FOUR BASICARITHMETIC PROCESSES• ADDITION is the processes of combining numbers of

the same kind. Measures like dollars and miles cannot be added because they are two different things.

• SUBTRACTION is the process of finding the difference between two like numbers by taking one (the subtrahend) from the other (the minuend). The answer is the remainder.

• MULTIPLICATION is the process of adding one number (the multiplicand) to itself as many times as there are units in another number (the multiplier). The answer is called the product. The multiplicand and multiplier are also called the factors of the product.

Arithmetic processes (cont…)• DIVISION is the process of finding how many times

one number (the divisor) is contained in another number (the dividend). The result is the quotient.

• Examples7 – 3 = 4 5 * 3 = 15

Minuend

Subtrahend

Remainder

Multiplicand

Multiplier

Product

4

024246

DivisorDividend

Quotient

Remainder

Common Fraction• A FRACTION is a portion of the whole; it signifies

division. The top number is called the numerator (same as the dividend), and the bottom number is called the denominator (same as the divisor). The line between the numbers is the division sign.

• A PROPER FRACTION has a numerator that is smaller than the denominator, as in ¾ or 2/7.

• An IMPROPER FRACTION has a numerator that is equal to or greater than the denominator, as in 6/6 or 9/4.

65 Numerator (or dividend)

Denominator (or divisor)

For convenience, this isoften written as 65

Common Fraction (cont…)• An improper fraction can be converted to a whole number

(6/6 =1) or a mixed number ( ) by dividing the denominator into the numerator. A mixed number contains both a whole number and a fraction.

• A complex fraction has one or more fractions in either the numerator or the denominator, or both. A complex fraction may be converted to a simple fraction by dividing the numerator by the denominator (remember the line in a fraction is a division sign). To divide by a fraction invert the denominator fraction (divisor) and multiply:

61

67 1

Common Fraction (Cont…)

NOTE: The answer to any example or problem should not be left as an improper fraction. It should be converted to a proper fraction or a mixed number, then reduce to lowest terms.

1514

1521

32

2115

32

211532

2117

235

257

527

527

1123

141

8314

83

1483

7

1

ADDITION OF FRACTIONS• To add fractions with the same denominators, add the

numerators.

• To add fractions with different denominators, select a common denominator into which all the denominators will divide evenly. The smallest number into which all denominators divide evenly is called the least (or lowest) common denominator (L.C.D) or least common multiple (L.C.M).

53

512

51

52

ADDITION OF FRACTIONS (cont…)• To find the L.C.M, spread out all the denominators in a short-

division box. Divide the denominators by a prime number (a number divisible only by itself and 1) or a multiple of a prime number that goes into at least two of the denominators. When a denominator cannot be divided by a divisor (prime number), it is brought down into the quotient. Repeat the process until no two parts of the quotient are divisible by the same prime number. Then multiply all divisors and all parts fo the quotient together. The result is the L.C.M.

• To find the L.C.M. of 207&

53,

85

4011245112418420585

40231

4063

40142425

4027835540

2040754038405207

53

85

L.C.M

PROJECT 1 FRACTIONSChange to the improper fractions.

75140

151320

3216

2916

8326

2717

1619

4381. 2. 3.

4. 5. 6.

7. 8.

PROJECT 1 (Cont…)Add the fractions and reduce the answers to the lowest terms.

3236,

2114,

5223

325,

658,

876

16728,

3291,

8364,

6537

214,

10719,

5226,

438

6112,

835,

327

413,

1654,

836

53,

107,

32,

65

95,

2411,

87,

61

1. 2.

3. 4.

5.

6.

7. 8.

SUBTRACTION OF FRACTIONS• To subtract fractions that have the same denominator,

simply subtract the numerators.

• To subtract fractions with different denominators, select the least common denominator, and proceed with the operation.

1711

17415

174

1715

256

5012

50152750

5312750

)1050(15)5050(27103

5027

25

6

MULTIPLICATION OF FRACTIONS• There are three steps in multiplying fractions: (a) Multiply the

numerators. (b) Multiply the denominators. (c) Reduce to the lowest terms.

• “Cancelling” saves time and work. In the above example:

• When multiplying by a whole number, think of the whole number as a fraction with a denominator of 1.

3 5 3 5 15 57 9 7 9 63 21

5

21

215

3751

35

71

95

73

1

3

123443

116

4316

4

1

DIVISION OF FRACTIONS• To divide fractions, invert the divisor, as explained earlier,

and multiply. An inverted fraction is a fraction with the numerator and denominator interchanged.

NOTE:When addition, subtraction, multiplication and division involves a mixed number, convert all mixed numbers to improper fractions. Then proceed with the operations.

65

3215

34

85

43

85

12

PROJECT 2 FRACTIONS• Perform the indicated operations. Reduce answers to the

lowest terms where necessary.

2127

5425

28516

7417

1636

217

4318

2156

652

836

655

43

181

513

3014

2013

2515

6548

3216

433

73

542

72

1613

4327

7341

323

878

617

4313

1. 2. 3.

4. 5. 6.

7. 8.

9. 10. 11.

12. 13.

Project 3 Fractions• Compute the total hours worked by each employee and on each day, and verify the totals.

434

417

216

439

41710

215

419

217

438

438

415

2110

416

217

217

418

6219

219

219

415

418

217

4325

215

416

219

433

214

2128

437

419

218

4138

415

2177

736

Mon. Tues. Wed. Thurs. Fri. Sat. Total.

Total.

Ali

Saeed

Aina

Sofia

Jawad

Iqrar

Azhar

Jamila

Sana

THE REAL NUMBER SYSTEM• A set is a collection of elements listed within braces. The

set {a, b, c, d, e} consists of five elements, namely a, b, c, d, and e. a set that contains no element is called an empty set (or null set). The symbol { } or the symbol Ø is used to represent the empty set.

• There are many different sets of real numbers, two important sets are the natural numbers and the whole numbers.

Natural numbers: {1, 2, 3, 4, 5, …. }Whole numbers: {0, 1, 2, 3, 4, 5, … }

• To understand the sets of numbers we introduce the real number line.

-4 -3 -2 -1 0 1 2 3 4

• The real number line continues indefinitely in both directions.

SETS OF REAL NUMBERS (cont…)• Another important set of numbers is the integers.

Integers: {… , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, … }

Negative integers Positive integers• The set of rational numbers consists of all numbers that

can be expressed as the quotient of two integers, with denominator not zero.

• Some numbers are not rational. Numbers such as square root of 2, written as , are not rational numbers. Any number that can be represented on the real number line that is not rational is called irrational number.

2

REAL NUMBERS• Any numbers that can be represented on the real number

line is real number.Real numbers

Rational numbers Irrational numbers

Integers Non-Integers

Negative 0 Positive numbers numbers

PROJECT 4 NUMBER SYSTEM1. List the elements of the set that are:a) Natural numbers b) Whole numbersc) Integers d) Rational numberse) Irrational numbers f) Real numbers

2. List the numbers that area) Positive integers b) Whole numbersc) Negative Integers d) Integerse) Rational numbers f) Irrational numbersg) Real numbers

,5,7,9.2,

74,9,0,3,96,

214,5.0,6

219,300,5,67.1,

125,

412,2,2,

21

ALGEBRA• Some Basic Definitions• Variable

A characteristic which changes from one individual to the other, e.g. the height of a student in your class, the temperature of different cities in Afghanistan, etc.

Variable is denoted by the lower case letters, e.g. x, y, z, etc.

Constant

A characteristic which does not change, e.g. the dimension of your class room, the height of the chair you are setting on, the number of papers in BBA 1st semester exam, etc.

A constant is denoted by the alphabets like a, b, c, d, etc.

Expression It is the combination of operands and operators,

e.g. x + y, 4 - 17, etc. Here x, y, 4 and 17 are called operands, while

the symbols +, - are called operators. The other operators are *and ÷.

The expression x + y is called algebraic expression and 4 - 17 is called the arithmetical expression.

Equation

An expression which involves the sign of equality is called equation.

Examples2x - 5y = 12

x2 + 3x - 5 = 0 x2 + y2 = r2(Equation of circle centered at the origin having radius r) 1/x + 7 = x/3

Operations on expressions• AdditionAdd x + 2y and 3x – 7y

x + 2y3x - 7y

____________ 4x – 5y

Continue….• Subtraction Subtract 9x – 2.5y from 4x + 5y

4x + 5y +9x – 2.5y

-5x + 7.5y In addition and subtraction we combine the similar terms of

the expressions.

+__________________

Continue….

Example:Add 7x + 3y + z, -3x + y, and 5x – 4z 7x + 3y + z

-3x + y + 0z5x + 0y – 4z

_____________9x + 4y – 3z

Distributive property• For any real numbers a, b, c

a (b + c) = ab + acExamples1. 2 (3 + 7) = 2 * 3 + 2 * 7 = 202. 4 (2x - 4y) = 4 * 2x – 4 * 4y = 8x - 16y3. -7 (3p - 5q) = -7(3p)-7(-5q)=-21p+35q4. x (2.3y + 1.2z) = 2.3xy + 1.2xz5. 3 (x + y + z) = 3x + 3y + 3z

Group Activity

Simplify1. 2x2 – 4x +8x2 – 3 (x + 2) – x2 - 2Sol: 2x2 -4x + 8x2 - 3x – 6 - x2 - 2 =(2x2 + 8x2 – x2) + (-4x - 3x) +(-6 - 2) = 9x2 - 7x - 82. x2 + 2y - y2 + 3x + 5x2 + 6y2 + 5y3. 2 [ 3 + 4 (x - 5) ] - [ 2 - (x - 3) ]

Linear Equation• An equation of the form

ax + b = c (1) where a, b and c, a≠0 are constants (real numbers) is called a linear

equation.

Examples1. x + 4 = 7 (a = 1, b = 4, c = 7)2. 2x – 3 = 5.4 (a = 2, b = -3, c = 5.4)3. -(x + 3) - (x - 6) = 3x - 4.5

Though this equation is not of the form (1), but it is still a linear equation, since it can be rewritten as:

5x - 4.5 = 3 (a = 5, b = -4.5, c = 3)

Solution of a linear equation

By the solution of a linear equation we mean to find that value of the unknown ‘x’ which satisfies the equation.

Ex1. x = 3, is the solution of x – 2 = 12. x = -6, is the solution of 2x + 1 = x - 53. x = 11/2 is the solution of 3 (x + 2) = 5 (x-1)

4. y = 9 is not a solution of 2y – 3 = 5

How to solve a linear equationWe can solve a linear equation by1. Substitution method2. Graphical method3. Algebraic method

(explanation by examples)

The substitution method In this method we make a guess for the solution of the

linear equation, we then put the guessed value in the equation and check whether the guess is correct or not?

Continue….

Examples Let us try to solve the equation ‘4x+2=4’, An

accurate guess for the solution is “x = ½”, we substitute this value in the given equation4 * ½ + 2 = 4 2 + 2 = 4 4 = 4

⁄⁄

Continue…. The substitution method is generally a time

consuming method, i.e. sometimes it takes very long time to find an accurate guess.

The graphical methodIn this method we sketch the graph of the linear equation, and then find the point where the graph cuts the x-axis, such a point is called the ‘x-intercept ’. This method is also not a very good method.

The algebraic methodIn this method we use the algebraic operations to find the solution. The method is described in the following examples.

Examples (Exercise Set 2.3, p#103)

Solve the following equations and check your solution.1. 2x=6Sol: Given the equation

2x=6Dividing both sides by 2

—x=22

—62

⁄⁄

⁄⁄3

Continue….Thus the solution is x=3.

Q16. x/3 = -2Sol: Multiplying both sides by 3, we get x = -6, which is the desired solution.Q19. -32x = -96Sol: Dividing both sides by -32, we get x=3 as the

required solution.Q24. -x = 9Sol: Multiplying both sides by -1, we get the solution x = -

9.

Continue….Q25. -2 = -ySol: Multiplying both sides by -1, we get

y = 2 as solution.Q33. 13x = 10Sol: Dividing both sides by 13, we get

x = 10/13 the required solution.Q52. -2x = 3/5Sol: Dividing both sides by -2, we get x = -3/10,

which is the required equation.

Continue….Q66. When solving the equation 3x = 5. Would you divide both

sides of the equation by 3 or by 5? Explain.Q67. When solving the equation -2x = 5. Would you add 2 to both

sides of the equation or divide both sides of the equation by -2? Explain.Q69. Consider the equation 4x = 3/5. Would it be easier to solve this

equation by dividing both sides of the equation by 4 or by multiplying both sides of the equation by ¼, the reciprocal of 4? Explain your answer. Find the solution to the problem.

Exercise Set 2.4 (9-65), p# 110Solve each equation.11. -2x-5=7Sol: Given

-2x-5=7Adding 5 to both sides-2x-5+5=7+5

-2x=12Dividing both sides by -2

-2x/2=12/-2 => x=-6The desired solution.

Continue….25. -4.2 = 2x + 1.6Sol: We have

-4.2 = 2x + 1.6Subtracting 1.6 from both sides -4.2 - 1.6 = 2x + 1.6 – 1.6

-5.8 = 2xNow dividing both sides by 2

-5.8/2 = 2x/2 => -2.9 = x (Using calculator)

Or x = -2.9 as desired.

Continue….Q33. x + 0.05x = 21Sol: 1.05x = 21

Dividing both sides by 1.05 (the coefficient of x)

x = 21/1.05 => x = 20 as required.Q36. -2.3 = -1.4 + 0.6xSol: Adding 1.4 to both sides

0.6x = -2.3 + 1.40.6x = -0.9Dividing through out by 0.6 x = -0.9/0.6 => x = -1.5

Continue….

Q38. 32.76 = 2.45x – 8.75xSol: Given

32.76 = 2.45x – 8.75x=> 32.76 = -6.30x

Dividing both sides by -6.30x = -32.76/6.30x = -5.2 the answer.

Continue….Q46. -2(x+4) + 5 = 1Sol: Using distributive property

-2x – 8 + 5 = 1 or -2x – 3 = 1

Adding 3 to both sides -2x -3 + 3 = 1 + 3

i.e. -2x = 4Now dividing both sides by -2

x = -4/2or x = -2 as desired.

Continue….Q58. 0.1(2.4x + 5) = 1.7Sol: given the equation

0.1(2.4x + 5) = 1.7Using distributive property0.24x + 0.5 = 1.7Subtracting 0.5 from both sides0.24x = 1.7 – 0.5

Þ 0.24x = 1.2Finally dividing both sides by 1.2x = 1.2/0.24x = 5the required answer.

Continue….Q65. 5.76 – 4.24x – 1.9x = 27.864Sol: Given that

5.76 – 4.24x – 1.9x = 27.864Þ 5.76 – 6.14x = 27.864

Now subtracting 5.76 from both sides 5.76 - 5.76 – 6.14x = 27.864 – 5.76

-6.14x = 22.104Finally dividing both sides by -6.14

-6.14x/-6.14 = -22.104/6.14x = 3.454

Which is the desired result.

Challenging problems

Solve each equation.1. 3(x-2) – (x+5) – 2(3-2x) = 182. -6 = -(x-5) – 3(5+2x) – 4(2x-4)3. 4[3 – 2(x+4)] – (x + 3) = 13

Solving linear equations with the variable on both sides of the equation

Hints:1. Use the distributive property to remove the parentheses.2. Combine like terms on the same side of the equal sign.3. Rewrite the equation with all terms containing the variable on one side of the

equation and all the terms not containing the variables on the other side of the equation.

4. Isolate the variable using the multiplication property, this gives the solution.5. Check your answer, by putting the value of ‘x’ in the original equation.

Exercise Set 2.5(9-54), p#118/119

Solve each equation.9. 4x = 3x + 5Sol: Given the equation

4x = 3x + 5Combining the terms involving variables on one side

if the equation4x – 3x = 5 x = 5

Which is the required answer.

Continue….15. 15 – 3x = 4x – 2xSol: Given that

15 – 3x = 4x – 2xShifting 3x to the right side of the equation.15 = 4x – 2x + 3x

Þ 15 = 5xIsolating ‘x’ using the multiplication property

15/5 = x or x = 3 as required.

Continue….25. x – 25 = 12x + 9 + 3xSol: We have

x – 25 = 12x + 9 + 3x Shifting ‘x’ to the right & 9 to left of the equation -25 – 9 = 12x + 3x – x -34 = 14x

Using multiplicative property-34/14 = x

or x = 2.428 as required.

Continue….28. 4r = 10 – 2(r-4)Sol: Given that

4r = 10 – 2(r-4)Using distributive property4r = 10 – 2r + 8Shifting 2r to the left of the equation4r + 2r = 10 + 8 6r = 18Dividing both sides by 6

r = 18/6or r = 3

as required.

Continue….34. 3y – 6y + 2 = 8y + 6 – 5ySol: Given the equation

3y – 6y + 2 = 8y + 6 – 5yÞ -3y + 2 = 3y + 6

Shifting -3y to the right & 6 to the left of the equation 2 – 6 = 3y + 3y -4 = 6yDividing both sides by 6 -4/6 = y

or y = -2/3 or y = -0.667

Continue….33. 0.1(x + 10) = 0.3x -4Sol: Given

0.1(x + 10) = 0.3x -4Using the distributive property 0.1x + 1 = 0.3x - 4Combining the terms involving the variable on one side of the

equation & constant on the other side0.3x – 0.1x = 4 + 1

0.2x = 5Using multiplicative property to isolate ‘x’

x = 5/0.2or x = 25

Which is the required solution.

Continue….36. 5(2.9x - 3) = 2(x +4)Sol: We are given that

5(2.9x - 3) = 2(x +4)By the use of distributive property, we have14.5x – 15 = 2x + 8

Shifting 2x to the left & -15 to the right of the equation14.5x – 2x = 8 + 1512.5x = 23 Isolating ‘x’ x = 23/12.5

or x = 1.84 as desired.

Continue….37. 9(-y + 3) = -6y + 15 – 3y + 12Sol: Given

9(-y + 3) = -6y + 15 – 3y + 12Using the distributive property-9y + 27 = -9y + 27Since the same expression appears on both

sides of the equation, therefore the statement is true for all real values of y. If we continue to solve this equation further, we arrive at

0 = 0.

A difference

Note that an equation is true only for a specific value(s) of the variable, while the identity is true for all values of the variable. In the previous question the given equation is an identity as it is true for all values of ‘x’. Further, every identity is an equation but not every equation is an identity.

Continue….39. -(3 - p) = -(2p + 3)Sol: Given that

-(3 - p) = -(2p + 3)Using distributive property-3 + p = -2p – 3Shifting 2p to the left(???) & -3 to the right of the

equationp + 2p = -3 + 3 3p = 0

=> p = 0Which was required.

Continue….51. 5 + 2x = 6(x + 1) – 5(x - 3)Sol: We have

5 + 2x = 6(x + 1) – 5(x - 3)Using distributive property, we have5 + 2x = 6x + 6 – 5x + 15

Or 5 + 2x = x + 21Shifting x to the left & 5 to the right2x – x = 21 – 5 x = 16 as required.

Challenging Problems

1. Solve-2(x+3)+5x=-3(x-3)+4x-(4-x)

2. Solve4(2x-3)-(x+7)-4x+6=5(x-2)-3x+7(2x+2)

3. Solve4 - [5 - 3(x + 3)]= x - 3

Ratio and ProportionDefinition

A Ratio is a quotient of two quantities. Ratios provide a way of comparing two numbers or quantities. The ratio of ‘a’ and ‘b’ is written as

a to b, a:b or a/ba and b are called the terms of the ratio.

EXAMPLES

1. If an algebra class consists of 32 males and 13 females, find

(a) The ratio of males to females.(b) The ratio of females to the entire class.Sol: (a) 32:13

(b) 13:45

Exercise Set 2.6, p#127The results of an English examination are 5 A’s, 6 B’s, 8 C’s, 4

D’s and 2 F’s. Write the following ratios.1. A’s to C’s.2. A’s to total grades.3. D’s to F’s.4. Grades better than C to total grades.5. Total grades to D’s.6. Grades better than C to grades less than C.

(Try these!)

Continue….Determine the following ratios. Write each ratio in lowest term.

7. 5 feet to 3 feet.8. 60 dollars to 80 dollars.9. 20 hours to 60 hours.10. 100 people to 80 people.11. 4 hours to 40 minutes.12. 6 feet to 4 yards. (1yard=3feet)13. 26 ounces to 4 pounds. (1pound=16 ounces)14. 7 dimes to 12 nickels. (1dime=10cents,

1nickle=5cents, 1dime=2nickle)

ProportionA proportion is a special type of equation. It is a statement of equality between two ratios.

How to denote a proportion?A proportion is denoted as a:b=c:d (read as ‘a is to b as c is to d’). We can also denote a proportion asa/b = c/d, a and d are referred as the extremes, and the b and c are referred as the means of the proportion.

Cross MultiplicationIf a/b = c/d then ad = bcNote that the product of means equal to the product of extremes.If any three of the four quantities of a proportion are known, the fourth can easily be found.

ExampleSolve for ‘x’ using cross multiplying x/3 = 25/15.

SolutionGiven that

x/3 = 25/15Using cross multiplication

x.15 = 3*25x = 75/15x = 5 (Check the answer!!!!!!!!)

Exercise Set 2.6 (21-32) p#128

Solve for the variable by cross multiplying.21. 4/x = 5/2022. x/4 = 12/4823. 5/3 = 75/x24. x/32 = -5/425. 90/x = -9/1026. -3/8 = x/4027. 1/9 = x/4528. y/6 = 7/4229. 3/z = 2/-2030. 3/12 = -14/z31. 15/20 = x/832. 12/3 = x/-100

To Solve Problem Using Proportion

1. Represent the unknown quantity by a variable (a letter).2. Set up the proportion by listing the given ratio on the left

side of the equal sign, and the unknown and other given quantities on the right side of the equal sign. When setting up the right side of the proportion, the same respective quantities should occupy the same respective positions on the left and right.

Exercise Set 2.6, P#127

18. In 1970 in the United States, 72,700 metric tons of aluminum was used for soft-drink and beer containers. In 1990 this amount had increased to 1,251,900 metric tons. Find the ratio of the amount of aluminum used for beer and soft-drink containers in 1990 to the amount used in 1970.

Sol: (Try!!!!!!!!).

Continue….33. A car can travel 32 miles on 1 gallon of gasoline. How far it can travel

on 12 gallon of gasoline?Ans: Let the distance covered in 12 gallon gasoline be x miles.

Nowmiles/gallon = miles/gallon

Or 32/1 = x/12Using cross multiplication

x . 1 = 32 * 12Or x = 384 as required.

Inequalities in One VariableAn inequality in one variable is the Mathematical statement in which one or more of the following symbol are used<, less than symbol>, greater than symbol≤, less than or equal to symbol≥, greater than or equal to symbolThe direction of the symbol is sometimes called the sense or order of the inequality.

Examples of inequalities in one variable

1. x+3<5 (x+3 is less than 5)2. x+4≥2x-6(x+4 is greater than or equal to 2x-6)3. 4>-x+3 (4 is greater than-x+3)4. 2x+6≤-2 (2x+6 is less than or equal to -2)Properties Used to Solve Inequalities

For real numbers a, b, and c5. If a>b, then a+c>b+c.6. If a>b, then a-c>b-c.7. If a>b and c>0, then ac>bc8. If a>b and c>0, then a/c>b/c9. If a>b and c<0, then ac<bc10. If a>b and c<0, then a/c<b/c

Similarly we can state the properties for the symbol ‘<‘

Exercise Set 2.7 (1-40), p#137

Solve the inequality and graph the solution on the real line.8. -4≤-x-3Sol: Given

-4≤-x-3Adding 3 to both sides-4+3≤-x-3+3 -1 ≤-xMultiplying both sides by -1

or x ≤ 1 (as -1<0, the inequality reverses the order).

Continue….12.6≥-2xSol: This inequality can also be written as

-2x≤6Dividing both sides by -2-2x/-2 ≥ 6/-3

Or x ≥ -3Which is the desired answer.

Continue….15. 12x + 24 < -1217. 4 – 6x > -519. 15 > -9x + 5024. -2x - 4 ≤ -5x + 1228. 2(x - 3) < 4x + 1029. -3(2x - 4) > 2(6x - 12)32. x + 5 ≥ x – 233. 6(3 - x) < 2x +1235. -21(2-x) + 3x > 4x + 438. -2(-5-x) > 3(x+2) +4 –x39. 5(2x + 3) ≥ 6 + (x + 2) - 2x40. -3(-2x +12) < -4(x+2) - 6

Think!!!!!!!!!!!!!41. When solving an inequality, if you obtain the result 3 < 5, what is the

solution?42. When solving an inequality, if you obtain the result 4 ≥ 2, what is the

solution?43. When solving an inequality, if you obtain the result 5 < 2, what is the

solution?44. When solving an inequality, if you obtain the result -4 ≥ -2, what is

the solution?45. When solving an inequality, under what conditions will it be

necessary to change the direction of the inequality symbol?

Practice Test (1-20), p#141

3.1 FormulasDefinition:

A formula is an equation commonly used to express a specific relationship mathematically.

Examples:1. The formula for the area of a rectangle is

area = length . Width or A = lw2. The formula for the perimeter of a square is

perimeter = 4. one side or P=4s3. The formula for the area of a triangle is

area = ½.base . Height or A=½bh

Simple Interest FormulaA formula used in banking is the simple interest formula, which is given byinterest = principal . rate . Time

Or i = prtHere ‘p’ is the principal (the amount invested or borrowed), ‘r’ is the interest rate, and ‘t’ is the amount of time of the investment or loan.This formula is used to determine the simple interest, i , earned on some savings accounts, or the simple interest an individual must pay on certain loan.

How to use the simple interest formula?

Example1: Avery borrows $2000 from a bank for 3 years. The bank charges 12% simple interest per year for the loan. How much the interest will Avery owe the bank?

Sol: Given thatThe principal, p, is $2000The rate, r , is 12% = 12/100=0.12

and the time, t , is 3 yearsUsing the formula i = prtPutting the corresponding values

i = 2000(0.12)(3)i = 720

The simple interest is $720. Thus Avery will pay $2720, after 3 years. (the principal, $2000 + the interest, $720).

Continue….Example: Amber invests $5000 in savings account which earns simple interest for 2

years. If the interest earned from the account is $800, find the rate.Sol: Here

Principal (investment) =p=$5000Time =t =2yearsInterest =i=$800rate =r=?Using the simple interest formula

i=prtSolving for ‘r’, we get r = i/ptPutting the values

r = 800/[(5000)(2)]r = 800/10000r = 0.08

Thus the simple interest rate is 0.08, or 8% per year.

Exercise Set 3.1, p#153In Exercise 73 through 76, use the simple interest formula.73. Mr. Thongsophapporr, borrowed $4000 for 3 years at 12% simple interest

rate per year. How much interest did he pay?74. Ms. Rodriguez lent her brother $4000 for a period of 2 years. At the end

of the 2 years, her brother repaid the $4000 plus $640 interest. What simple interest rate did her brother pay?

75. Ms. Levy invested a certain amount of money in a savings account paying 7% interest per year. When she withdrew her money at the end of 3 years, she received $1050 in interest. How much money did Ms. Levy place in the savings account?

76. Mr. O’Connor borrowed $6000 at 7½% simple interest per year. When he withdrew his money, he received $1800 in interest. How long had he left his money in the account?

Continue….Solve each equation for y; then find the value of y for the given value of x.

25. 2x + y = 8, when x = 228. -3x – 5y = -10, when x = 030. 15 = 3y – x, when x = 334. -12 = -2x – 3y, when x = -436. 2x + 5y = 20, when x = -5

Solve for the variable indicated.38. d = rt, for r42. V = lwh for w47. 4n + 3 = m, for n50. Y = mx + b, for x57. ax + by = c, for y

Continue….

Use the formula d = (1/2)n2 – (3/2)n, to find the number of diagonals in a figure with the given number of sides.

61. 10 sides62. 6 sides

Use the formula C = (5/9)(F - 32) to find the Celsius temperature (C) equivalent the given Fahrenheit temperature (F).

63. F = 500

64. F = 860

Continue….In chemistry the ideal gas law is P = KT/V, where P is the pressure, T is the temperature, V is the volume, and K is a constant. Find the missing quantity.

67. T = 10, K = 1, V = 168. T= 30, P = 3, K = 0.569. P = 80, T = 100, V = 570. P = 30, K = 2, V = 6

The sum of the first n even numbers can be found by the formula S = n2 + n. Find the sum of the numbers indicated.

71. First 5 even numbers.72. First 10 even numbers.

Changing Application Problems into Equations

Verbal Algebraic5 more than a number x+5A number increased by 3 x+37 less than a number x-7A number decreased by 12 x-12Twice a number 2xThe product of 6 and a number 6xOne-eighth of a number (1/8)x or x/8A number divided by 3 (1/3)x or x/34 more than twice a number 2x+45 less than three times a number 3x-53 times the sum of a number and 8 3(x+8)Twice the difference of a number and 4 2(x-4)

Express Relationships between Two Related Quantities

Verbal One number Second numberTwo Numbers differ by 3 x x + 3John’s age now and john’s age in 6 years

x x + 6

One number is six times the other

x 6x

One number is 12% less than the other

x x – 0.12x

A 25 foot length of wood cut in two pieces

x 25 - x

The sum of two numbers is 10

x 10 - x

84

Exercise Set 3.2, p#162/163Write as an algebraic expression.1. Five more than a number.2. Seven less than a number.3. Four times a number.4. The product of a number and eight.5. 70% of a number x.6. 8% of a number y.8. A 7½% sales tax on a car costing p dollars.9. The 16% of the U.S population, p, who do not receive adequate nourishment.10. Only 7% of all U.S tires, t, are recycled.11. Three less than six times a number.12. Six times the difference of a number and 3.13. Seven plus three-fourths of a number.14. Four times a number decreased by two.15. Twice the sum of a number and 8.16. Seventeen decreased by x.

Continue….17. The cost of purchasing x rolls of electrical tape at $4 each.18. The rental fee for subscribing to home box office for x months at $12 per month.19. The cost in dollars of traveling x miles at 23 cents per mile.20. The distance covered in t hours when traveling 30 miles per hour.22. The cost of waste disposal for y months at $16 per month.23. The population growth of a city in n years if the city is growing at a rate of 300

persons per year.24. The number of calories in x gram of carbohydrates if each gram of carbohydrates

contains 4 calories.25. The number of cents in x quarters. (Quarter=a fourth of a dollar)26. The number of cents in x quarters and y dimes.27. The number of inches in x feet.28. The number of inches in x feet and y inches.29. The number of ounces in e pounds.31. An average chicken egg contains 275 milligram of cholesterol and an ounce of

chicken contains about 25 mg of cholesterol. Write the amount of cholesterol in x chicken eggs and y ounces of chicken.

Continue….Express as a verbal statement. (There are many acceptable answers.)

33. x – 6 (six less than a number)34. x + 3 (three more than a number)35. 4x + 1 (one more than four times a number)36. 3x – 4 (four less than three times a number)37. 5x – 7 (seven less than five times a number)38. 2x – 3 (three less than twice a number)39. 4x – 2 (two less than four times a number)40. 5 – x (a number subtracted from five)41. 2 – 3x (three times a number subtracted from

two)42. 4 + 6x (four more than six times a number)43. 2(x - 1) (twice the difference of a number and one)44. 3(x + 2)(three times the sum of a number and two)

Continue….Select a variable to represent one quantity and state what that variable represents. Then express the second quantity in terms of the variable.

45. Eileen’s salary is $45 more than Martin’s salary.46. A boy is 12 years older than his brother.47. A number is one-third of another.48. Two consecutive integers.49. Two consecutive even integers.50. One hundred dollars divided between two people.51. Two numbers differ by 12.52. A number is 5 less than the four times another number.53. A number is 3 more than one-half of another number.54. A Cadilac costs 1.7 times as much as a Ford.55. A number is 4 less than three times another number.56. An 80-foot tree cut into two pieces.57. Two consecutive odd integers.58. A number and the number increased by 12%.59. A number and the number decreased by 15%.60. The cost of an item and the cost increased by a 7% sales tax.

Home WorkFrom

Question 61to

Question 68 On page # 163

Continue….Express as an equation.

69. One number is five times another. The sum of two numbers is 18.70. Marie is 6 years older than Denise. The sum of their ages is 48.71. The sum of two consecutive integers is 47.72. The product of two consecutive even integers is 48.73. Twice a number decreased by 8 is 12.74. For two consecutive integers, the sum of the smaller and twice the larger is 29.75. One-fifth of the sum of a number and 10 is 150.76. One train travels six times as far as another. The total distance traveled by both trains

is 700 miles.77. One train travels 8 miles less than twice the other. The total distance traveled by both

the trains is 1000 miles.78. A number increased by 8% is 92.79. The cost of a car plus a 7% tax is $13,600.80. The cost of a jacket at a 25%-off sale is $65.81. The cost of a meal plus 15% tip is $18.82. The cost of a videocassette recorder reduced by 20% is $215.83. The product of a number and the number plus 5% is 120.

Practice

In Exercises,(p#164)

express each equation as a verbal statement.

3.3 Solving Application problems

To solve the word problem1. Read the question carefully.2. If possible draw a sketch to help visualize the problem.3. Determine which quantity you are being asked to find. Choose a letter

to represent this unknown quantity. Write down exactly what this letter represents. If there is more than one quantity, represent all unknown quantities in terms of this variable.

4. Write the word problem as an equation.5. Solve the equation for the unknown quantity.6. Answer the question or questions asked.7. Check the solution in the original stated problem.

Exercise Set 3.3 p# 174

Discussion on Exercises

Group Activity/Challenging Problems

1. To find the average of a set of values, you find the sum of the values and divide the sum by the number of values.

(a) If Paul's first three test grades are 74, 88, and 76, write an equation that can be used to find the grade that Paul must get on his fourth exam to have an 80 average.

(b) Solve the equation from part (a) and determine the grade Paul must receive.

ExponentsIn the expression xn, x is called the base and n is called the exponent. xn is read as “x to the nth power.”

xnEXPONENT

BASE

What an exponent represents??

An exponent tells us how many times a number is multiplied with itself.For example 105 means105 = 10 x 10 x 10 x 10 x 10

= 100000Similarly

73 = 7 x 7 x 7 = 343

Continue….

64

= 6 x 6 x 6 x 6

36

216

1296

EXPONENTIAL FORM

EXPANDED FORM

STANDARD FORM

LAWS OF EXPONENTSCertain mathematical operation can be done whenever we have variables which contain exponents and equal bases. These operations are called “laws of exponent”.These laws are

Product Rule for ExponentsQuotient Rule for ExponentsZero Exponent RulePower Rule for Exponent

Each one is explained as under

Product rule for exponents

xn . xm = xn+m

Thus when same bases are multiplied, the exponents are added

xn . x

m

Bases are same

x=n+m

Exponents are Added

Examples (Exercise Set 4.1, p#

195)Simplify.1. x2 . x4 = x6

2. x5 . x4 = x9

3. y . y2 = y3

4. 42 . 4 = 43 = 645. 32 . 33 = 35 = 2436. x4 . x2 = x6

7. y3 . y2 = y5

8. x3 . x4 = x7

9. y4 . y = y5

Quotient Rule for Exponents

0, xxxx nm

n

m

Bases are same

Exponents are subtracted

Since division by “0” is not allowed

101

Continue….

EXAMPLES ON QUOTIENT RULE(EXERCISE SET 4.1, P#195)

(10 - 20)

Power Rule for Exponents

(xm)n = xm.n

The power rule indicates that when we raise an exponential expression to a power, we keep the base and multiply the exponents.

Exponential expression raise to a power

Keeping the base & multiplying the exponents

Expanded Power Rule for Exponents

0,0,

yb

ybxa

byax

mm

mmm

This rules illustrates that every factor within the parenthesis is raised to the power outside the parenthesis.

(As division by zero is not allowed)

104

Zero Exponent Rule

0,10 xxBy the zero exponent rule, any real number, except 0, raised to the zero power equals 1.

NOTE:00 is not a real number.This is called the undeterminedform.

Continue….DISCUSSION ON EXERCISE 4.1 page # 195/196

Negative Exponents

2535

3 xx

xx

25

3 1....

..xxxxxx

xxxxx

5

3

xx

We will develop this rule as follows:Using Quotient rule we have

Again by dividing out common factors,

1.

2.

We see that is equal to both & . Therefore

must equal

2x2

1x

2x2

1x 107

Negative Exponents Rule

0,1 x

xx m

m

It means that when a variable or a number is raised to a negative exponent, the expressionmay be rewritten as 1 divided by the variableor number to the positive exponent.

Note: When we are asked to simplify an exponentialexpression, our final answer should contain no negativeexponent.Also when a factor is moved from the denominator to thenumerator or from numerator to the denominator, the signOf the exponent changes.

Continue….

DISCUSSION ON EXERCISE 4.2, P # 202/203

System of Linear equationBy the system of linear equations we mean a group of linear equations.Sometimes in business we deal with many variables and unknown quantities. For example a company consider overhead cost, cost of material, labor cost, maximum possible production, selling price of the item, and a host of other items when seeking to maximize their profit. The business may express the relationship between the variables in equation or inequalities. These equations or inequalities form a system of linear equations or inequalities. The solution of the system of equations or inequalities gives the values of the variables for which the company can maximize profit.

Examples

631072

yxyx

12306174

zyxzyx

105541243

31

32

321

xxxxxxx

23653

074132

21

21

21

21

xxxxxxxx

(2 variables, 2 equations) (3variables, 2 equations)

(3 variables, 3 equations) (2 variables, 4 equations)In this course we will study linear system of equations in two variables111

Solution of System of Equations

By the solution of linear system we mean the order pair(s), which satisfy all the equations in the system simultaneously.e.g. Consider the system

It has the solution (1,6), since this order pair satisfy both the equations in the system. But (2,7) is not a solution to the system as it satisfy the first equation but not the second.

425

xy

xy

112

Continue…A system can be solved by the following methods:

1. The graphical method2. The substitution method3. The addition method

While solving a linear system, we face one of the following three situations:

a) The system is consistent (sol: exists)b) The system is inconsistent (no sol:)c) The system is dependent (infinite sol:)

Continue….Consistent, Inconsistent, Dependent system

Line 1 Line 2

y

xLine 1

Line 2

y

x

Line 1

Line 2

x

y

(a)

(b)

(c)

Consistent System

Inconsistent SystemDependent System

. Solution

Solution by Graphical MethodTo solve a linear system of equations graphically, we graph each equation and determine the point(s) of intersection.We will not present here the solution by graphical method, because often the solution by this method may be incorrect since we have to estimate the coordinates of the point of intersection. Thus we follow the other two methods.

Solution of System of equations by Substitution

To solve a system of equations by substitution1. Solve for a variable in either equation. (To avoid working with

fractions, prefer to solve for the variable with the numerical coefficient of 1).

2. Substitute the expression found for the variable in step 1 into the other equation.

3. Solve the equation in step 2 to find the value of one variable.4. Substitute the value found in step 3 into the equation found in step1 to

find the other variable.

Exercise Set 8.2 (p# 454)

1.

13242

yxyx

x

Solve the system of equations by substitution

Sol:

(1)

(2)Solving (1) for

42 yx (3)Putting this value of in (2)x

1178134813)24(2

yyyyyy

Putting in (3), we get1y 2xThus the solution is (2,1) 117

Continue….

DISCUSSION ON EXERCISE SET 8.2 (p#454/455)

Solution by Addition Method• This method is also called the elimination method.• This is often an easier method to solve a linear system of

equations.• In this method we add or subtract the given equations to

get a third equation which contains only one variable.• Note that our immediate goal is to obtain one equation

containing only one unknown.

The Method1. If necessary, shift all the terms containing variables to the left side of

the equal sign and the constants to the right side of the equal sign.2. If necessary, multiply one or both equations by a constant(s) so that

when the equations are added the resulting sum will contain only one variable.

3. Add the equations. This gives a single equation which contains only one variable.

4. Solve the equation obtained in step 3.5. Put the value obtained from step 3 into either of the original

equation. Solve this equation for the remaining variable.

Continue….

DISCUSSION ON EXERCISE SET 8.3, p#464

CHAPTER NO. 09

ROOTS AND RADICALS

• In this chapter the concentration will be on the radical equations and their solutions.

Definition• An equation that contains a variable in a radicand is called a radical equation.

n bax )( RADICAND

RADICAL SIGNINDEX

EXAMPLES

14)43(14

3859

5274

237

3

xx

xx

xx

x

x1.

2.

3.

4.

5.

HOW TO SOLVE A RADICAL EQUATION? (square root term)• Use the appropriate properties to write the equation

with the square root term by itself on one side of the equation. We call this isolating the radical.

• Combine like terms.• Square both sides of the equation to remove the square

root.• Solve the equation for the variable.• Check the solution in the original equation for the

extraneous roots ( a number obtained when solving an equation that is not a

solution to the original equation).

Exercises

2

1. 8 2. 5

3. 5 4. 3 6

5. 4 8 6. 3 5

7. 3 4 2 8. 6 4

9. 2 8 10. 5 3

11. 2 4 2 12. 4 9

13. 8 2 14. 2 5 4

x x

x x

x x

x x x

x x

x x

x x x x

TO SOLVE A RADICAL EQUATION WITH TWO SQUARE ROOT TERM

• Rewrite the equation, when necessary, so that there is only one square root term on each side of the equation.

• Square both sides of the equation and solve to find the solution.

Exercises1. 2 6 5 27

2. 3 9 2

3. 3 3 5 1

4. 4 5 9

5. 3 8

6. 2 3 2 3 2

x x

x x

x x

x x

x x

x x

CHAPTER # 10

QUADRATIC EQUATIONS AND THEIR SOLUTIONS

Continue…• An equation of the form

is called a quadratic equation.• Since this is a polynomial equation of degree 2,

therefore it has two solutions.• These solutions may or may not equal, may or

may not real.

0,02 acbxax

Continue…• Such equations can be solved using

a) Factorizationb) Completing squarec) Quadratic formulaWe will solve these equations using (b) &(c). EXAMPLES of Quadratic Equations:

079

072

0543

2

2

2

xx

x

xx

Completing Squarei. Make the numerical coefficient of the squared term equal to

1.ii. Shift the constant term of the equation to the right side of the

equal sign.iii. Take one-half of the numerical coefficient of the 1st degree

term, square it, and add this quantity to both sides of the equation.

iv. Replace the trinomial with its equivalent squared binomial.v. Use the square root property to eliminate the square.vi. Solve for the variable.vii. Check your answer in the original equation.

Application of the method

DISCUSSION ONEXERCISE 10.2,P# 544

Solution of Quadratic Equation by Quadratic Formula

• Given the quadratic equation

The solution to this equation is given by:

known as the Quadratic formula (Muhammad Bin Musa formula).

0,02 acbxax

aacbbx

242

Continue…• The radicand is known as the

‘Discriminant’.• Three cases arise while solving a Quadratic

equation by the Quadratic formula.a) When (two distinct & real

solutions exist)b) When (repeated solutions are

there)c) When (no real solution exist)

acb 42

042 acb

042 acb

042 acb

Remember that• If the discriminant is a negative number

We can’t go further. The equation has no real number solution. For such type of equations, your answer should

be “no real number solution”. Don’t leave the answer blank. Don’t write 0 for the answer.

Application of the formula

Discussion on Exercise Set 10.3, p# 553