in the year. 

The above problems all had one thing in common.  The number of times compounded was a finite number.  We can also havecontinuous compounding.  That is, compounding basically every second on the second.  This would be rather cumbersome to calculate because the compounding is extremely large.  We can use a similar formula if the compounding is continuous.P(t) = Poert

Notice the appearance of the number e.  If you look closely at the compound interest formula, you will see imbedded the definition of the number e.  Only use this formula if you are sure the compounding is continuous.

Problems  1)  Php. 500 is invested in an account paying 8% interest compounded continuously.  They leave it in the account for 3 years.  How much interest is accumulated?                  Solution:  Initial amount Php 500, with r = .08 and t = 3.                       P(3) = 500(e.08(3)) = 635.62.  This means the interest is Php.135.62.  2)  A population of insects rapidly increases so that the population after t days from now is given by A(t) = 5000e.02t.  Answer the following questions:      a)  What is the initial population?      b)  How many will there be after a week?      c)  How many will there be after a month? (30 days) 

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             Solutions:              a)  The initial population is 5000 from the formula.              b)  A(7) = 5000e.14 = 5751              c)  A(30) = 5000e.6 = 9111Logarithmic Functions Common Logarithm Demo: Log Funtion Appletlog x = a if and only if 10a = xThe important thing to remember is the log represents the exponent.  In the case of common logs, the base is always base 10.  Study the following examples.1)  log 100 = 2 because 102 = 100.2)  log 1000 = 3 because 103 = 1000.3)  log 1 = 0 because 100 = 1.4)  log .1 = -1 because 10-1 = .15)  log .01 = -2 because 10-2 = .01The log function is the inverse function of the exponential function and as such their graphs are reflections about the y = x line.  Here is the graph of the common log and the inverse.

Some important facts you need to understand from the log graph.  The domain of the log is x > 0.  The range is all real numbers.  The zero is at x = 1.  You can only find the log of positive numbers.  Logs of numbers less than one are negative and logs of numbers greater than one are positive.

We can find the log of other bases by using the following formula similar to the common log definition.logb x = n if and only if x = bn.Here are some examples:1) log2 8 = 3 because 23 = 82)  log3 81 = 4 because 34 = 81.3)  log4 1/16 = -2 because 4-2 = 1/16

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4) log8 1 = 0 because 80 = 1

One of the most important log function is called the natural log which has the number e as the base.  When e is used as a base we use the following notation:ln x = a if and only if ea = xMost natural logs need to be calculated on your calculator.  The graph of the natural log is shown below:

Solving Simple Log Equations1)  Log x = 3                  Solution:  To solve an equation of this type, rewrite the equation in exponential form.  x = 103

 = 1000 2)  Log |x| = 2                  Solution:  To solve an equation of this type, again rewrite the equation in exponential form and solve for x.|x| = 102

 = 100x = 100 or -1003)  Log (x2 + 19) = 2                  Solution:  Again, rewrite as an exponential equation and solve for x.x2 + 19 = 102

x2 + 19 = 100 x2 = 81 x = 9 or -94)  Log x = .3                  Again, rewrite exponentially.x = 10.3  Use your calculator and round to hundredths.x = 2.00

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5)  Ln x = -1.2                  Solution:  Same as above.x = e-1.2

x = .3

Laws of Logarithms

    1)  Logb MN = Logb M + Logb N     2)  Logb M/N = Logb M - Logb N     3)  Logb M = Logb N if and only if M = N     4)  Logb Mk = k Logb M     5)  Logb b = 1     6)  Logb 1 = 0     7)  Logb bk = k     8)  bLogb x =  x

 Sample problems Write each log in expanded form. 1)  Log5 xy2 =                          Solution:  Log5 x + Log5 y2 = Log5 x + 2 Log5 y 2)  Log7 (xy/z2) =                         Solution:  Log7 x + Log7 y - 2 Log7 z 

3)  Lo g8 √xy=Lo g8 ( xy )12=

12(Lo g8 x+Lo g8 y)

Express each as a single log. 1)  Log x + Log y - Log z =                      Solution:  Log (xy)/z 2)  2 Ln x + 3 Ln y =                      Solution:  Ln x2y3 3)  (1/2) Ln x - (1/3) Ln y = 

                     Solution: lnx1 /2

y1/3 =ln√ x3√ y

Writing logs as single logs can be helpful in solving many log equations. 1)  Log2 (x + 1) + Log2 3 = 4                      Solution: First combine the logs as a single log.                                          Log2 3(x + 1) = 4                                       Now rewrite as an exponential equation.                                          3(x + 1) = 24                                        Now solve for x.                                          3x + 3 = 16                                          3x = 13                                          x = 13/3 

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Since this doesn't make the number inside the log zero or negative, the answer is acceptable. 2)  Log (x + 3) + Log x = 1                      Solution:  Again, combine the logs as a single log.                                          Log x(x + 3) = 1                                        Rewrite as an exponential.                                            x(x + 3) = 10                                        Solve for x.                                            x2 + 3x = 10                                            x2 + 3x - 10 = 0                                            (x + 5)(x - 2) = 0                                            x = -5 or x = 2 

We have to throw out 5.  Why?  Because it makes (x + 3) negative and we can't take the log of a negative number.  So the only answer is x = 2. 3)  Ln (x - 4) + Ln x = Ln 21                      Solution:  Notice, this time we have a log on both sides.  If we write the left side as a single log, we can use the rule that if the logs are equal, the quantity inside must be equal.                                      Ln x(x - 4) = Ln 21                                    Since the logs are equal, what is inside must be equal.                                      x(x - 4) = 21                                    Solve for x.                                      x2 - 4x = 21                                      x2 - 4x - 21 = 0                                      (x - 7)(x + 3) = 0                                       x = 7 or x = -3 

Again, we need to throw out one of the answers because it makes both quantities negative.  Throw out -3 and keep 7.  Thus, the answer is x = 7.

Simplify each log .1)  ln e5                  Solution:  This is rule number 7.  The answer is 5!

2)  Log 10-3                  Solution:  This is again rule #7.  The answer:  -3  (This answers the question:  what power do you raise 10 to get 10 to the third?

3)  eln 7                  Solution:  This is rule #8.  The answer is 7.   4)  e2ln 5                  Solution:  We can use rule #8 as soon as we simplify the problem.  Rewrite as:  eln 25 = 25  The 25 came from 52.   5)  10Log 6 

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                 Solution:  Rule #8 again.  Answer:  6   6)  102 + log 5                  Solution:  We need to simplify before we can apply one of the rules.  Rewrite as:  (102)(10log 5)  Adding exponents means you are multiplying the bases.                                 = 100(5)  Use rule #8 again.                                 = 50

Change of Base Formula

 An exponential equation is an equation that contains a variable in the exponent.  We solved problems of this type in a previous chapter by putting the problem into the same base.  Unfortunately, it is not always possible to do this.  Take for example, the equation 2x = 17.  We cannot put this equation in the same base.  So, how do we solve the problem?  We use thechange of base formula!!  We can change any base to a different base any time we want.  The most used bases are obviously base 10 and base e because they are the only bases that appear on your calculator!!Change of base formula Logb x = Loga x/Loga b

Pick a new base and the formula says it is equal to the log of the number in the new base divided by the log of the old base in the new base.

Examples1)  Log2 37 =                  Solution:  Change to base 10 and use your calculator.                     =  Log 37/log 2                                   Now use your calculator and round to hundredths.                     = 5.21          This seems reasonable, as the log2 32 = 5 and log2 64 = 6.   2)  Log7 99 =                  Solution:  Change to either base 10 or base e.  Both will give you the same answer.  Try it both ways and see.                      = Log 99/Log 7    or  Ln 99/ Ln 7                          Use your calculator on both of the above and prove to yourself that you get the same answer.  Both ways give you 2.36.

Solving Exponential Equations using change of baseNow, let's go back up and try the original equation: 2x = 17 To put these in the same base, take the log of both sides.  Either in base 10 or base e.  Hint.  Use base e only if the problem contains e. Log 2x = Log 17 Using the log rules, we can write as: x Log 2 = Log 17 Now isolate for x and use your calculator. 

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x = Log 17/Log 2 x = 4.09 To check your answer, type in 24.09 and see what you get!  The answer will come out slightly larger than 17 do to rounding. 

Sample Problems1)  e3x = 23                      Solution:  Use natural log this time.                          Ln e3x = Ln 23                          3x Ln e = Ln 23                          3x = Ln 23  ( Ln e = 1)                          x = (Ln 23)/3                          x = 1.05 2)  How long does it take $100 to become $1000 if invested at 10% compounded quarterly?                      Solution:  Ao = 100, A(t) = 1000, r = .1,  n = 4                          1000 = 100(1 + .1/4)4t                           10 = 1.0254t   Use the change of base formula                          Log 10 = Log 1.0254t                          1 = 4t Log 1.025  (Log 10 = 1)                          1/(4Log 1.025) = t                          t = 23.31                         It will take 23.3 years to have $1000 from the $100                         investment. 

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ASSESSMENT 

1)  Simplify each:                      a)  4 . 2-4 =                     b)  (4 . 2)-4 =              c)  (a-1 + b-1)-2 =   2)  Simplify each:   

     3)  Solve the equation:  93x = 81x + 1 4)  The half life of a radioactive isotope is 7 days.  If 5.6 kg are present now, how much will be present in t days?  In 13 days?   5)  A bacteria colony triples every 6 days.  The population currently is 5000 bacteria.  What will be the population in 21 days?   6)  $1000 is invested at 8% compounded quarterly for 3 years.  How much interest will you receive at the end of the time period?   7)  You invest $3000 at 7% compounded continuously for 2 years.  How much money will be in the account at the end of the time period?   8)  Solve each log equation:                  a)  Log x = 21                    b)  Log |x| = 15                    c)  Ln (x2 - 1) = 3                    d)  Log x = 1.6  (Use calculator and round to hundredths.)   9)  Write each log as a single log:                    a)  Log x + log y + 2Log z                    b)  Ln x + Ln y - 3Ln z 

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  10)  Simplify each log:                    a)  Ln e5 =                    b)  102log4 =                    c)  101 + log 5 =   11)  Solve the equation:  Log (x + 2) + Log 5 = 4   12)  Graph y = 3x and y = log3 x on the same axis.   13)  Use a calculator to find Log7 58 and round to hundredths place.   14)  solve the equation:  15x + 1 = 29 and round to hundredths place.   15)  How long will it take to double $1000 if invested at 6% compounded monthly? 

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Chapter VI

System of Equations

And Inequalities

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.

SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES

Many applications of mathematics lead to more than one equation in several variables. The resulting equations are called a system of equations. The solution set of a system of equations consists of all solutions that are common to the equation in the system. ax + by = cWe proved that the graph of an equation of the form is a line and that all ordered pairs (x , y) satisfying the equation are coordinates of points in the line. A system of two linear equations in two variables x and y can be written as. {a1x + b1y = c1} {a2x + b2y = c2}Where a1,b1, c1, a2, b2,and c2 are real numbers. The left brace is used to indicate that the two equations form a system. If an ordered pair (x , y) is to satisfy the system of two linear equations, the corresponding points (x , y) must lie on the two lines that are the graph of the equations.ILLUSTRATION 1

A particular system of two linear equations is.

{ 2x+ y=35x+3 y=10

The solution set of each of the equations in the system is an infinite set of ordered pairs of real numbers, and the graphs of these sets are lines. Recall that to draw a sketch of a line we need to find two points on the line; usually we plot the points where the line intersects the coordinate axis. On the line 2x + y = 3 we have the points (3/2 , 0) and (0,3) while on the line 5x + 3y = 10 we have the points (2,0) and(0, 10/3). Shows on the same coordinate system sketches of two lines. It is apparent that two lines intersect at exactly one point. This point, (-1 , 5) can be verified by substituting into the equations as follows:

2(-1) + 5 = 3

5(-1) + 3(5) = 10

The only ordered pair that is common to the solution sets of the two equations is (-1,5). Hence the solution set of the system is {(-1,5)}.

ILLUSTRATION 2

Consider the system

{6 x−3 y=52 x− y=4

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The lines having these equations appear to be parallel. It can easily to be proved that the lines are indeed parallel by writing each of the equation, we have

6x - 3y = 5 2x – y = 4

-3y = -6x + 5 -y = -2x + 4

y = 2x – 5/3 y = 2x – 4

Example 1

Use the substitution method to find the solution set of the system.

Illustration 1: { 2x+ y=35x+3 y=10

Solution:

We solve the first equation for y and get the equivalent system

{ y=3−2 x5x+3 y=10

We replace y in the second equation by its equal, 3 – 2 x, from the first equation. We then have the equivalent system

{ y=3−2 x5 x+3 (3−2 x )=10

Simplifying the second equation, we have

{ y=3−2x−x+9=10

Solving the second equation for x, we get

{y=3−2 xx=−1

Finally, we substitute the value of x from the second equation into the first equation and we have

{ y=5x=−1

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This system is equivalent to the given one. Hence the solution set is ( -1 , 5)

Example 2

Use the elimination method to find the solution set of the system of equations in Example 1.

{ 2x+ y=35x+3 y=10

Remember that our goal is to eliminate one of the variables. Observe that the coefficient of y is 1 in the first equation and 3 in the second equation. To obtain an equation not involving y, we therefore replace the second equation by the sum of the second equation and -3 times the first. We begin by multiplying the first equation by -3 and writing the equivalent system.

{−6−3 y=−95 x+3 y=10

Adding the equations given the following computations:

−6−3 y=−95 x+3 y=10

−x=1 With this equation and the first equations in the given system, we can write the following equivalent system

{2x+ y=3−x=1

If we now multiply both sides of the second equation by -1, we have the equivalent system

{2x+ y=3x=−1

We next substitute -1 for x in the first equation to obtain

{2 (−1 )+ y=3x=−1

{ y=5x=−1

SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES

So far the linear (first degree) equations we have discussed have contained at most two variables. In this section we introduce systems of linear equations in three variables.

Consider the equation

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2 x− y+4 z=10

For which the replacement set of each of the three variables x, y and z is the set R of real numbers. This equation is linear in the three variables. A solution of a linear equation in the three variables x, y and z is the ordered triple of real numbers (r,s,z) such that if x is replaced by r, y by s, and z by t, the resulting statement is true. The set of all solutions is the solution set of the equation

Illustration 1

For the equation

2 x− y+4 z=10

The ordered triple pair (3,4,2) is a solution because

2(3) – 4 +4 (2 )=10

Some other ordered triples that satisfy this equation are (-1, 8, 5) , (2, -6, 0), (1,0,2), (5,0,0) , (0,-6,1) , (8, 2 1) ,and (7,2 - ½). It appears that the solution set is infinite.

The graph of an equation in three variables is a set of points represented by ordered triples of real numbers. Such points appear in a three dimensional coordinate system, which we do not discuss. You should, however, be aware that the graph of a linear equation in a three variables is a plane.

Suppose that we have the following system of linear equations in the variables x, y and z.

{¿a1 x+b1 y+c 1 z=d 1a2 x+b 2 y+c2 z=d 2a3 x+b 3 y+c3 z=d 3

The solution set of this system is the intersection of the solution sets of the three equations. Because the graph of each equation is a plane, the solution set can be interpreted geometrically as the intersection of three planes. When this intersection consist and independent.

Algebraic methods for finding the solution set of a system of three linear equations in three variables are analogous to those used to solve linear systems in two variables. The following examples shows the substitution method.

Example Find the solution set of the system

{ x− y−4 z=32x−3 y+2 z=02 x− y+2 z=2

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Solution, we solve the first set of the system

{ x= y+4 z+32x−3 y+2 z=02 x− y+2 z=2

We now substitute the value of x from the first equation into the other two equations , and we obtain the equivalent system

{ x= y+4 z+32 ( y+4 z+3 )−3 y+2 z=02 ( y+4 z+3 )− y+2 z=2

{ x= y+4 z+3− y+10 z=−6

y+10 z=−4

We next solve the second equation for y and get

x= y+4 z+3

y=10 z+6

y+10 z=−4

Substituting the value of y from the second equation into the third gives the equivalent system.

¿

{x= y+4 z+3y=10 z+620 z=−10

{x= y+4 z+3y=10 z+6z=−1/2

Substituting the value of z from the third equation into the second equation, we obtain

{x= y+4 z+3y=1

z=−1/2

Substituting the values of y and z from the second and third equations into the first equation, we get

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{ x=2y=1

z=1/2

The latter system is equivalent to the given system. Hence the solution set of the given system is (2,1 ,1/2).The solution can be checked by substituting into each of the given equations. Doing this we have

{2−1+2=34−3−1=04−1−1=2

The equations of the given system are consistent and independent.

Exercise

1.) {4 x+3 y+z=15x− y−2 z=2

2 x−2 y+z=4 2.) { 2x+3 y+z=8

5x+2 y+3 z=−13x−2 y+5 z=15

3.) { x− y+3 z=22x+2 y−z=5

5 x+2 z=7 4.) { 3 x+2 y−z=4

3 x+ y+3 z=−26 x−3 y−2 z=−6

5.) {2x−3 y−5 z=4x+7 y+6 z=−77 x+2 y−9 z=6

6.) {3x−2 y+4 z=47 x−5 y−z=9x+9 y−9 z=1

7.) {3x−5 y+2 z=−22 x+3 z=−34 y−3 z=8

8.) { x− y=23 y+z=1x−2 z=7

9.) {3x−2 y=1z− y=5

z−2 x=5 10.) { x− y+5 z=2

4 x−3 y+5 z=33 x−2 y+4 z=1

SYSTEMS INVOLVING QUADRATIC EQUATIONS

In sections 6.1 and 6.2 our discussions of systems of equations was confined to linear systems. However, a number of applications lead to nonlinear systems as illustrated in exercises 25 through 36. The word problems in these exercises use concepts presented previously, but the resulting systems involve at least one quadratic equation. In this section we discussed methods of solving such systems of two equations in two variables.

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We consider first a system that contains a linear equation and quadratic equation. In this case the system can be solved for one variable in terms of the other, and the resulting expression can be substituted into the quadratic equation, as shown in the following example.

Example 1

Find the solution set of the system.

{ y2=4 xx+ y=3

Solution

We solve the second equation for x and obtain the equivalent system.

{ y2=4 xx=3− y

Replacing x in the first equation by its equal from the second, we have the equivalent system

{y2=4 (3− y)x=3− y

{y2+4 y−12=0x=3− y

We now solve the first equation.

( y−2 )( y+6)=0

y−2=0 y+6=0

y¿2 y=−6

Because the first equation of system (II) is equivalent to the equations y=2 and y=−6, system (II) is equivalent to the systems

{ y=2x=3− y

and { y=−6x=3− y

In each of the latter two systems we have substitute into the second equation the value of y from the first , and we have

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{y=2x=1

and {y=−6x=9

These two systems are equivalent to system (I). Thus, the solution set of (I) is (1,2) , (9,-6).

SYSTEMS OF LINEAR INEQUALITIES AND INTRODUCTION TO LINEAR PROGRAMMING

Systems of linear inequalities are important in economics, business, statistics, science, engineering, and other fields. With electronic computers performing most of the computation, large numbers of inequalities with many unknowns are usually involved. In this section we briefly discuss how to solve system of linear inequalities. We then give an introduction to linear programming, a related approach to decision making problems.

Statement of the form

Ax+By+C>0 Ax+By+C<0

Ax+By+C ≥ 0 Ax+By+C ≤ 0

Where A,B and C are constants, A and B are not both zero, are inequalities of first degree in two variables. By the graph of such an inequality, we mean the (x, y) in the rectangular Cartesian coordinate system for which (x, y) is an ordered pair satisfying the inequality.

Every line in a plane divides the plane into two regions, one on each side of the line. Each of these regions is called a half plane. The graphs of inequalities of the forms.

Ax+By+C>0 and Ax+By+C<0

Are half planes. We shall show this for the particular inequalities

2 x− y−4>0∧Ax+By+C<0

Let L be the line having the equation2 x− y−4=0. If we solve this equation for y, we obtain y=2 x−4. If (x, y) is any point in the plane, exactly one of the following statements holds:

y=2 x−4 y>2 x−4 y<2 x−4

Now,y>2 x−4 if and only if the point (x, y) is any point (x, 2x - 4) on line L; Furthermore, y<2 x−4 if and only if the point (x, y) is below the point (x, 2x - 4) on L; therefore the line L divides the plane into two regions. One region is the half plane above L, which is the graph of inequality y>2 x−4, and the other region is the half plane above

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L, which is the graph of the inequalities y>2 x−4, and the region is the half plane below L, which is the graph of inequality y<2 x−4. A similar discussion holds for any line L having an equation of the form Ax+By+C=0 where B ≠ 0.

If B¿0, an equation of line L is Ax+C=0 , and L is a vertical line whose equationx=4. Then if (x, y) is any point in the plane, exactly one of the following statements is true:

The point (x, y) is to the right of the point (4, y) if and only if x¿4. Showing the graph of inequality x¿4 as the half plane lying to the right of the line x¿ 4. Similarly, the graph of x ¿4 if, and only if the point (x, y) is to the left of the point (4, y). The discussion can be extended to any line having an equation of the form Ax +C=0.

By generalizing the above arguments to any line, we can prove this theorem.

THEOREM

(I) the graph of y¿mx+b is the half plane lying above the line y¿mx+b .

(II) the graph of y¿mx+b is the half plane lying below the line y¿mx+b .

(III) the graph of (y¿mx+b¿ x¿a is the half plane lying to the right of line x¿a .

(IV) The graph of x¿a is the half plane lying to the left of the line x¿a.

Example 1

Draw a sketch of the graph of the inequality

2 x−4 y+5>0

Solution

The given inequality is equivalent to

−4 y>−2 x−5

y ¿12+5 /4

The graph of inequality is the half plane below the line having the equation y=1 /2 x+5/4. A sketch of this graph is the shaded half plane.

A closed half plain is a half plane together with the line bounding it is the graph of an inequality of the form.

Ax+By+C=0∨Ax+By+C ≤ 0

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Illustration

The inequality

4 x+5 y−20 ≥0

Is equivalent to

5 y ≥−4 x+20

y ≥−4/5 x+4

Therefore the graph of this inequality is the closed half plane consisting of the line y=−3/5 x+4 and the half plane above it. A sketch of the graph.

Two intersecting lines divide the points of the plane into four regions. Each of these regions is the intersection of two half planes and is defined by a system of two linear inequalities.

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ASSESSMENT

Solve each equations.

1.) { x− y=82x+ y=1

2.) { y=8+2x6 x+3 y=0

3.) { 2 x+ y=68 x=6 y+9

4.) {9 x−3 y=7y=3 x−5/2

5.) { y=2 x−46 x−3 y−12=0

6.) {2x−3 y=−15 x−4 y=8

7.) {4 x−2 y−7=0

x=12

y+5 8.) { 3x− y=16 x+5 y=2

9.) {2x+6 y=−114 x−3 y=−2

10.) { y=3 x−56 x−2 y=10

Find the solution set of the system.

1.) { x2+ y2=25x− y+1=0

2.) { x2+ y2=25x−2 y=−2

3.) { x2− y=1x2−2 y=−1

4.) {x2− y2=92 x+ y=6

5.) {x2− y2=9x+ y=5

6.) {4 x2+ y2=252x+ y+1=0

7.) {x2− y−4=0x− y−3=0

8.) {4 x2+ y−3=08 x+ y−7=0

9.) {x2−2 y2=2x+2 y=2

10.) {4 x2+ y2=17x2+ y=5

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Chapter VII

Selected Topics in

Algebra

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SEQUENCE FUNCTION Sequence is a set of numbers written in a given order. Function whose domain is the set of natural numbers and whose range is a

subset of the real numbers Terms are the numbers or values in the sequence

Examples:1. 1,2,3,4,5,6,...,n2. 2,4,6,8,10,…,n3. 5,10,15,20,…,nA Sequence Function as a function whose domain is a subset of positive integers.

Example:A ball is dropped from the height of 10 feet. Each time that it bounces, it reaches

a height that is half of its previous height. We can list the height to which the ball bounces in order until it finally comes to rest.

After Bounce 2 4 6 5 8Height (ft) 2 5 7 6 5

The numbers 2,4,6,8,10 form a sequence which is also an example of a sequence. It can be in a form of ordered pairs and we let x as the after bounce and y as the height.

{(2,2),(4,5),(6,7),(5,6),(8,5)}Finite Sequence is a function whose domain is the set of integers and has the last

term {1,2,3,…,n}. Infinite sequence is a list that continues without ends.

Example:Identify whether the sequence is finite or infinite.

1. The first 5 natural numbers: 1, 2, 3 ,4, 52. The set of even whole numbers: 2, 4, 6, 8 , 10 ...3. The multiples of 4 up to the number 40: 4, 8, 12, 16,…, 40.4. The sequence of the multiples of 3: 3, 6, 9, 12...

Solution:1 and 3 are examples of finite sequence while the examples in 2 and 4 are infinite

sequence. The 3 dots at the end of the infinite sequence indicates that the sequence goes on without stopping.

FINDING THE TERM IN THE SEQUENCE

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Sequence is usually denoted by their general or nth term.For example,

t n=n+2

n

The equation provides how we will be able to compute for the nth term of the sequence. To get the value of a specific term , replace n with the number of the term.For example:

To get the first 3 terms of t n=n+2

n , replace n with 2,4 and 6.

Solution:

t n=n+2

n=2+2

2=2

t n=n+2

n=4+2

2=3

t n=n+2

n=6+2

2=4

The domain of the sequence is the set {2,4,6}.The range is denoted as {2,3,4}.

Example:

1. Find the first 5 terms of the sequence:t n=2 n−3

The domain is the set {1, 2, 3, 4, 5} Substituting this into t n=2 n−3 , we have:

t n=2(1)−3 = -1

t n=2(2)−3 = 1

t n=2(3)−3 = 3

t n=2(4)−3 = 5

t n=2(5)−3 = 7

The first 5 terms of the sequence t n=2 n−3 is {-1,1,3,5,7}2. Find the first 3 terms of the sequence:

t n=n2

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The domain is {1,2,3}

Substituting this intot n=n2, we have:

t n=12=1

t n=22=4

t n=32 = 9

The first 3 terms of the sequence {1,4,9}FINDING THE GENERAL TERM

When given the first few terms of the sequence, we can look at the pattern in order to obtain the next terms or a general statement regarding the nth term.Example:1. Find the next term and describe the pattern.

1,4,9,16,25…The next term is 36. The terms are all squares.1st Term 12= 12nd Term 22= 43rd Term 32= 94th Term 42= 165th Term 52= 256th Term 62= 36

SERIES

Series is the sum of the termst1 + t2 + t3 + t4 +,…, + tn

Note: The sum of the infinite sequence is an infinite series. A partial sum is the sum of the first n terms. A partial sum is also called the sum of finite series, and is denoted as sn , where n denotes the number of terms in the sum.For example:

If 4, 7, 10, 13, 16, 19…. Is a sequence, then 4 + 7 + 10 + 13 + 16 +19 is the corresponding sum or series

If only we want to get the sum of only 4 terms in the sequence, it is a partial sum written as S4 , which is equal to 4 + 7 + 10 + 13.

Example:

For the sequence 9, 13, 17, 21, 25, 29, 33, 37, find 1. S4

2. S1

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Solution:1. S4= 9+13+17+21 = 602. S1=9+13+17+21+25+29+33 = 147

SIGMA NOTATION

Use the symbol ∑, a capital sigma of the Greek alphabet which corresponds to

the letter S. The symbol i ,is called the index of summation. Used as summation or the total sum of the sequence.

Example:1

∑i=1

5

i2=12+22+32+42+52

What we have observed here is that, the i =1 is located under the sigma symbol. So, it indicates that the first term on the right-hand side is the value of i2 when i is equal to 1. The next terms are the values of i2 when i is 2,3,4 and 5. We stop here because in the upper sigma notation there is number 5.

The sigma notation can be defined by the equation:

∑i=m

n

F (i )=F (m )+F (m+1 )+F (m+2 )+…+F (n)

Where m and n are integers and m ≤ n. The right hand sign of the equation consist of the sum of n – m + 1terms, the first of which is obtained by replacing i by m + 1 and so on until the last term is obtained.

The upper limit will be m The lower limit will be n The symbol i is the “dummy” symbol because any other symbol could be used

without changing the right side.

Example:

∑i=2

7i3

=23+32+43+53+63+73

Sometimes the term of a sum involve subscripts . For instance the sumn1+n2+n3+n4+…+nn

Can also be written in sigma notation

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∑i=1

n

ni

PRINCIPLE MATHEMATICAL INDUCTION

Definition Let P(n )be a proposition on an integer variable n. Then P(n )is true for

all integers n≥s if and only if the following two conditions are both satisfied :

(i) P( s ) is true ,

ii) If P(k ), where k s , is assumed to be true, then P(k+1 ) is true .

Example:We use the mathematical induction to prove that

1+3+5+7+….+(2n – 1) =n2

STEP 1. We apply that the formula is true for n = 1. If n = 1 then the formula becomes1=12

1=1which is true.

STEP 2. Show that Pk is true.

1+3+5+7+…+(2k−1 )=k 2

If the equation 1 is true, then

1+3+5+…+(2k−1 )+[2 (k+1 )−1 ]=(k+1 )2

This is also true.Then we add 2k +1 – 1 to the left side of the Equation 1 and its equivalent 2k+1 to the right side and we have:

1+3+5+…+(2k−1 )+[2 (k+1 )−1 ]=k 2+ (2k+1 )

1+3+5+…+(2k−1 )+[2 (k+1 )−1 ]=(k+1 )2

Which is Equation 2.

Example:Use mathematical induction to prove that n3 + 3n2 + 2n is divisible by 3 for n 1.

Basis case: n = 1 n3 + 3n2 + 2n = 13 + 3 ·12 + 2·1 = 1 + 3 + 2 = 6 and 3 | 6.

Hypothesis: Assume k3 + 3k2 + 2k is divisible by 3 for k 1.(i.e. k3 + 3k2 + 2k = 3s for some integer s.)

Induction step: (k + 1) 3 + 3(k + 1)2 + 2(k + 1) = (k3 + 3k2 + 3k + 1) + 3(k2 + 2k + 1) + 2(k + 1)

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= (k3 + 3k2 + 2k) + k + 1 + 3k2 + 6k + 3 + 2k + 2 = 3s + 3k2 + 9k + 6 hypothesis and algebra = 3s + 3(k2 + 3k + 2) factoring = 3[s + k2 + 3k + 2] factoring Therefore 3 | [(k + 1)3 + 3(k + 1)2 + 2(k + 1)]. Therefore, by the principle of mathematical induction, the statement is true for all n.Example:

Use mathematical induction to prove that 2n < n! for n ≥ 4

Basis: n = 4 lhs 24 = 16 rhs 4! = 24 Since 16 < 24 the basis holds.

Hypothesis: Assume 2k < k! for k ≥ 4

Induction step:

Proof: 2k+1 = 2 • 2k < 2 • k! < (k + 1) • k! since k >=4 = (k + 1)!

Therefore, by the principle of mathematical induction, the statement is true for all n ≥ 4.

STATISTICSStatistics is derived from the Latin word status which means “state”. In the word

statistics, it refers to the actual numbers derived from data and a method of analysis.Definition

Statistics is a branch of mathematics which concerned with the methods for collecting, organizing, presenting, analyzing and interpreting quantitative data to aid in drawing valid conclusions and in decision making.

TYPES OF STATISTICS

Descriptive statistics A type of statistics, which focuses on the collecting, summarizing, and presenting a

mass of data so as to yield meaningful information.Examples:

1. A math teacher wants to determine the percentage of students who passed the examination.2. Lula is a bowler wants to find her bowling average for the past 10 games.

Inferential statistic

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A type of statistics, that deals with making generalizations and analyzing sample data to draw conclusions about a population. This is a process of obtaining information about a large group from the study of a smaller group.

Examples:1. Kim is a basketball player wants to estimate his chance of winning the MVP award based on his current season averages and averages of his opponents.2. A manager would like to predict base on previous year’s sales, the sales performance of company for the next six years.

BASIC STATISTICAL TERMS

1. Population is consisting of all elements such as events, objects, and individuals whose characteristic is being studied.

Example:The researcher would like to determine the number of male BSE students in

CvSU-Imus campus. 2. Variable is a characteristic of an item or individual that will be analyzed using statistics.

Variables are usually denoted by any capital letter.

3. Sample- is a portion of population selected for study.4. Parameter- a numerical measure that describes a characteristic of a population. 5. Statistics- a numerical measure that describes a characteristic of a sample.

TYPES OF VARIABLE

Qualitative variable- a variable, that cannot be measured numerically but can be classified into different categories.

Examples: Names, Gender, Hair color, subjects enrolled in a semester, species.

Quantitative variable- consists of variable that can be measured numerically. Examples: price of a house, height, gross sales, numbers of cars own.

CLASSIFICATION OF QUANTITATIVE VARIABLE

Discrete Quantitative Variable – a variable results from either a finite number of possible values or a countable number of possible values. In other words, a discrete variable can assume only certain values no intermediate values.

Examples: number of patient, number of sold cars, number of book

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Continuous Quantitative Variable- a variable results from infinitely many possible values that can be associated with points on a continuous scale in such a way that there are no gaps of interruptions.

Examples: time, height of a person, weight

LEVELS OF MEASUREMENTThe level of measurement of data determines the algebraic operations that can be

performed and th statistical tools that can be applied to the data set.LEVEL 1. Nominal is characterized by data that consist of names, labels, or categories only.

Examples: gender, marital status, employment, religion, address, degree program LEVEL 2. Ordinal it involves data that may be arranged in some order, but differences between data values either cannot be determined or are meaningless.The data measured can be ordered or rank.

Examples: grades of the students, military rank, job position, year level LEVEL 3. Interval In this level has precise differences between measures but there is no true zero.

Examples: temperature, IQ scoreLEVEL 4. Ratio is the interval level modified to include the inherent zero starting point. For values at this level, differences and ratios are meaningful.

Examples: weekly allowance, area, and volume

APPLICATION IN PROBABILITYProbability is a way to measure the chances that something will occur in relation to the possible alternatives. For example, the probability is not a guarantee. A couple might have six children and all are boys, or they might have six children and all are girls.

Now you might think that a couple with six girls would not expect to have another girl if they decided to have a seventh child. In fact, the probability that the seventh child is a girl is still ½ , since the gender of this child is not affected by the gender of the previous six children.

What if you know that a family has seven children and six of them are girls? Is the probability that the seventh child is a girl still1/2? There are eight possibilities: all seven children are girls, or either the first, second, third, fourth, fifth, sixth, or seventh child is a boy. Since in only one of these cases the seventh child is a girl, the probability is 1/8.

STATISTICS AND LINE PLOTS

Objectives After studying this lesson, you should be able to:

Interpret numerical data from a table, and

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Display and interpret statistical data on a line plot.

Each day when you read news papers or magazines, watch television, or listen to the radio, you are bombarded with numerical information about the national economy, sports, politics, and so on. Interpreting this numerical information, or data, is important to your understanding of the world around you. A branch of mathematics called statistics helps provide you with methods for collecting, organizing

and interpreting data.

Statistical data can be organized and

presented in numerous ways. One of the most

common ways is to use a table or chart. The

chart at the right shows the hourly wages

earned by the principal wage earner in ten

families.

Using tables or charts like the one at the right should enable you to more easily analyze the given data.

Example 1

Use the information in the chart above to answer each question.

a. What are the maximum and minimum hourly wages of the principal wage earner for the ten families.

The families.

b. What percent of the families have a principal wage earner that makes less than $ 10.00 per hour.

In some instances, statistical data can be presented on a number line. Numerical information displayed on a number line is called a line plot. For example, the data in the table above can presented in a line plot.

Family Hourly wageA $ 8.00B $ 10.50C $ 20.25D $ 9.40E $11.00F $ 13.75G $ 8.50

H $ 10.50I $ 9.00J $ 11.00

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For example 1, you know that the data in the chart range from $ 8.00 per hour to $ 20.25 per hour. In order to represent each hourly wage on a number line, the scale used must includes these values. A “w” is use to represent each hourly wage. When more than one “w” s has the same location on the number line, additional “w” are placed one above the other. A line plot for the hourly wages is shown below.

Exercises

Practice

1. Use the line plot at the right to answer each question.

a. What was the highest score on the test?

b. What was the lowest score on the test?

c. How many students took the test?

d. How many students scored in the 40’s?

e. What score was received by the most students?

STEM AND LEAF PLOTS

Objective : After studying this lesson, you should be able to:

Display and interpret data on a stem-leaf-plot.

Application: Mr.Juaez wants to study the distribution of the scores for a 100-point unit exam given in his first-period Biology class. The scores of the 35 students in the class are listed below.

82 77 49 84 44 98 93 71 76 65 89 95 78 6989 64

88 54 96 87 92 80 44 85 93 89 55 62 79 90 86 75

74 99 62

20-25 325-30 1030-35 835-40 240-45 745-50 5

50 1

100

He can organize and display the scores in a compact way using a stem-leaf-plot.

In a stem-leaf-plot, the greatest common place value of the data is used to form the stems. The numbers in the next common place-value position are then used to form the leaves. In the list above, the greatest place value is tens. Thus, the number 82 would have stem 8 and leaf 2.

To make the stem-and-leaf-plot, first make a vertical list of the stems. Since the test scores range from 44-99, the stems range from 4-9. Then, plot each number by placing leaf 2 to the right of the stem 8. Theright. A second stem-and-leaf-plot can be made To arrange the leaves in numerical order fromLeast to greatest as shown at the right. This will make it easier for Mr. Juarez to analyze the data.

STEM LEAF

4 9 4 45 4 56 5 9 4 2 27 7 1 6 8 9 5 48 2 4 9 9 8 7 0 5 9 6

9 8 3 5 6 2 3 0 98/2 Represents a score of 82

STEM LEAF4 4 4 95 4 56 2 2 4 5 97 1 4 5 6 7 8 98 0 2 4 5 6 7 8 9 9

99 0 2 3 3 5 6 8 9

Use the information in the stem-and-leaf-plots above to answer each question.a. What were the highest and lowest scores on the test? 99 and 44b. Which test scores occurred most frequently? 89 (3times)c. In which 10-point interval did the most students score?80-89 (10 students)d. How many students received a score of 70 or better? 25 students

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Example 1 Sometimes the data for a stem-and-leaf-plot are numbers that have more than two digits. Before plotting these numbers, they may need to be rounded or truncated to determine each stem and leaf. Suppose you wanted to plot 356 using the hundreds digit for the stem. ROUNDED TRUNCATEDRounded 356 to 360. Thus, you To truncate means to cut off, so Would plot 356 using stem 3 truncate 356 as 350. Thus, youAnd leaf 6. What would be the would plot 356 using stem 3 and leafStem and leaf of 499? 5 and 0 5. What would be the stem and leaf of 499? 4 and 9A back-to-back stem-and-leaf plot is sometimes used to compare two sets of data or rounded and truncated values of the same set of data. In a back-to-back plot, the same stem is used for the leaves of both plot.

ASSESSMENT

TEST IA. For each sequence, find the indicated partial sum.1. 3,5,7,9,11…S5

2. 2,4,8,16,32…S7

3. 1,2,3,4,5,6,7,8,9,10,11,12,13,14,…,S6

4. 5,10,15,20,25,30,35,40,…,S7

B. Find the indicated term of the sequence.

1.t n=4 n−7 : t 5

2.t n=4 n−2

3: t 6

3.t n=5+2n : t 10

4.t n=4 n2

: t 4

5.t n=4 (n−7)2: t 5

C. write the following in sigma notation

1. 2+4+6+8+102. 1+3+5+7+9+11+13+15+173. 52+62+72

102

4. 1+ 12+ 1

3+ 1

4+ 1

55. (n+1) + (n+2) +n+3)… (n+n)TEST II Applications1. Football The stem-and-leaf plot below gives the number of catches of the NFL’s leading pass receiver for each season through 1990. a. What was the greatest number of

catches during the season?b. What was the least number ofcatches during the season?c. How many seasons are listed?d. What number of catches occurredmost frequently?e. How many times did the leading

STEM LEAF6 0 1 2 6 7 7 1 1 1 2 2 3 3 3 4 5

88 0 2 6 8 8 99 0 1 2 2 510 0 0 1 69/2 Represents 92 catches