algebra

1. Chapter I
Review topics in
Algebra 1
1

2. 2
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 ab 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)
1
3
4)
10
9
5)
9
6

3. If the numbers are not repeating or terminating decimals. They called irrational number.
3
For examples:
1) π
2) √2
3) 0.61351
4) √8
5) √11
Properties of real numbers
Let us denote the set of real numbers by 푅. These properties are statement derived from
the basic axioms of the real numbers system. Axioms are assumptions on operation with
numbers.
Axioms of Equality
Let a, b, c, d ∈ R
1. Reflexive Law
If a=a
2. Symmetric Law
If b=c then c=b
3. Transitive Law
If b=c and c=d then b=d
4. Additional Law of Equality
If a=b then a+c=b+c
5. Multiplication Law of Equality
If a=b then a.c=b.c
Axioms for Addition and Multiplication
Let a, b, c, d, ∈ R
1) A. Closure property for addition
a+b ∈ R
Examples:
1) 3+3=6
2) 7+(-4)=3
3) -8+4=-4
B. Closure property for multiplication
a.b ∈ R
Examples:
1) 3(7)=21
2) -8(3)=-24
3) 0.11=0

4. 2) A. Commutative prroperty for addition
4
a+b=b+a
Examples:
1)
1
2
+ 7 = 7 + 1
2
2) 0.3 + (− 5
6
) = − 5
6
+ 0.3
3)
1
3
+ 21 = 21 + 1
3
B. Commutative prroperty for multiplication
a.b=b.a
Examples:
1)
4
5
(22) = 22 (4
5
)
2) 6.3=3.6
3)
10
9
(−25) = −25 (10
9
)
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)+
1
2
= 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)] 1
4
= 5 [7 (1
4
)]
3) [3푥(6푥)]]5 = 3푥[6푥(5)]
4) Identity property for multiplication
a.1=a
Examples:
1) 1.a3=a3
2)
3
7
(1) = 3
7
3) 3.1=3
5) A. Inverse property for addition
a+(-a)=0
Examples:
1) 6+(-6)=0
2) 10+(-10)=0
3) -3+3=0

5. B. Inverse property for multiplication
5
푎. 1
푎
= 1
Examples:
1) -2(− 1
)=1
2
1
8
2) 8(
)=1
3) -6(-
1
6
)=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 훼푛 , α is the base and 혯 is the exponent. The expression 훼푛
means that the value α is multiplied 혯 times by itself.
Examples:
1) 63= 6.6.6
=216
2) 56= 5.5.5.5.5
=15625
3) 42= 4.4
=16
Integral and zero exponents
Laws of Integral and Zero Exponents
Theorem 1:
For any real number α, (α≠ 0)
푎0 = 1
Examples:
1) (6푎0 + 3)0=1
2) 6α0+70=6(1)+1=7
3) 2α0+70=2(1)+1=3
Theorem 2:
For any real numbers α,
αm. α혯= αm+n
where m and n are integers.
Examples:
1) α5.α4=푎5+4 = 푎9
2) 4푥푦2(2푥2푦2) = 8푥 1+2푦2+2 = 8푥 3푦4
3) 푥 푎+3. 푥 푎+4 = 푥 2푎 +7

6. 6
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),
(
푎
푏
)푛 =
푎푛
푏푛
where n is any integer.
Examples:
1) (
푎2
푏3 )2 = 푎4
푏6
3
4
2) (
)3 = 33
43 = 27
64
푥
푦 +2
3) (
)2=
푥2
2
= 푥(푦 +2)2 푦2 +4푦 +4
Theorem 6:
For any real numbers a(a≠0),
푎푚
푎푛 = 푎푚−푛
where m and n are integers.
Examples:
1)
푎7
푎5 = 푎7−5=푎2
2)
푥3푦4 푧5
푥푦푧
= 푥 3−1푦4−1 푧5−1 = 푥 2푦3푧4
3)
푥4푦4
푥4푦4 = 푥 4−4푦4− 4 = 푥 0푦0 = 1(1) = 1

7. 7
Theorem 7:
For any real numbers a(a≠0),
푎−푛 =
1
푎푛
Where n is any positive integer.
Examples:
1) 3푥 3푦−2=
3푥3
푦2
2) (4푥 2푦)−2 = 1
(4푥2푦)2 = 1
8푥4 푦2
3) (푥 2 + 푦)−2 = 1
(푥2+푦)2 = 1
푥4+푦2
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 푥
1
2 is the same as √2 (read as square root of 2), and 푥
2
3 is
the same as 3√푥2
(read as cube root of x squared). The expression 푛√푎푚 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.
푎
푚
푛
=푛√푎푚
Laws of Radicals
Theorem 1:
For any real numbers a,
√푎푛 = 푎 푛
Examples:
1) √42 = 4
2) 3√(푥2푦)3=푥 2푦
3
3) √33 =3
Theorem 2:
For any real numbers a,and b.
√푎 푛 . √푏 푛 = √푎푏 푛
Examples:
1) √3. √3 = √3.3 = √9=3
2) √4. √3 = √4.3 = √12
3) √푎. √푏 = √푎. 푏

8. 8
Theorem 3:
For any real numbers a,and b, (b≠0)
√푎 푛
√푏 푛 = √
푎
푏
푛
Examples:
1) 3√푎
√푏 3 = √
푎
푏
3
2) √4
√5
= √4
5
3) 4√푥
√푦 4 = √
푥
푦
4
Theorem 4:
For any real numbers a ,
푚
√푎 푚푛 = √ √푎 푛
푛
= √ √푎 푚
Examples:
3
1) √64 6 = √√64
= √8 3 = 2
2
2) √16 4 = √√16
= √4 2 =2
3) √100 3 = √100 3 =√100 = 10
Theorem 5:
For any real numbers a
k √푎푘푚 푛 = 푛√푎푚
Examples:
1) √24 6
= √22.2 2.3 = 3√22 = √4 3
2) √93 6 = 3.2√93.1 = √9 2 =3
3)
Addition and Sutraction of Radicals
To add and subtract radicals, first we need to combine the like terms with
similar radicals.
Examples:
1) √2 + 3√2 − 2√2 = 2√2
2) √8 + √18 + √32 = √4.2 + √9.2+√16.2 = 2√2 + 3√2 + 4√2 = 9√2
3) 푦√푥 3푦 − √푥 3푦3 + 푥√푥푦3 = 푦√푥 2. 푥푦 − √푥 2. 푥. 푦2. 푦 + 푥√푥. 푦2. 푦 =
푥푦√푥푦 − 푥푦√푥푦 + 푥푦√푥푦 = 푥푦√푥푦

9. Multiplication and Division of Radicals
To 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√푥푦. √푥 2푦 3 . √푥푧 3 =√푥푦. 푥 2푦. 푥푧 3 = √푥 4푦2푧 3 = 푥 √푥푦2푧 3
3) √16 3 ÷ √−2 3 =√16 ÷ (−2 3 )= √−8 3 = −2
9
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 x2+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)
=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

10. 10
Multiplication of Polynomials
Examples:
1) 푥 푚.푥 푛 = 푥 푚+푛
2) 푥 −2.푥 2=푥 0 = 1
3) Multiply a+2b+3c by 5m.
= a+2b+3c(5m) in multiplication, we apply the
=5am+10bm+15cm distributive property
Division of Polynomials
To divide a polynomial by a monomial, divide each term of the polynomial by the
monomial.
푥 푚
푥 푛 = 푥 푚−푛 푎푛푑 푥 −푛 =
1
푥 푛
Examples:
1)
푥5
푥2 = 푥 3
1
푥5
2) 푥 −5=
3) Divide 7푥 2 − 5푥 푏푦 푥
푥 is the divisor and 7푥 2 − 5푥 as the dividend, we have
7푥2 −5푥
푥
=
7푥2
푥
-
5푥
푥
=7푥 − 5
1.4 Factoring
Factors 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=2.9
18=3.6
18=18.1
A 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:
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 3
All 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 20a2b
20a2b=2.10.a.a.b
=2.2.5.a.a.b

11. The greatest common factor of two or more integer is the product of the prime
11
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 factors
The GCF of 54, 63, and 180 is 3.3 or 9.
2) 8푎2푏 푎푛푑 18푎2푏2푐
8푎2푏=②. 2.2. 푎 ⃝ . 푎 ⃝ . 푏 ⃝
18푎2푏2 푐=②.3.3. 푎 ⃝ . 푎 ⃝ . 푏 ⃝. 푏. 푐
= 2푎2푏
The GCF of 8푎2 푏 푎푛푑 18푎2푏2푐 is 2푎2푏.
Factoring Using the Distributive Property
To multiplied a polynomial by a monomial by using the distributive property.
Multiplying Polynomials Factoring Polynomials
3(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 10푦2 + 15푦
10푦2=2.⑤. 푦 ⃝. 푦
15푦 =3.⑤.푦
The GCF is 5y
10푦2 + 15푦=5y(2y)+5y(3)
=5y(2y+3) distributive property
2) Factor 21푎푏2 − 33푎2푏푐
21푎푏2=③.7. a ⃝ . 푏 ⃝. 푏
33푎2푏푐=③.11. a ⃝ . 푎. 푏 .⃝c the GCF is 3ab
21푎푏2 − 33푎2푏푐= 3ab(7b)-3ab(11ac)
=3ab(7b-11ac) distributive property
Factoring 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.

12. 12
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-35
2) Factor 8푚2 푛 − 5푚 − 24푚푛 + 15
8푚2 푛 − 5푚 − 24푚푛 + 15=(8푚2푛 − 5푚) + (−24푚푛 + 15
= 푚(8푚푛 − 5) + (−3)(8푚푛 − 5)
=푚 − 3(8푚푛 − 5)
Check:
푚 − 3(8푚푛 − 5) = 푚(8푚푛) + 푚(−5) + (−3)(8푚푛) + (−3)(-5)
= 8푚2 푛 − 5푚 − 24푚푛 + 15
Factoring 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 2푦2 + 7푦 + 6
Factors of 12 Sum of Factors
1.12 1+12=13 no
2.6 2+6=8 no
3.4 3+4=7 yes
2푦2 + (3 + 4)푦 + 6 Select the factors 3 and 4.
2푦2 + 3푦 + 4푦 + 6
(2푦2 + 3푦) + (4푦 + 6) Group terms that have a
푦(2푦 + 3) + 2(2푦 + 3) common monomials factor
(푦 + 2)(2푦 + 3) Factor (use the distirbutive property)
Therefore 2푦2 + 7푦 + 6= (푦 + 2)(2푦 + 3)

13. 13
Example:
Factor 5푥 − 6 + 푥 2
The trinomials 5푥 − 6 + 푥 2 can be written as 푥 2 + 5푥 − 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, 푥 2 + 5푥 − 6 = (푥 − 1)(푥 + 6)
Factoring Differences of Square
The 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.
푎2 − 푏2 = (푎 − 푏)(푎 + 푏)
Examples:
1) 푓푎푐푡표푟 푎2 − 64
푎2 − 64 = (푎)2 − (8)2
= (푎 − 8)(푎 + 8)
푎. 푎 = 푎2 푎푛푑 8.8 = 64 use the difference of square
2) 푓푎푐푡표푟 푎푥 2 − 100푦2
푎푥 2 − 100푦2 = (3푥)2 − 10푦2
= (3푥 − 10푦)(3푥 + 10푦)
3푥. 3푥 = 9푥 2 푎푛푑 10푦. 10푦 = 100푦2
Perfect Square and Factoring
Numbers 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 (푎 + 푏)2 푎푛푑 (푎 − 푏)2 are called perfect
squares and the expansions of these products are called perfect square trinomials.
(푎 + 푏)2 = 푎2 + 2푎푏 + 푏2
(푎 − 푏)2 = 푎2 − 2푎푏 + 푏2
Finding a Product Factoring
(푦 + 8)2 = 푦2 + 2(푦)(8) + 82 푦2 + 16푦 + 64 = (푦)2 + 2(푦)(8) +
(8)2
= 푦2 + 16푦 + 64 = (푦 + 8)2

14. Examples:
Determine whether 16푎2 + 81 − 72푎 is a perfect square trinomial.
1) 16푎2 + 81 − 72푎 = 16푎2 − 72푎 + 81
= (4푎)2 − 2(4푎)(푎) + (푎)2
= (4푎 − 9)2
2) 푥 2 + 22푥 + 121 = (푥)2 + 2(푥)(11) + (11)2
14
= (푥 + 11)2
Solving Equations by Factoring Zero Product Property
For 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)=0
16t(9-t)=0, then 16t=0 or 9-t zero product property
16t=0 or 9-t=0 solve each equation
t=0 9=t
check: Substitute 0and 9 for t in the original.
16t(9-t)=0
16(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=0
y+2=0 or 3y+5=0
y=-2 3y=-5
푦 = − 5
3
Check: (y+2)(3y+5)=0
(-2+2)[(3)(-2)+5]=0 or (− 5
3
+ 2) [(3) (− 5
3
) + 5] = 0
0(-1)=0
1
3
(0) = 0
0=0 0=0
SOLUTION SET: (-2, − 5
3
)
1.5 Rational Expressions
A 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 Terms
We need to lowest term the fraction, if the numerator and denominator have no
common factor.

15. 푥2−2푥푦+4푦2
3푥(푥+푦)−2푦 (푥+푦)+2푥2 −푦2
15
Examples:
1)
4푎2 푏푐3
6푎푏3 푐4 = 2.2.푎.푎 .푏.푐.푐.푐
2.3.푎.푏.푏.푏.푐.푐.푐 .푐
= 2푎
3푏2푐
2)
푥2+2푥푦 +푦2
푥2 −푦2 = (푥+푦)(푥+푦 )
=
(푥+푦)−(푥−푦 )
푥+ 푦
푥− 푦
3)
푥3+8푦 3
4푥+8푦
= 푥+2푦 (푥2−2푥푦 +4푦2
4(푥+2푦 )
=
4
Multiplying and Dividing Rational Expressions
In multiplication if
푝
푞
푎푛푑 푟
푠
are rational expressions and q and s are real numbers
not equal to 0, then
푝
푞
. 푟
푠
= 푝푟
푞푠
.
Examples:
1)
4
3
. 1
5
= 4
15
2)
푐
. (푎 + 2푏)(푎 − 푏)
푎2−푏2 푐
=
(푎 + 푏)(푎 − 푏)
. (푎 + 2푏)(푎 − 푏)
푐(푎+2푏)
=
푎+푏
In dividing algebraic fractions, multiply the dividend by the reciprocal of the divisor. The
reciprocal of a fraction is its multiplicative inverse.
Examples:
1)
4
3
÷
6
5
= 4
3
. 5
6
= 20
18
표푟 10
9
2)
8
7
÷ 3 = 8
7
. 1
2
= 8
14
표푟 4
7
3)
푦2 −16
푦 −5
÷ 2푦 −8
푥푦−5푥
(푦−4)(푦+4)
=
푦 −5
. 푥(푦−5)
2(푦−4)
= 푥푦 +4푥
2
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)
5
6
− 2
3
+ 1
8
= 20−16+3
24
= 7
24
2)
4
5
+ 3
5
+ 2
5
= 4+3+2
5
= 9
5
3) 3푥 − 2푦 + 2푥2 −푦2
푥 +푦
=
푥+푦
=
3푥2 +3푥푦 −2푥푦 +2푦2 +2푥2 −푦2
푥 +푦
=
5푥2 +푥푦− 3푦2
푥+푦

16. Simplifying Complex Rational Expressions
A factor which contains one or more fractions either in the numerator or
The 푛푡ℎ root of a real number
If n is a positive integer greater than
1 푎푛푑 푎 푎푛푑 푏 are real number such that
푏푛 = 푎, then b is an 푛푡ℎ root of a.
The principal 푛푡ℎ root of a real number. If n is a
positive integer greater than 1, a is a real number,
and √푎 푛 denotes the princial 푛푡ℎ root of a, then
If a>0, √푎 푛 is the positive 푛푡ℎ root of a.
If a<0, and n is odd, √푎 푛 is the negative 푛푡ℎ
root of a.
√0 푛 = 0
16
denominator or in both.
Examples:
1)
4
31
3
= 4
3
. 3
1
= 12
3
표푟 4
2)
3
2+
1
3
= 3
6+1
3
= 3
7
3
= 3. 3
7
= 9
7
1.6 Rational Exponents
We defined 푎푛 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.
1
That is, we wish to attach a meaning to 푎
⁄푛 푎푛푑 푎
푚
⁄푛, 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 = 4
2) 3 is a fourth root of 81 because 34 = 81
3) 4 is a cube root of 64 because 43 = 64
Definition
.
The symbol √ is called a radical sign. The entire expression √푎 푛 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.

17. 17
Examples 2:
1) √4 = 2
2) √81 4 = 3
3) √64 3 = 4
Definition
Examples 3:
1) 25
1
⁄2 = √25 = 5
2) −8
1
⁄3= √−8 3 = −2
3) ( 1
81
)1/4=√ 1
81
4
= 1
3
Definition
Examples 4:
1) 9
3
⁄2=(√9)3=33=27
2) 8
2
⁄3 = (√8 3 )2=22 = 4
3) −27
4
⁄3 = (√−27 3 )4=(-3)4=81
It can be shown that the commutative law holds for rational
exponents, and therefore
(푎푚)1/n=(푎
1
⁄푛)m
From which it follows that 푛√푎푚 = ( √푎 푛 )m
The next theorem follows from this equality and the definition of 푎
푚
⁄푛
If n is a positive integer greater than 1, and a is
a real number, then if √푎 푛 is a real number
푎
1
⁄푛 = √푎 푛
If m and n are positive integers that are
relatively prime, and a is a real number,
then if √푎 푛 is a real number
푎
푚
⁄푛 = (√푎 푛 )m ⇔ 푎
푚
⁄푛 = (푎
1
⁄푛)m

18. If m and n are positive integrers that are
relatively prime, and a is a real number,
then if √푎 푛 is a real number
푚
⁄푛 = (푎푚)1/n
18
Theorem 1
Examples 5:
푚
⁄푛 = 푛√푚 Theorem 1 푎is applied ⇔ 푎
in the following:
1) 9
3
⁄2=√93=729 =27
2) 8
2
⁄3 = √8 3 2=√64 3 = 4
3) −27
4
⁄3 = (√−27 3 )4=√531441 3 =81
푎
Observe that 푎
푚
⁄푛 can be evaluated by finding either ( √푎 푛 )m or 푛√푎푚. Compare
example 4 and 5 and you will see the computation of ( √푎 푛 )m in example 4 is simpler than
that for 푛√푎푚 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 (푎푚 )1/n=│a│m/n
A particular case of this equality occurs when m=n. We then have (푎푛)1/n=│a│
(if n is a positive even integer) or, equivalently, 푛√푎푛 = │a│ (if n is even)
If n is 2, we have √푎2 = │a│
Examples 7:
1) [(-9)2]1/2=│-9│=9
2) [(-9)2]1/4=│-9│2/4=91/2=3

19. Definition
If m and n positiv e integer that are
relatively prime and a is a real number and
1
⁄3)2=(1
19
Example:8
1) 8
−2
⁄3 = 1
8
2
⁄3=
1
( 3√8)2 =
1
22=
1
4
2) 8
−2
⁄3 = (8
−1
⁄3)2=( 1
8
2
1
4
)2=
3)
1
푥
⁄3
푥
1
⁄4 =푥
1
⁄3.
1
푥
1
⁄4=푥
1
⁄3.푥
−1
⁄4 = 푥(1
⁄3)−1
⁄4 = 푥
1
⁄12
a≠0, then if √푎 푛 is a real number.
푎
−푚
⁄푛 =
1
푎
푚
⁄푛

20. ASSESSMENT
20
TEST I
Name the property that justifies each step.
1. Simple 6a+(8b+2a)
a. 6a+(8b+2a)=6a+(2a+8b)
b. =(6a+2a)+8b
c. =(6+2)a+8b
d. =8a+8b
2. Simplify 6푎2 + (6푎 + 푎2 ) + 9푎
a. 6푎2 + (6푎 + 푎2 ) + 9푎 = 6푎2 + (푎2 + 6푎)+9a
b. = (6푎2 + 푎2 ) + (6푎+9a)
c. = (6푎2 + 1푎2) + (6푎+9a)
d. =6 + 1(푎2) + (6 + 9)푎
e. = 7푎2 + 15푎
Simplify and express the following.
1.
푎푚+5
푎푚−2
2. [
(푥+푦)0+푎0+ 푏0
푎+푏+푐
]-2
3. (푎−2 + 푦)-2
4. (37푥+5 )(34푥−4)
5. (9푥푦2)(4푥 3푦)
Rational Expression (simplify)
1.
9푛
63푛
÷ 9푛
2.
−15푚3 푛2 푝2
−35푚2 푛5 푝
3.
푥+푦
푥2−푦2
4.
3푚− 1
9(푚−1)2−4
5.
4푚푛+6
10푚 +8푛
6.
2푥2 +3푥 −5
10푥 +25
7.
푥2−5푥 −24
4푥2 −27푥 −40
8.
25푎2 +70푎 +49
25푎 2 −49
Factor each polynomial into two binomials
1. a2+ 12a+ 27
2. y2+ 21y+ 110
3. n2-4n+ 4
4.x2-12x + 20
5. x2+ 11x -12

21. Answer the following word problemsand multiple choice questions
1. 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
21
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 II
A. 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. √4
6. √5
7. 0
8. 3
9. 0.63586358
10. √866
C.Determine which real number property is shown by each of the following
1. − 1
4
+ 1
4
= 0
2. 2(1)=2
3.
1
4
(4)=1
4. -7+(-4)=-4+(-7)
5. 0.3(0)=0.3

22. 22
6. 5[3+(-1)]=5(3-1)
7. (8+
9
8
9
8
)+0.45=8+(
+0.45)
8. 5(8+8)=5(8)+5(8)
9. 6x+(8x+10)=(6x+8x)+10
10. 5a+2b=2b+5a
D.Simplify and express the following expressions with positive and negative integrals
only.
1. 50
2.
10푚4
30m
3.
16푏4 푐
−4푏푐3
4. 푦3. 푦4
5. (5푥푦)6
6. (푎푏)3
7. (푥 3푦2)3
8. [(−5)2]2
9.
푥5푦6
푥푦
=
10.
푎7
푎3
E.Simplify the following radicals
1.(5√2)(3√6)
2.(3푎3√4푥2)(4 √3푥푦 3 )
3.4√ 9
16
4.√2(3+√3)
5.5√2+3√2
6.√18 − 2√27 + 3√3 − 6√8
7.√16푏 + √4푏
8.
−12√24
3√2
9.√8 + √50
10. 4√푥 7푦10
F.Simplify the following polynomials.
1. (5푥 − 1) + (10푥 2 + 7푥)
2.(20푥 2 + 2) + (15푥2 − 8) + (3푥 2 − 4)
3.(푥 2 + 푦2 + 8) + (4푥 2 − 2푦2 − 9)
4.(−3푥 2 + 5푦 − 4푥푦 + 푦2) 푓푟표푚(2푥2 − 4푦 + 7푥푦 − 6푦2)
5.2푥 2 + 6푥 + 5 푎푛푑 3푥2 − 2푥 − 1

23. 23
6.(푥 + 2)(푥2 − 2푥 + 3)
7.푎푏(2푎 + 1)
푥2−3푥 −10
8.
푥+ 2
푥6+2푥4 +6푥−9
9.
푥3+3
10. (3푥 3 − 11푥 2푦 + 11푥푦2 − 2푦3) ÷ (푥 − 2푦)
G.Find the factors and its GCF
1. 21
2. 4, 12
3. 63
4. 304
5. 18, 35
6. 12푎푛2, 40푎4
7. 7푦2, 14푦2
8. 15, 10
9. 6푎2, 18푏2, 9푏3
10. 18푥 2푦2, 6푦2, 42푥 2푦3
H.Factor each trinomials. If possible. If the trinomial cannot be factored using integers.
Write prime.
1. 6푚푥 − 4푚 + 3푥 − 2푟
2. 3푚푦 − 푎푏 + 푎푚 − 30푦
3. 푎2 − 2푎푏 + 푎 − 20
4. 3푚2 − 5푚2 푝 + 3푝2 − 5푝3
5. 4푎푥 − 14푏푥 + 35푏푦 − 10푎푦
6. 푦2 + 12푦 + 27
7. 푐2 + 2푐 − 3
8. 푥 2 − 5푥 − 24
9. 7푎2 + 22푎 + 3
10. 6푦2 − 11푦 + 4
I.Factor out each polynomials
1. 푥 2 − 49
2. 푥 2 − 36푦2
3. 16푎2 − 9푏2
4. 2푎2 − 25
5. 2푧2 − 98
6. 푛2 − 8푛 + 16
7. 4푘2 − 4푘 + 1
8. 푥 2 + 6푥 − 9
9. 1 − 10푧 + 25푧2
10. 50푥 240푥 + 8

24. 24
J. Solve and simplify each polynomials
1.
푎+1
푎3 − 푎 +2
푎2 + 푎 +3
푎
2.
5푥3
7푦4 . 21푦 2
10푥2
3.
9푥5
36푥2
4.
5−푎
푎2− 25
5.
10푎2 −29푎 +10
6푎2 −29푎 +10
÷ 10푎2 −19푎 +6
12푎 2 −28푎 +15
6.
1
푥+ℎ
−
1
푥
ℎ
7.
푥6−7푥3 − 8
4푥2−4푥−8 ÷ (2푥 2 + 4푥 + 8)
8.
푎
푏
−
푏
푎
푎
푏
+
푏
푎
9.
푡2 −2푡 −15
푡2 −9
. 푡2− 6푡 +9
12−4푡
10.
푎−1+푏−1
푎−2−푏−2

25. Chapter II
Equations and
Inequalities
25

26. 26
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=-7
r+16+(-16)=-7+(-16) add -16 to each side
r=-23 the sum of -16 and 16 is 0
check:
r+16=-7
-23+16=-7
-7=-7
2) x+(3.28)=-17.56
x+(3.28)=-17.56
x+(3.28)+(3.28)=-17.56+3.28
x=-14.28
check: x+(3.28)=-17.56
-14.28+(-3.28)=-17.56
-17.56=-17.56
3) y+21=-7
y+21+(-21)=-7+(-21)
y=-28
check: y+21=-7
-28+21=-7
-7=-7
Equations by Using Subtraction
The 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=-6
x+15-15=-6-15
x=-21
check:
x+15=-6
-21+15=-6
-6=-6

27. 27
2) b-(-8)=23
b+8=23
b+8-8=23-8
b=15
check:
b-(-8)=23
15-(-8)=23
23=23
Equations by Using Multiplication and Division
To 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=bc
Eamples:
1)
5
12
= 푟
24
24( 5
12
) = ( 푟
24
)24 multiply each side by 24
10 = 푟
Check:
5
= 푟
12
24
replace r with 10
5
12
=
10
24
5
12
=
5
12
2) 24=-2a
24=-2a
−
1
2
(24) = −
1
2
(2푎)
−12=a
푐ℎ푒푐푘:
24 =−2a
24 =−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
푎
푐
= 푏
푐
.
Examples:
1. -6x=11
−6푥
−6
= 11
−6
divide each side by -6
푥 = −
11
6

28. 28
Check: -6x=11
−6(− 11
6
)=11
11 = 11
2. 4x=24
4푥
4
=
24
4
X=6
Check: 4x=24
4(6)=24
24=24
2.2 Appplication of Linear Equations
In 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
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.

29. 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 1
If 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 rectangle
4w-3: the number of centimeters in the length of the rectangle
(4w-3)cm
w cm w cm
(4w-3)cm
w+(4w-3)+ w+(4w-3)=19
10w-6=19
10w=25
29
w=
5
2
4w-3=4(5
)-3
2
=7
Example 2
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?
Solution
x: the number of dollars invested at 12 percent
15,000-x: the number of dollars invested at 8 percent

30. Number of Dollar × Rate = Number of Dollars
invested in Interest
12 percent investment x 0.12 0.12x
8 percent investment 15,000-x 0.08 0.08(15,000-x)
0.12x+0.08(15,000-x)=1456
30
0.12x+1200-0.08x=1456
0.04x=256
x=6400 15,000-x=15,000-6400
=8600
Thus 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 rate
r-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
Daughter 28 r
28
푟
Father 24 r-12
24
푟−12
Equation:
28
=
푟
24
푟 − 12
Solve the equation by first multiplying on both sides by the LCD:
푟(푟 − 12)
28
푟
= 푟(푟 − 12)
24
푟 − 12
(r-12)28=r(24)
(r-12)7=r(6)
7r-84=6r
7r-6r=84
r=84 r-12=84-12
=72
Therefore, the daughter’s average rate is 84km/hr and the father’s average rate is
72km/hr.

31. 2.3 Quadratic Equation in One Variable
ax2 + bx + c =0, a≠0
where 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
31
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=0
This 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 = 0
We 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 if
X+5 = 0 or x-2 = 0
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 /3
6+ 5x =4x2
-4x2 +5x +6 = 0
4x 2-5x -6 = 0
( 4x+3 ) (x-2 ) = 0
4x=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 form
X 2 = d
That is, there is no first degree term. Then an equivalent equation is
X2 –d = 0

32. 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 = 0
x = √d x = -√d
therefore 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
32
Example 2:
a. X2 = 25 b. x2 =13 c. x2 =16
X = ±√ 25 x =√13 x = ± √-16
X = ± √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
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
푥 = −푏±√푏2−4푎푐
2푎
Example 3: Use quadratic formula to find the solution set of the equation
6x2= 10 + 11x
×=
−푏 ± √푏2 − 4푎푐
2푎
=
−(−11) ± √(−11)2 − 4(6)(−10)
2(6)
=
11 ± √121 + 240
12

33. 33
=
11 ± √361
12
=
11 ± 19
12
=
11 + 19
12
=
11 − 19
12
= 30
12
= −8
12
= 5
2
= − 2
3
The solution set is {− 2
3
, 5
2
}.
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 order
If 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
An example of a quadratic inequality having domain R is
X2 + 5x > 13
The inequality
4x/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 numbers
a. Addition Property
If a < b, then a + c < b + c
b. Subtraction Property

34. 34
If a < b, then a – c < b – c
c. Multiplication Property
If a < b and c > 0, then ac < bc
If a < b and c < 0, then ac > bc
Example 1
Find and show on the real number line the solution set of the inequality.
3x – 8 < 7
3x – 8 + 8 < 7 + 8
3x < 15
1/3 (3x) < 1/3 (15)
X < 5
Example 2
Find and show on the real number line the solution set of the inequality
X – 7/4 ≤ x
(4)x – 7/4 ≤ (4)x
X – 7 ≤ 4x
X – 7 – 4x + 7 ≤ 4x – 4x + 7
-3x ≤ 7
-1/3 (-3x) ≥ (-1/3)7
X ≥ -7/3
2.5 POLYNOMIAL AND RATIONAL INEQUALITIES
A quadratic and inequalities is one of the form ax2 + bx + c < 0
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 < 2x
We 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) < 0
Example 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) ≥ 0
Figure 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 , +∞)

35. 35
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
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)

36. 36
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 / <1/3
The 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
= 9 = 9

37. ASSESSMENT
37
TEST I
A. 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 – 2
4. (x – 4) ( x + 2) = 7
5. 2p2 – 4p – 5 = 0
6. √2x + 5 + x = 5
7. √y + 2 + √y + 5 - √8 – y = 0
8. / r + 1 / = 6
r – 1
9. / 8x - 2 / = x + 7
10. / 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 ≤ 20
12. x2 < 64
13. 2x + 1 > 1
x – 5
14. 3x – 4 ≤ 12
2x – 3
15. 2x2 – 3 x ≥ 5
16. 5x + 8 ≥ 2x – 2
17. -7 ≤ 8 – 4x ≤ 10
-5
18. / x – 10 / ≤ 17
19. 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.

38. 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?
38
TEST II
A. Find the solution set of the equation.
1. X2 = 64
2. 6t2 – 11 = 0
3. 9x2 = x
4. 3t/3t + 4 + 2/5 – 1/3t – 4
5. 64y2 – 80y + 25 = 0
B. Find the solution set of the equation by completing the square.
6. X2 + 6x + 8 = 0
7. 4x2 + 4x – 3 = 0
8. 3y2 + 4y + 2 = 0
C. Find the solution set of the equation by using the quadratic form.
9. X2 + 1 = 6x
10. 5y2 – 4y – 2 = 0
D.Find the solution of the inequality and write it with interval notation.
1. 2x – 1 < 6
2. -3 > 4x + ¾
3. 11 ≥ 5x – 4 > 1
4. 2x – 4 < 6
5. 1 < 4x – 1/3 < 5
6. 5x – 2 < 5x – 2/4
7. 10 – 3x > 4x – 5/-3
8. 5 ≤ 3x – 4 < 14
9. -1 < 7 – 2x/5 ≤ 5
10. 6 ≤ 2 – x ≤ 8
E.Find the solution set of the equation.
11. / x – 8 / = 9
12. / 4y – 10 / = 5

39. 39
13. / 3t -2 / = t2
14. / x – 1 / = x2
15. / 5t – 15 / = 20
CHAPTER III
Points and
Equations

40. 40
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 Theorem
In 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
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 2
Solution
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 Equations
The 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 = 0
Solution

41. y2 = 4x
y = ± 2√x
y = 2 √x and y = -2 √x
41
Symmetry 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
ℎ =
−2 + 4
2
= 1
푘 = 3+5
2
=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

42. = 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.
42
Example 1
Determine the slope of a line.
A (3,7) and B (-2,-4)
m = -4 – 7 /-2 -3
= -11 / -5
= 11 / 5
A (-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

43. ASSESSMENT
TEST I
A.Find the center and radius of the circle.
43
1. x2 + y2 + 4x – 6y -3 = 0
2. 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 = 0
12. 2x + 3y + 12 = 0
13. 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)
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 = 0
2. x2 + y2 – 10x – 10y + 25 = 0
3. 3x2 + 3y2 + 4y – 7 = 0
4. x2 + y2 - 6x - 8y + 9 = 0

44. 5. 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)
44
(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-axis
G. 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

45. Chapter IV
Functions and their
Plane
45

46. 46
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 R2 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 + c
If the function f is the defined by
F ( x ) = -2 x2 + 8x – 5
The graph of f is the same as the graph of an equation
Y = 2x2 + 8x – 5
This equation is equivalent to
2(x2 – 4x) = -y – 5
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

47. 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
47
f (x) = ax2 + bx + c
-1 = ax2 + bx + c
ax2 + bx = y – c
a (x2 +b/a x) = y – c
a (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 value
Example use theorem 1 to find either a maximum or minimum value of the function g if
g (x) = -5/2 x2 + 8x -10
For 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
= -18/5
Rational Functions
A rational function is of the form 푓(푥) = 푝(푥)
, where p(x) and q(x)are polynomial
푞(푥)
functions and 푞(푥) ≠ 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

48. 48
Connection:
Mathematical History
Mathematician 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 푥 2푦 = 푎2 (푎 − 푦), called the “curve of
Agnesi”. The equation can be expressed as 푦 = 푎2
푥2+푎2.
Because the function described above is the ratio of two polynomial expression a3 and
푥 2 + 푎2 is called a rational function. A rational function is function of the form 푓 (푥) =
푝(푥)
, where p(x) and q(x) are polynomial functions and q(x)≠ 0
푞(푥)
Examples of Rational Function:
풇(풙) = 푥
푥− 1
푔(푥) = 3
푥−3
ℎ(푥) = 푥 +1
(푥+2) (푥−5)
The lines that graph of the rational function approaches is called
Asymptote. If the function is not define when 푥 = 푎, then either there is
a “hole” in the graph 푥 = 푎.
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.

49. 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
12푥푦2
5푥푦 + 3푧2 푣
49
4푥 2푦 − 3푥푦2 − 푥푦
3푥 − 5푦 + 푎 + 10푏
5푥 4 + 2푥 3 − 푥 2 − 6푥 + 8
1
2
3
4
5
Monomial
Binomial
Trinomial
Multinomial
Multinomial
Example:
Determine whether or not each expression is a polynomial. Then state the degree of each
polynomial.
a.
2
7
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 + √푥 − 3
This expression is not polynomials because √푥 is not a monomial.
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

50. 50
Example:
Find (k2 +3k +9) (k +3)
(k2 +3k +9) (k +3)
= 푘2 (k+3) + 3k (k +3) +9(k +3) distributive property
=k2∙k+k2∙ 3 + 3푘 ∙ +9 ∙ 푘 + 3 ∙ 9 distributive property
= 푘2∙ 푘 + 푘2∙ 3푘2+9푘 + 9푘 + 27
=k2+6푘2+18푘 + 27 combined like terms
Dividing Polynomials
You 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: c2 –c –30
c –6
c
푐 − 6√푐2 − 푐 − 30
푐2− 6
5푐 −30
−푐 − (−6푐) = −푐 + 6푐 표푟 5푐 푐 + 5
푐 − 6√푐2 − 푐 − 30
푐2−6
5푐 −30
5푐 −30
0
Polynomial 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
푓(푥) = 푎푛푥 푛 + 푎푛−1푥 푛− 1 + ⋯ + 푎2푥 2 + 푎1푥 + 푎0
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
푓(푥) = 푥 3
is a polynomial function of one variable. Polynomial functions of multiple variables can
also be defined, using polynomials in multiple indeterminates, as in
푓(푥, 푦) = 2푥 3 + 4푥 2푦 + 푥푦5 + 푦2 − 7
An example is also the function 푓(푥) = cos(2푎푟푐표푠(푥)) 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 푓(푥) = 2푥 2 −1
Polynomial functions are a class of functions having many important properties. They are
all continuous, smooth, entire, computable, etc

51. Graphs of Polynomial Function
A polynomial function in one real variable can be represented by a graph.
51
 The graph of the zero polynomial
f(x) = 0
is the x-axis.
 The graph of a degree 0 polynomial
f(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 polynomial
f(x) = a0 + a1x + a2x2, where a2 ≠ 0
is a parabola.
 The graph of a degree 3 polynomial
f(x) = a0 + a1x + a2x2, + a3x3, where a3 ≠ 0
is a cubic curve.
 The graph of any polynomial with degree 2 or greater
f(x) = ao + a1x + a2x2 + ... + anxn , where an ≠ 0 and n ≥ 2
is 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.
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)

52. 52
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
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)

53. 53
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,
(푓 ∘ 푔) = 푥 and (푔 ∘ 푓)(푥) = 푥
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.

54. 54
Property of Inverse
Function
Suppose 푓 and푓−1 are inverse function. Then 푓(푎) = 푏 and
only if 푓−1(푏) = 푎
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 )

55. ASSESSMENT
TEST I.
A. Draw a sketch of the graph of the function and determine its domain and range.
55
1. f = {(x,y) / y = 3x – 1 }
2. F = { ( x, y / y = 2x2 }
3. F = { ( x, y) / y = / 3x + 2 / }
4. G = { ( x, y} = x2 – 4 }
x-2
5. g = { ( x, y /y = (x2 – 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 – 3
2. f(x) = 2x2 – 2x – 1
3. f(x) = x2 – 3x + 1
4. 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.

56. Chapter V
Exponential and
Logarithmic
Function
56
c

57. 57
EXPONENTIAL FUNCTION
An equation of the form 푦 = 푎 ∙ 푏푥 , 푤ℎ푒푟푒 푎 ≠ 0, 푏 > 0, and 푏 ≠ 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 푏 > 0 and 푏 ≠ 1. for 푛 > 0, there is a number p such
that 퐿표푔푏 푛 = 푝 if and only if 푏푝 = 푛.
The chart below shows some equivalent exponential and logarithmic equations.
Exponential Equation Logarithmic Equation
52 = 25
105 = 100,000
80 = 1
2−4 =
1
16
9
1
2 = 3
log5 25 = 2
log10 100,000 = 5
log8 1 = 0
log2
1
16
= −4
log9 3 =
1
2
Integral Exponents
Basic Laws of Exponents
1. 풃풙 ∙ 풃풚 = 풃풙+풚
2.
풃풙
풃풚 = 풃풙−풚 (풃 ≠ ퟎ)
3. 풊풇 풃 ≠ ퟎ, ퟏ, −ퟏ 풕풉풆풏 풃풙 = 풃풚 풂풏풅 풐풏풍풚 풊풇 풙 = 풚
4. (풂풃)풙 퐜퐱 퐛퐲
5. (풂
풃
)
풙
= 풂풙
풃풙 (풃 ≠ ퟎ)
6. 풊풇 풙 ≠ ퟎ, 풂 > ퟎ, 풃 > ퟎ, 풕풉풆풏 풂풙 = 풃풙 풊풇 풂풏풅 풐풏풍풚 풊풇 풂 = 풃
7. (풃풙)풚 = 푩풙풚
8. 풃ퟎ = ퟏ (풃 ≠ ퟎ)
9. 풃−풙 = ퟏ
풃풙
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

58. 58
then raises the power.
c) = 7/8 d) = 14-3 = 1/143 = 1/2744
e) (3−2 + 3−3)−1 =
= ( 1
9
+ 1
27
)
−1
= ( 3
27
+ 1
27
)
−1
= ( 4
27
)
1
= 27
4
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) (푎−2 − 푏−2)−1
= ( 1
푎 −2 − 1
푏2)
−1
= ( 푏2
푎2 푏2 − 푎2
푎2푏2)
−1
= (푏2−푎2
푎2 푏2 )
−1
= 푎2 푏2
(푏−푎)(푏+ 푎)
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)
(3푎−1)
2
(3푎−1)−2
= (3푎−1 )4
= 81푎−4 = 81
푎4
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 = 0
r is the rate (as a decimal)
t is the time
A(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.

59. 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?
59
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.
1).
4−4
4−2 + 4−3
= 4−4
4−2+4−3 ∙ 44
44
= 1
42 + 41
= 1
16 +4
= 1
20
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) 푏
1
푛 = √푏 푛
2)푏
푚
푛 = ( 푛√푏) = 푛√푏푚
Example:
1) 8
1
3
=3√8=2
2) 2) 8−
1
3 = 1
= 1
3√8 2
3
= 23 + 8
3) 163/4 = ( √16 4 )
3
4 = 1
4) 16−
(4√16 )
3 = 1
23 = 1
8
5) ( 8
27
)
1
3 = √ 8
27
3
= 2
3

60. 60
6) ( 8
27
)
−
2
3 = (27
8
)
2
3 = ( √27
8
3
2
= (3
)
2
)
2
= 9
4
7) (1001/2 - 361/2)2 = (10 - 6)2 = 42 = 16
8) x1/2(x3/2 + 2x1/2) = x2 + 2x
9)
( √4푎2) 3 2
6√4푎2
=
(4푎2 )
2훽
(4푎2)
1
6
1
6 = (4푎2)
= (4푎2)2훽−
1
2 = 2푎
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:
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

61. Exponential Functions
Any 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 8
2 4
1 2
0 1
-1 1/2
-2 1/4
-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.
61
Now let's look at the graph of
f(x) = (1/2)x
X f(x)
3 1/8
2 1/4
1 1/2
0 1
-1 2
-2 4
-3 8

62. 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.
62

63. 63
Do you see the similarities of each graph?
How about this one?
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 72
If 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!!

64. 2) Suppose that t hours from now the population of a bacteria colony is given by: P(t) =
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 =
64
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

65. 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
the year. Study the following table of values and use your calculator to double check the
results:
65
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
퐴(푡) = 퐴0 (1 + 푟
푛
)
푥푡
Where:
A(t) = amount after time t.
Ao = Initial amount
r = rate in decimal
n = number of times compounded in a year.
t = time in years
Thus, if the interest was paid semiannually, n = 2. Paid quarterly would make n = 4,
Paid monthly, n = 12, etc.

66. Sample Problems
1) 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
66
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)
Solutions:
a) The initial population is 5000 from the formula.

67. 67
b) A(7) = 5000e.14 = 5751
c) A(30) = 5000e.6 = 9111
Logarithmic Functions
Common Logarithm
Demo: Log Funtion Applet
log x = a if and only if 10a = x
The 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 = .1
5) log .01 = -2 because 10-2 = .01
The 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 = 8
2) log3 81 = 4 because 34 = 81.
3) log4 1/16 = -2 because 4-2 = 1/16
4) log8 1 = 0 because 80 = 1

68. 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 = x
Most natural logs need to be calculated on your calculator. The graph of the natural log
is shown below:
68
Solving Simple Log Equations
1) 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
= 100
x = 100 or -100
3) 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 -9
4) Log x = .3
Again, rewrite exponentially.
x = 10.3 Use your calculator and round to hundredths.
x = 2.00

69. 69
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) 퐿표푔8√푥푦 = 퐿표푔8(푥푦)
1
2 = 1
2
(퐿표푔8푥 + 퐿표푔8푦)
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: 퐿푛 푥1/2
푦1/3 = 퐿푛 √푥
√푦 3
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
Since this doesn't make the number inside the log zero or negative, the answer is
acceptable.

70. 70
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
Solution: Rule #8 again. Answer: 6

71. 71
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.
Examples
1) 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 base
Now, 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.
x = Log 17/Log 2
x = 4.09

72. 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.
72
Sample Problems
1) 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.

73. ASSESSMENT
73
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
10) Simplify each log:

74. 74
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?

75. Chapter VI
System of Equations
And Inequalities
75
.

76. 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.
2푥 + 푦 = 3
5푥 + 3푦 = 10
76
ax + by = c
We 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.
{
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
6푥 − 3푦 = 5
2푥 − 푦 = 4
{

77. 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
2푥 + 푦 = 3
5푥 + 3푦 = 10
푦 = 3 − 2푥
5푥 + 3푦 = 10
77
Example 1
Use the substitution method to find the solution set of the system.
Illustration 1: {
Solution:
We solve the first equation for y and get the equivalent system
{
We replace y in the second equation by its equal, 3 – 2 x, from the first equation. We then
have the equivalent system
{
푦 = 3 − 2푥
5푥 + 3(3 − 2푥 ) = 10
Simplifying the second equation, we have
{
푦 = 3 − 2푥
−푥 + 9 = 10
Solving the second equation for x, we get
{
푦 = 3 − 2푥
푥 = −1
Finally, we substitute the value of x from the second equation into the first equation and
we have
{
푦 = 5
푥 = −1
This system is equivalent to the given one. Hence the solution set is ( -1 , 5)

78. 78
Example 2
Use the elimination method to find the solution set of the system of equations in Example
1.
2푥 + 푦 = 3
5푥 + 3푦 = 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 − 3푦 = −9
5푥 + 3푦 = 10
Adding the equations given the following computations:
−6−3푦 = −9
5푥 +3푦 =10
−푥 =1
With this equation and the first equations in the given system, we can write the following
equivalent system
{
2푥 + 푦 = 3
−푥 = 1
If we now multiply both sides of the second equation by -1, we have the equivalent
system
{
2푥 + 푦 = 3
푥 = −1
We next substitute -1 for x in the first equation to obtain
{
2(−1) + 푦 = 3
푥 = −1
{
푦 = 5
푥 = −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
2푥 − 푦 + 4푧 = 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

79. 79
Illustration 1
For the equation
2푥 − 푦 + 4푧 = 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.
{
푎1푥 + 푏1푦 + 푐1푧 = 푑1
푎2푥 + 푏2푦 + 푐2푧 = 푑2
푎3푥 + 푏3푦 + 푐3푧 = 푑3
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
푥 − 푦 − 4푧 = 3
2푥 − 3푦 + 2푧 = 0
2푥 − 푦 + 2푧 = 2
{
Solution, we solve the first set of the system
푥 = 푦 + 4푧 + 3
2푥 − 3푦 + 2푧 = 0
2푥 − 푦 + 2푧 = 2
{

80. We now substitute the value of x from the first equation into the other two equations , and
we obtain the equivalent system
푥 = 푦 + 4푧 + 3
−푦 + 10푧 = −6
푦 + 10푧 = −4
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
푥 = 푦 + 4푧 + 3
80
{
푥 = 푦 + 4푧 + 3
2(푦 + 4푧 + 3) − 3푦 + 2푧 = 0
2(푦 + 4푧 + 3) − 푦 + 2푧 = 2
{
We next solve the second equation for y and get
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
푦 + 10푧 = −4
Substituting the value of y from the second equation into the third gives the equivalent
system.
{
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
(10푧 + 6 +) + 10 = −4
푥 = 푦 + 4푧 + 3
푦 = 10푧 + 6
20푧 = −10
{
{
푧 = −1/2
Substituting the value of z from the third equation into the second equation, we obtain
{
푦 = 1
푧 = −1/2
Substituting the values of y and z from the second and third equations into the first
equation, we get
푥 = 2
푦 = 1
푧 = 1/2
{

81. 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
81
2 − 1 + 2 = 3
4 − 3 − 1 = 0
4 − 1 − 1 = 2
{
The equations of the given system are consistent and independent.
Exercise
1.) {
4푥 + 3푦 + 푧 = 15
푥 − 푦 − 2푧 = 2
2푥 − 2푦 + 푧 = 4
2.) {
2푥 + 3푦 + 푧 = 8
5푥 + 2푦 + 3푧 = −13
푥 − 2푦 + 5푧 = 15
3.) {
푥 − 푦 + 3푧 = 2
2푥 + 2푦 − 푧 = 5
5푥 + 2푧 = 7
3푥 + 2푦 − 푧 = 4
3푥 + 푦 + 3푧 = −2
6푥 − 3푦 − 2푧 = −6
4.) {
5.) {
2푥 − 3푦 − 5푧 = 4
푥 + 7푦 + 6푧 = −7
7푥 + 2푦 − 9푧 = 6
3푥 − 2푦 + 4푧 = 4
7푥 − 5푦 − 푧 = 9
푥 + 9푦 − 9푧 = 1
6.) {
7.) {
3푥 − 5푦 + 2푧 = −2
2푥 + 3푧 = −3
4푦 − 3푧 = 8
푥 − 푦 = 2
3푦 + 푧 = 1
푥 − 2푧 = 7
8.) {
9.) {
3푥 − 2푦 = 1
푧 − 푦 = 5
푧 − 2푥 = 5
푥 − 푦 + 5푧 = 2
4푥 − 3푦 + 5푧 = 3
3푥 − 2푦 + 4푧 = 1
10.) {
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.
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.

82. 푦2 = 4(3 − 푦)
푥 = 3 − 푦
푦2 + 4푦 − 12 = 0
푦 = −6
푥 = 3 − 푦
82
Example 1
Find the solution set of the system.
푦2 = 4푥
푥 + 푦 = 3
{
Solution
We solve the second equation for x and obtain the equivalent system.
푦2 = 4푥
푥 = 3 − 푦
{
Replacing x in the first equation by its equal from the second, we have the equivalent
system
{
{
푥 = 3 − 푦
We now solve the first equation.
(푦 − 2)(푦 + 6) = 0
푦 − 2 = 0 푦 + 6 = 0
y= 2 푦 = −6
Because the first equation of system (II) is equivalent to the equations 푦 = 2 and 푦 =
−6, system (II) is equivalent to the systems
푦 = 2
푥 = 3 − 푦
{
and {
In each of the latter two systems we have substitute into the second equation the value of
y from the first , and we have
{
푦 = 2
푥 = 1
and {
푦 = −6
푥 = 9
These two systems are equivalent to system (I). Thus, the solution set of (I) is (1,2) , (9,-
6).

83. 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.
83
Statement of the form
퐴푥 + 퐵푦 + 퐶 > 0 퐴푥 + 퐵푦 + 퐶 < 0
퐴푥 + 퐵푦 + 퐶 ≥ 0 퐴푥 + 퐵푦 + 퐶 ≤ 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.
퐴푥 + 퐵푦 + 퐶 > 0 and 퐴푥 + 퐵푦 + 퐶 < 0
Are half planes. We shall show this for the particular inequalities
2푥 − 푦 − 4 > 0 푎푛푑 퐴푥 + 퐵푦 + 퐶 < 0
Let L be the line having the equation 2푥 − 푦 − 4 = 0. If we solve this equation for y, we
obtain 푦 = 2푥 − 4. If (x, y) is any point in the plane, exactly one of the following
statements holds:
푦 = 2푥 − 4 푦 > 2푥 − 4 푦 <
2푥 − 4
Now, 푦 > 2푥 − 4 if and only if the point (x, y) is any point (x, 2x - 4) on line L;
Furthermore, 푦 < 2푥 − 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 푦 > 2푥 − 4, and the other region is the half plane
above L, which is the graph of the inequalities 푦 > 2푥 − 4, and the region is the half
plane below L, which is the graph of inequality 푦 < 2푥 − 4. A similar discussion holds
for any line L having an equation of the form 퐴푥 + 퐵푦 + 퐶 = 0 푤ℎ푒푟푒 퐵 ≠ 0.

84. If B= 0, an equation of line L is 퐴푥 + 퐶 = 0, and L is a vertical line whose
equation 푥 = 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 +퐶 = 0.
By generalizing the above arguments to any line, we can prove this theorem.
84
THEOREM
(I) the graph of y> 푚푥 + 푏 is the half plane lying above the line y= 푚푥 + 푏.
(II) the graph of y< 푚푥 + 푏 is the half plane lying below the line y= 푚푥 + 푏.
(III) the graph of (y< 푚푥 + 푏) x> 푎 is the half plane lying to the right of line x= 푎.
(IV) The graph of x< 푎 is the half plane lying to the left of the line x= 푎.
Example 1
Draw a sketch of the graph of the inequality
2푥 − 4푦 + 5 > 0
Solution
The given inequality is equivalent to
−4푦 > −2푥 − 5
푦 > 1
2
+ 5/4
The graph of inequality is the half plane below the line having the equation 푦 = 1/2푥 +
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.
퐴푥 + 퐵푦 + 퐶 = 0 표푟 퐴푥 + 퐵푦 + 퐶 ≤ 0

85. 85
Illustration
The inequality
4푥 + 5푦 − 20 ≥ 0
Is equivalent to
5푦 ≥ −4푥 + 20
푦 ≥ −4/5푥 + 4
Therefore the graph of this inequality is the closed half plane consisting of the line 푦 =
−3/5푥 + 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.