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Philippines Nonformal Education Project(ADS-Assisted)

Bureau of Nonformal EducationDEPARTMENT OF EDUCATION,CULTURE AND SPORTS

Unit Conversions (Equivalents)

Length Time

1 in = 2.54 em1 ern = 0.394 in1 ft = 30.5 em1 m = 39.37 in = 3.28 ft1 mi = 5280 ft = 1.61 km1 km = 0.621 mi1 nautical mile (U.S.) = 1.15 mi = 6076 ft = 1.852 killI

1 fermi = 1 femtometer (fm) = 10-15m1 light year (ly) = 9.46 X 1015m1 parsec = 3.261y = 3.09 x 1016m

Volume

1 liter (L) = 1000 mL = 1000 ern' = 1.0 x 10-3 m3 =1.057 quart (U.S.) = 54.6 in '

1 gallon (U.S.) = 4 qt (U.S.) = 231 in' = 3.78 L= 0.83 gal (Imperial)

1 m3= 35.31 ft3

Speed

1 mi/h = 1.47 ftls = 1.609 km/h = 0.447 rnIs1 km/h = 0.278 m/s = 0.621 mi/h1 ft/s = 0.305 rnIs = 0.682 mi/h1 rnIs = 3.28 ft/s = 3.60 krnlh1 knot = 1.151 mi/h = 0.5144 rnIs

Angle

1 radian (rad) = 57.300 = 57°18'10 = 0.0l745 rad

1 day = 8.64 x 104s1 year = 3.156 x 107s

Mass

1 atomic mass unit (u) = 1.6605 x 1O-27kg1 kg = 0.0685 slug[1 kg has a weight of 2.20 Ib where g = 9.81 m/s"]

Force

lIb = 4.45 N1 N = 105 dyne = 0.225 lb

Energy and Work

1 J = 107 ergs = 0.738 ft·lb = N'rn = 1 kg m/s?1 ft·lb = 1.36 J1 kcal = 4.18 x 103 J1 eV = 1.602 x 10-19J

Power

1 W = 1 J/s = 0.738 ft· IbislkW = 1000W1 hp = 550 ft-lb/s = 746 W

Pressure

1 atm = 1.013 bar = 1.013 x 105N1m2 = 14.71b/in2

1 lb/in? = 6.90 x 103 N/m2

1 Pa = 1 N/m2 = 1.45 x LOr'Ib/in?

Newton's Laws of Motion Statistics

Law of inertia

A body continues in its state of rest or of uniformvelocity in a straight line unless acted by a non zeronet force.

Law of acceleration

Acceleration of a body is directly proportional to thenet force acting on it, and inversely proportional to itsmass

Lf=ma

Laws of action and reaction

Whenever a body exerts a force on another body, thesecond exerts an equal and opposite force on the first.

range: x - Xhighest lowest

_ LXmean: X=-

n

median: (odd n): middle term

(even n): sum of the 2 middle terms2

mode: most frequent value

standard deviation:n -1

OEP ED CENTRAL LIBRARYHERALCO AVE. PASIG un

Inequalities

Copyright 2001Bureau of Nonformal Education

DEPARTMENT OF EDUCATION,CULTURE AND SPORTS

All rights reserved. No part of this module may be reproduced in any form or by any means,electronic or mechanical, including photocopying, recording or by any information storageand retrieval system, without the prior written permission from the publisher.

Published in the Philippines by:

Bureau of Non formal EducationDepartment of Education, Culture and Sports

3/P Mabini Bldg., DECS ComplexMeralco Avenue, Pasig City, Philippines

Tel. No.: (02) 635 - 5189 Fax No.: (02) 635 - 5191

,,

Office of the Secretary

REPUBLIC OF THE PHILIPPINESDEPARTMENT OF EDUCATION, CULTURE AND SPORTS

DECS Complex, Meralco AvenuePasig City, Philippines

Sarna-SarnasaDECS

Mv dear NFE A&E Learners,

Welcome to the Nonformal Education Accreditation and Equivalency(NFE A&E) Academic-Focussed Bridging Program of the Department ofEducation, Culture and Sports (DECS). The Academic-Focussed BridgingProgram is a new nonformal education approach to learning that offers NFEA&E Secondary Level test passers who are interested in entering college, anopportunity to prepare for the intellectual challenges of a university or collegeeducation. The Program was developed by the Bureau of Non formalEducation (BNFE) to strengthen the NFE A&E as a truly alternative systemoflearning to formal schooling by expanding the pathways oflearning and lifeopportunities of out-of-school (OSY) and adults.

The Academic-Focussed Bridging Program is built around a specialcurriculum comprising skills and competencies felt to be essential for NFEA&E Secondary Level test passers to cope with the demands and entryrequirements of college education. This includes skills and competencies inthe traditional academic disciplines ofhigher level Mathematics, AdvanceScience, and English and Filipino Communication Skills. It also covers non-traditional college preparation skills such as study skills, report and essaywriting, library and research skills, skills in giving seminar presentations andother college and university survival skills. Based on this curriculum, we haveprepared a range of self-instructional learning modules. By studying themodules and completing the various self-assessment activities, andassignments, we hope that you will be able to prepare for the academic rigorsand requirement of college and university life.

We hope you find the modules interesting, informative and challenging.They have been specially prepared to help you learn by yourself throughinquiry, investigation, problem solving and application of skills and concepts toeveryday life situations. Studying by using self-instructional modules meansyou will need to take on responsibility for your own learning and will require ahigh level of commitment, motivation and self-discipline. These, however, areskills which are essential for you to succeed at higher levels of learning atcollege or university.

We hope that through this Academic-Focussed Bridging Program, youand other qualified out-of-school youth and adults will have an alternativemeans to move up a continuous ladder of learning as a pathway to a bettertomorrow.

Good luck and mabuhayi

G<avcJ1.~~RAUL ROCO

Secretary

What Is This Module About?

6 > 4, m - 5 n > 3 n7 X + 14 =, 9y < 5 X - Y

Inone ofthe modules in mathematics, the set of real numbers and its properties werediscussed. It was mentioned that the set of real numbers is closed under addition andmultiplication. Moreover, real numbers can be represented by points on a line called acoordinate or number line. Likewise, points on a coordinate line are labeled with real numbers.Points to the right ofthe origin on the number line correspond to positive real numbers, whilepoints to the left ofthe origin correspond to negative real numbers.

Besides these properties, the real number system has other important features that will bedescribed in this module. They are orderrelations, inequalities and absolute value. You will alsolearn how to graph an inequality and solve word problems involving inequalities.

This module contains four lessons. These are:

Lesson 1 - Inequalities and Their Properties

Lesson 2 - Solutions of Linear Inequalities

Lesson 3 - Inequalities Involving Absolute Values

Lesson 4 - Systems of Inequalities

What Will You Learn From This Module?

After studying this module, you should be able to:

• illustrate orderrelations;

• define inequality and sketch a graph of an inequality;

• illustrate and apply the properties of inequalities;

• solve inequalities with one unknown;

• define absolute value;

• solve and graph an absolute inequality;

• solve systems oflinear inequalities by graphing; and

• solve word problems involving inequalities.

• Wait!

Before studying this module, please be sure that you have completed the module entitledEquations.

I Let's See What You Already Know

Before you start studying this module, take the following test first to find out how much youalready know the topics to be discussed.

1. Which of the following statements is correct?

1> 0a.

2c. -5 = 5

1b. -8<-6 d. - > -

4 3

2. If8 > 2, wmch ofthe following statements is the correct conclusion?

a. 2 is located to the right of 8 on the number line.

b. 2 is located to the left of8 on the number line.

c. 8 is located to the left of2 on the Number line.

d 8,is located to the left of 0 on the number line.

3. An inequality is a mathematical statement involving any of the following symbolsexeept:

a. c. >

b. =1= d. <

4. The graph of the inequality -2 < x < 1 is:

a. ~_"~r I'H r , , F , " )-4 -3 -2 -1 0 1 2 3 4

c. ( III, , T , , , , , I )

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

1 ,~d. < I I , , , , I )

-4 -3 -2 -1 o 1 2 3 4

b. ( I I t I I t I I I )-4 -3 -2 -1 0 1 2 3 4

5. Which of the following statements illustrates transitive property?

a. If 2 < 3 and 3 < 7 then 2 < 7. c. If2 < 3 then 3 > 2.

d. If2 < 3 then 2 + 4 > 3 + 4.b If2 < 3 then 2(3) < 7(3).

2

6. If5 < 10 and ifboth sides of the inequality are multiplied by -3, then the conclusionmust be:

a. -15 < -30 c. -15 > -30

b. -15 2: -30 d. -15 < -30

7. The solution set of the inequality 1- 4x < 9 is:

a. x <-2 c. x >-2

b. x 2:-2 d. x..2:-2

8. The absolute value of a number is always __ .

a. the number itself c. the negative of the given number

d. zerob. non-negative

9. The solution set ofthe inequality 15x-11 < 5 is:

a. -4 <x < 6 c. -5 <x < 5

b.4 6--<x<-5 5

d.4--<x<55

10. The graph ofthe solution set ofthe inequality 12x-11> 3 is:

r r r r R1a. ( I I I I I I I ) c. ( I I I I )

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

b. ( I I r I I r I I I ) d. ( IE I I r I I 1Tt )

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

11. The graph of the inequality 2x < 1 is:

y I~a. (I I I I I I) c. (I I I-2.0 -1.5 -1.0 -0.5 0 0.5 1.0 1.5 2.0 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0

f ~) (~ ~b. (I 1 1 d. I 1 I)-2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0

3

12. The solution set ofthe system of inequalities x +y > 1 andx-2y > 2 is:

a.

b.

13. Whatisthevalueofxin2x+4 2: x-2?

a. x 2:-6

b. x 2:-2

c.

d.2-

c. x 2: 2

14. The length of a-rectangle is 5 centimeters (em) longer than the width. How long will thewidth be if the perimeter of the rectangle is at least 40 cII?-?

a. x 2: 7.5

b. x s 7.5

c .. x 2: 12.5

d. x ~ 12.5

4

Well, how was it? Do you think you fared well? Compare your answers with those in theAnswer Key on page 33 to find out.

If all your answers are correct, very good! This shows that you already know much aboutthe topic. You may still study the module to review what you already know. Who knows, youmight learn a few more new things as well.

If you got a low score, don't feel bad. This means that this module is for you. Itwill help youunderstand important concepts that you can apply in your daily life. If you study this modulecarefully, you will learn the answers to all the items in the test and a lot more! Are you ready?

You may go now to the next page to begin Lesson 1.

5

LESSON 1

Inequalities and Their Properties

We experience and see inequalities around us. There are inequalities in social status, such asbetween a king and a beggar; inequalities in knowledge and health, such as between a healthyschoolchild and sickly street child. There are also inequalities among different objects.

Each of the relations mentioned above can be described using symbols in mathematics.There are a lot of symbols which show inequality between quantities. Some of these symbols are:f:., ~, S .The following are examples of relations showing inequalities:

• 5<-2

• x+l>3

• m ~ 8

• t2:f:. 4You will learn more about such relations in this lesson.

~ Let's Learn

The statement "a is greater than b" (that is, a > b where a and b are real numbers) wouldmean that a - b is a positive number.

For example, 5 > 2 means that 5 - 2 =3, which is positive, meaning it is greater than O.

Similarly, "a is less than b" (that is, a < b where a and b are real numbers) would mean thata - b is negative.

6

Moreover, if you have an axis, as in the figure below, on which the positive direction is to theright and the negative direction is to the left, then the positive value of a - b is to the right ofO.For example, since 5 - 2 = 3, then 5 is found 3 units to the right of 2.

< I I I I I I I I)-6 -5 -4 -3 -2 -1 0 2 3 4 5 6 7

So, the relation 2 < 5 means that 2 is found 3 units to the left of 5 since 2 - 5 =-3.

Just like equations, inequalities have properties, too. They will help you understand betterhow to solve problems involving inequalities.

• Trichotomy. Only one of the following is true for any real numbers a and b.

• a> b

• a =b

• a 2, then 5 "* 2 and 5 < 2.

• Addition property. The sense of inequality is not changed if the same number isadded to or subtracted from both sides ofthe inequality.

If a > b, then a ± c > b ± c for all values of c.

For example, in the relation:

5>2

Adding (-1) to one side would mean adding the same quantity to the other side, that is:

5+(-1»2+(-1)

4>1

• Multiplication property (where c > 0). The sense of inequality is not changed ifbothsides of the inequality are multiplied or divided by the same positive number.

a bIf a > b, then (a )(c) > (b)(c) and - > - for any positive number c.

c c


-3 <2

Dividing both sides of the inequality by 3,2

-1<-3

7

• Multiplication property (where c < 0). The sense of an inequality is changed ifbothsides of the inequality are multiplied or divided by the same negative number.

a bIf a < b, then (a)(c) > (b)(c) and - > -.

c c


4>2

Multiplying both sides of the inequality by -1,

(4)(-1) < (2)(-1)

-4 <-2

Were you able to understand each of the properties of inequalities? Tell which property isapplied in the following.

• If2 < 3, then 6 < 7. _

• Ifx>y,then3x>3y. _

• If7-x<x+3,thenx-7>-x-3. _

• Ifx - 2> 3, then x > 5. _

• If-3x<-6,thenx>2. _

Are your answers the same as these? The relation 6 < 7 was obtained by adding 4 to bothsides of2 < 3. Thus, the addition property was applied.

The relation 3x > 3ywas obtained by multiplying both sides ofx > y by 3. Thus, themultiplication property by a positive number was applied.

The relation x - 7 > -x - 3 was obtained by multiplying both sides of7 - x < x + 3 by -1.Thus, the multiplication property by a negative number was applied.

The relation x> 5 was obtained by adding 2 to both sides of x - 2 > 3. Thus, the additionproperty was applied.

The relation x > 2 was obtained by dividing both sides of - 3x < -6 by -3. Thus, themultiplication property by a negative number was applied.

Adding to or multiplying by different quantities each number in an inequality may change theorder of inequality.

For example, in the inequality:

4<5

Adding 6 to the left side and 1 to the right side ofthe inequality will give

4+6<5+1

10 < 6, which is false.

8

~ Let's Review

Tell which property is used in each of the following.

1. If-m > 3,thenm <-3. ---------------------2. If 4x < 3x - 6, then x < -6. _

3. If 5 + x> 3 +y, then 7 + x > 5 + y. ------------------------4. Ifx < 3, then 4x < 12. _

5. If 5x < 5, then 5x +y < 5 +y. _

Compare your answers with those in the Answer Key on page 33.

aLet's Study and Analyze

Is it true that if x2 > 4, then x> 2? -----------------------------Explain. _

Compare your answer with that in the Answer Key on page 33.

~ Let's See What You Have Learned

Tell what property is shown in each of the following.

1. If3 > -1, then -4 > -8.

2. If 4x + 1 > 1, then 4x > o.3. If4x+4< 10, then-2x-2>-5.

4. If2x > -1, then 4x > -2.

5. If5x<20, thenx<4.

Compare your answers with those found in the Answer Key on page 33. If you got aperfect score, congratulations! You understood the lesson well. If you did not, don't worry. Justreview the lesson before moving on to Lesson 2.

9

, Let's Remember

• An inequality is a mathematical statement showing that a quantity is less than, greaterthan or not equal to another quantity.

• The properties of inequalities are the following:

• Trichotomy

For all real numbers a and b, one and only one ofthe following is true.

a>b;

a = b; ora b, then a + c > b + c.

For any real number c, if a < b, then a + c b, thenac> bc.

For any positive real number c, if a < b, then ac < be.

Multiplication property by a negative number c

For any negative real number c, if a> b, then ac < be.

For any negative real number c, if a < b, then ac > be.

• Ifyou add to or multiply by different quantities each member of an inequality, thesense or order of the inequality may change.

•

10

LeSSON 2

Solutions of Linear Inequalities

You have been exposed to solving linear as well as quadratic equations in one and twounknowns. Ina linear equation in one unknown, the solution is unique (meaning there is only onesolution) while in a linear equation in two unknowns, there may be (a) a unique solution, (b) nosolution or (c) an infinite number of solutions. Similarly, for quadratic equations, you have learnedthat solutions may either be real or imaginary.

Inthis lesson, you will learn the methods of solving linear inequalities.

This lesson will enable you to plot and solve an inequality algebraically as well as solveproblems involving linear equalities.

~ Let's Learn--=-

The graph of an inequality is similar to that of a linear equation with some modifications. Thefollowing are graphs of some inequalities.

• x < 1 (read as "x is less than 1")

x ( I I-7 -6 -5 -4 -3 -2 -1 0

I I 1 I I2 3 4 5 6

I)7

Note that the hollow circle means that the point x = 1 is not included in the set of possiblevalues.

11

• y 2: -3 (read as ''y is greater than or equal to -3")

y 6 7-7 -6 -5 -4 -3 -2 -1

Note that the shaded circle means that the point x =-3 is included.

Consider the inequality x + 3 > 7.

We solve this in a way similar to how linear or quadratic equations are solved.

Add (-3) to both sides of the inequality, then

x+3>7

x + 3 + (-3) > 7 + (-3)

x>4

This means that all numbers greater than 4 are possible values ofx.

Drawing the graph of the solution, we have:

x( ,-4 -3 -2 -1 o

, , r I , I I2 3 4 5 678

IT>9 10

To check the solution, take any number on the number line, say 5, then:

x+3>7

5+3>7

8>7

8 is indeed greater than 7. Therefore, the solution and the graph are correct.

Let's try another example.

What if 4x- 3 ~ 1?

• What value should be added to both sides of the inequality to get 4x:s 4?

• To get x :s 1, what property is used? _

• What value has to be added to/multiplied by both sides of the inequality?

12

• The solution is -------------------------Let's solve the inequality4x-3:S 1together.

Adding 3 to both sides ofthe inequality, you get

4x:s 4

MUltiplying both sides of the inequality by Y4, you get

x:::; 1

Thus, the graph ofthe solution is

x ( 1 I· 1-6 -5 -4 -3 -2 -1

1 r 1 1 1 1 1 101234567

I)8

To check your solution, take any value in the solution, say -2, then

4x - 3:::; 1

4(-2) - 3:::; 1

-8-3:::;1

-11 < 1

Therefore, the solution and the graph are correct.

-I} Let's Try This

Solve and graph the inequality: 7r - 5 2: 16.

• What property or properties will be used to solve this inequality? Give the value to beadded to or multiplied by the inequality. _

• The solution to the inequality is _

• The graph of the solution is

x <: I I I I I I I-6 -5 -4 -3 -2 -1 0

I I I I I I I>2345678


13

~g Let's Learn

Just like in the case oflinear and quadratic equations, there are also word problems thatinvolve linear inequalities. You can deal with these problems in the same way you deal withproblems involving linear and quadratic equations.

Consider the following problems.

PROBLEM 1. Three times a number is at most 27. What numbers are possible?

The word "at most" means the highest or the largest value while the word "at least" meansthe lowest or the smallest value.

Thus, the statement "three times a number is at most 27" means that three times a numberhas 27 as the largest possible value. This also means that values less than 27 are permissible.

Therefore, if you letx be the number, then the inequality representing the above condition is:

3x S27

Solving for x, you get:

x.s:9

Check if your solution is correct. Take a number less than 9, say 8, and then substitute thisin the inequality.

Thus, 3(8).s: 27

24.s: 27

Therefore, the solution is correct.

PROBLEM 2. Two more than four times a number is between 2 and 18. What numbersare possible?

Note that when you say that a certain value is between two numbers, a and b, this is anotherway of saying that the value is greater than a and less than b. That is;

If a <x< b, then x > a and x < b.

• If you let x be the number, express the following in symbols:

• Four times a number ----------------------------------------Two more than four times a number -----------------------------

• The inequality is _

• This inequality can also be written as _

• The solution is------------------------------------------------

14

Compare your answers with these:

The inequality representing the problem is:

2 < 4x + 2 < 18

This can be written as:

4x + 2 > 2 and 4x + 2 < 18

Solving each of these ineon» !:~~es,you get:

4x+2>2 and 4x+2<18

4x> 0 4x< 16

x>O and x< 4

Graphing these, you will get the following:

x< If I I I I t I I I I I I I 1 >-6 -5 -4 -3 -2 -1 0 2345678

Since the intersection ofthe graphs is 0 to 4 exclusive of the numbers 0 and 4, the solution

O<x<4

Taking only the intersection of the two inequalities, the fmal graph of the solution will looklike this:

< r I I 1 I >-5 -4 -3 -2 -1 0 2 3 4 5

To check, take any value between 0 and 4, say 1, and substitute it in the inequality:

2 < 4x + 2 < 18

Thus,

2 < 4( 1) + 2 < 18

2 < 4 + 2 < 18

2 < 6 < 18

The solution is therefore correct.

15

.. Let's Study and Analyze

Given a > b and c >d, tell whether the following statements are correct or not.

1. Is a + c > b + d? _Explain. _

2. Isac>bd? -----------------------------Explain. ~ _



A. Find the solution of each of the following inequalities and plot the graph of the solution.

1. 15:5 3x

2. 2x+1>5

3. 3x-1 < 2x + 4

4. 4x + 3:::: 2x + 11

5. 5x+ 6 <x-2

B. Solve the following:

I. The perimeter of a rectangle must not be greater than 40 em and the length mustbe 8 em. What is the set of values for the width?

2. Ifsix is added to a number and the sum is multiplied by two, the product isgreater than 40. What are the possible numbers?

Compare your answers with those in the Answer Key on pages 34 and 35. If you got aperfect score, that's very good! Ifnot,just review this lesson before moving on to the next.

16

~ Let's Remember

• Linear inequalities are solved the same way linear equations are solved.

• The phrases "at most" and "at least" are oftentimes used in linear inequalities. Thephrase "at most" means" the largest or the highest value is __ " while the phrase "atleast" means" the smallest value is "__

For example, if "xis at most 3," then the greatest value for x is 3.

• If a> b and c > d, then a + c > b +d. Since these inequalities are ofthe same senseor order, adding two bigger quantities results in a quantity bigger than the sum of thelesser values.

For example, if6 > 4 and 8 > 5, then 6 + 8 > 4 + 5.Thus, 14>9.

• If a> b and c > d, then ac> bd. As mentioned above, multiplying two biggerquantities results in a product bigger than the product of the lesser values.

For example, if6 >4 and 8 > 5, then (6) (8) > (4) (5).Thus, 48 > 20.

17

LESSON 3

Inequalities Involving Absolute ValuesThings are becomingcomplicated, how will Isolve inequalitiesi nvolvi ng absolutevalues?

If you are asked to get the distance of an obj ect from another obj ect, how will you do it?

To find the distance between two objects, you try to get any straight-edged object or ameterstick to measure the distance. Distance is associated with a positive number. Note that nodistance is negative.

Recall that the absolute value of a number a (denoted by laD is geometrically interpreted asthe distance of a point from the origin.

This lesson will enable you to graph and solve absolute inequalities.

~ Let's Learn

The concept of the absolute value of a number serves as a steppingstone in learning theconcept of the absolute inequality.

Recall that if you have ~I= 3, then this means that x =3 or x = -3. This can also beinterpreted as the distance of a point 3 units away from the origin as shown below.

Inthe figure, the point which is 3 units away from the origin can be represented as point A,and the point with a coordinate of -3 can be represented as point B.

-6 -5 -4 -3 -2 -1 0 2 3 4 5 6 7

:1:'I'

I

1

I:

Suppose you have ~I< 3. You can interpret this to be the set of all points with a distance3 units away from the origin and these are the points whose coordinates are between -3 and 3.That is,

-3 <x < 3

18

The figure below shows the graph of Ixl< 3.

< I r I 1 I I I I )-6 -5 -4 -3 -2 -1 o 2 3 4 5 6 7

Thus, we say that:

Ixl < a is equivalent to -a < x < a.

Consider the inequality: Ix + 31< 4.

To solve this inequality,

-4<x+3<4

Adding (-3) to all parts ofthe inequality,

-4 + (-3) < x + 3 + (-3) < 4 + (-3)

Simplifying,

-7 <x < 1

The graph of the solution set is:

< I r I 1-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

I I I )2 3 4

To check whether the obtained solution set is correct or not, take any value from the set,say-4, then

1x+31<4

1-4+31<4

1-11<4

1<4

Let's try another example.

Suppose you have Ix - 11s 7.

• Rewrite the inequality without the absolute value sign.

• What value has to be added to/multiplied by all the parts ofthe inequality?

• What is the solution set of the given absolute inequality?

19

Are your answers the same as these?

~ - 11:s7 is equivalent to -7:S x-I :s 7.

Add 1 to all the parts of the inequality. Thus,

-6:Sx:sS

The graph is shown below:

< I I I I r )-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16

.. Let's Try This

Solve the inequality I3x - 41 <2 and graph the solution.

• The given absolute inequality is equivalent to

• What value should be added to/multiplied by all the parts ofthe inequality? Show theprocess.

• The solution set to the absolute inequality is _

• The sketch ofthe graph is:

(I )

Compare your answers with those found in the Answer Key on page 35.

20

~ Let's Learn

Consider the absolute inequality ~12:2.

What does this inequality geometrically mean?

If you answered, "all the points whose distance from the origin is greater than or equal to 2units," then you are correct.

The graph of the inequality is shown below:

< I~ I I '111·'1 )-7 -6 -5 -4 -3 -2 -1 0 2 3 4 5 6

Thus, if~1 >a, then x > aor x <-a.

If you have the inequality ~ + 31>4, solving for x will give you:

Ix + 31> 4 is equivalent to x + 3 > 4 or x + 3 <-4.

Thus, adding (-3) to both sides of the inequality,

x + 3 > 4 or x + 3 <-4

x + 3 + (-3) > 4 + (-3) or x + 3 + (-3) < -4 + (-3)

Therefore,

x> I or x <-7

The sketch of the graph is shown below.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 o 123

To check whether your solution set is correct or not, take any value from the set, say -8,then

~+ 31>4

1-8 + 31> 4

I-51 >4

5>4

Suppose you are given the absolute inequality 17- 2x12:9.

• This absolute inequality is equivalent to _

• What value( s) must be added to/multiplied by both sides ofthe inequalities?

21 OEP ED CENTRAL L"~RA'"NERALCO AVE. PAS'G tn,

• The solution set of the absolute inequality is _

Compare your answers with the following

17- 2xl2: 9 is equivalent to 7 - 2x 2: 9 or 7 - 2x :::;-9

Adding (-7) to both sides of the inequalities, you get:

-2x 2: 2 or -2x S -16

Dividing both sides ofthe inequalities by (-2), the sense of the inequalities changes, thus,

x S --lor x 2: 8

The graph ofthe inequality is shown below:

f I I I I-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 o 2 3 4 5 6 7 8 9 10 11

I}Let's Try This

Solve and graph the absolute inequality 15 - 3xl :::;10.

• This absolute inequality is equivalent to _

• What value( s) must be added to/multiplied by each member ofthe inequality?

• The solution set ofthe given absolute inequality is" _

• Graph the absolute inequality:

(I >Compare your answers with those in the Answer Key on page 36 .

• Let's Study and Analyze

If Ixl > 1, then -1 > x> 1. Is this true or false? _

VVhy? _

Compare your answer with that in the Answer Key on page 36.

22


Solve and graph the following absolute inequalities:

1. Ix + 41 < 8

2. Ix + 21 ::: 3

3. 15x - 61< 1

4. Ix - 31> 2

5. 12- 3xl ~ 10

Compare your answers with those given in the Answer Key on pages 36 and 37. If you goteverything right, that's very good! If not,just review this lesson before moving on to the nextlesson.

, Let's Remember

• The absolute inequalities Ixl> a and Ixl<a are interpreted in the same way as theabsolute value of a number is interpreted. They are geometrically interpreted as thedistance of a point from the origin of a number line.

• The absolute inequality Ixl < a is equivalent to -a <x < a.

• The absolute inequality Ixl >a is equivalent to x> a or x <-a.

• The absolute inequality Ixl > 1 is not equivalent to -1> x> 1 since there is no valuethat is less than -1 that is at the same time greater than 1.

23

LESSON 4

Systems of Inequalities

Inthis lesson, you will be concerned with the solution sets of simple systems oflinearequations as well as systems oflinear inequalities in two unknowns or variables, The skills youdeveloped in Lesson 3 will also be used here in solving for the values of x in the given systems oflinear inequalities.

~ Let's Learn

A pair of inequalities in two variables makes up a system of inequalities. The inequalities,when graphed on the same coordinate plane, show overlapping regions where you can pick acoordinate that satisfies both inequalities.

As you go along the study of the systems oflinear inequalities, it is advisable that you recallsome important points about !faphing linear equations.

Consider the system of inequalities: x + 2y - 42: 0

24

2x-y-32:0

Using the slope-intercept form of a linear equation, first determine the graphs of theequations: x + 2y - 4 = 0 and 2x - y - 3 = O.

Solving for their slope-intercept forms, you get:

1y=--x+2 andy=2x-32

Hence, to locate the region x + 2y - 4 2: 0, draw the line corresponding to x + 2y - 4 = 0and then shade the right side ofthe line.

The same procedure has to be done to locate the region corresponding to 2x - y - 3 2: O.Draw the line whose equation is 2x - y - 3 =0, then shade the region to the right of the line.

Why is the right side shaded? This is because the relation symbol is >. Note that thesymbol denotes that the boundary line is included in the solution set.

The graphs of the equations are shown below:

2x- y- 3 = 0

Thus, the double-shaded region represents the solution to the given system oflinearinequalities.

To check, take any point within the intersection (but not on the boundary line), say (6, 2),then substitute this in the inequalities. That is,

x + 2y - 4 2: 0 and x - y - 3 2: 0

6 + 2 (2) - 4 2: 0 and 6 - 2 - 3 2: 0

Thus, 62:0 and 12:0.

To check further if your solution set is correct, take at least 5 points in the region ofintersection.

Suppose you are given the system: x +y < I and

x-y<3

25

• Write the slope-intercept form of the boundary lines.

• The region described by x + y < 1 is the region to the ofthe line x + y =1. Is the boundary line included? _

Why? __

.• The region described by x - y < 3 is the region to the ofthe line x-y = 3.Is the boundary line included? __

Why? __

Are your answers the same as these?

Writing the slope-intercept form of the equations of the boundary lines, you get:

y=-x + 1

y=x-3

The left sides ofthe lines are to be shaded because the symbol < is used.

Thus, the graph ofthe system of inequalities win look like this:

Note that the boundary lines are dotted. This is to show that they are not to be included inthe solution set.

Three possible points in the intersection are (-2,1), (-3, -2) and (-2, -1).

26

.• Let's Try This

Solve and graph the system of inequalities: x + y - 2 < 0 andx-y-l>O

• Write in slope-intercept form the equations ofthe boundary lines. _

• The region described by x + Y - 2 < 0 is to the _

of the boundary line. Why? _

• The region described by x - y - 1 > 0 is to the _

of the boundary line. Why? _

• Graph the system of linear inequalities.

y

• Name at least 3 points in the region of intersection.


27

(!p Let's Think About This

1. Is it possible for systems oflinear inequalities to have a unique solution?

Explrun. ___

2. Is it possible for systems of linear inequalities to have no solution?

Explain. _

?Let's See What You Have LearnedI'Solve the following systems oflinear inequalities using the graphical method.

1. x + Y :S 5 and x - y 2: 4

2. 2x +Y :S -3 and 5x + 1 2: 2y

3. 4y :S 2x - 10 and y - x 2: 6

4. 2x - 3y > -I and x :s y

5. x + 2y:s 7 and 2x + y > -2

Compare your answers with those found in the Answer Key on pages 38 and 39.

, Let's Remember

• Systems oflinear inequalities can be best solved using the graphical method.

• Insolving systems oflinear inequalities, follow these steps:

Express in slope-intercept form the equations of the boundary lines.

Draw the lines corresponding to the boundary lines.

The region to be shaded depends on the relation described by the inequality.Shade the right side when the" greater than" symbol is used; shade the left sidewhen the "less than" symbol is used.

Identify at least 5 points in the intersection ofthe regions to check whetherthe solution set obtained is correct or not.

28

• It 'isnot possible for a system oflinear inequalities to have a unique solutionbecause two rectangular regions cannot just meet at a point. However, it is possible forit to have no solution. This is the case where the shaded regions fall on different sidesof two parallel boundary lines.

You have now reached the end of the module. Congratulations! Did you enjoy studying thismodule? Did you learn a lot from it? The following is a summary of its main points to help youremember them better .

.Pr Let's Sum Up

• An inequality is a mathematical statement showing that a quantity is less than, greaterthan or not equal to another quantity.

• When solving for the unknown in a linear inequality, we follow the same steps as insolving for the unknown in a linear equality.

• illgraphing a system oflinear inequalities we follow these steps:

• Express the equations of the boundary lines in slope-intercept form.

• Graph the boundary lines and shade the regions denoted by the relations in theinequalities.

Shade the region in which two regions of the boundary lines intersect. This is thesolution set for the system of inequalities.

&.What Have You Learned?

Read each problem carefully then encircle the letter ofthe correct answer.

1. The statement "Ten is greater than negative one" can be written as _

a. 10<-1b. 102:-1

c. 10>-1d. 10 *-1

2. Which of the following sets of numbers is arranged in ascending order?

a. -2,0,-3 c. -4, -3,-6

b.1 3 52' 4'12 d. -1,0,3

3. Which of the following statements does not define an inequality?1 5

a. 3> 7 c. 2*6

b. :_2 < 0 d.

29

4 The graph ofl-3x > 10 is:

0 ..< I I ) c. < I I )a.

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

If 0 • •b. -y

c. All are true.

b. Exactly one is true. d. Not one is true.

6. If -6 <-1 and ifboth sides of the inequality are multiplied by 2, the conclusion mustbe:

a. -12<-2

b. -12S-2

c. -12 >-2

d. -122-2

7. The solution set ofthe inequality 1 S 2x - 3 S 5 is:

a. 2 SX <4 c. 2 <x < 4

b. 2 <x S 4

8. The absolute value of-_!' is:2

a.12

c.12

b. -2 d. o

9. The solution set of the inequality Ix + 11< 2 is:

-3 < x < 11a. c. -ISxs--3

b. -3 Sx S 1 d.1

-1 <x<--3

10. The graph ofthe solution set ofthe inequality 14x + 31< 1 is:

a. -1 0 -1 0

30

11. The graph of the solution ofthe inequality 3 x - y > 2 is __ .

• l 0 ..a( I I I I ) c.( I I I I I )

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

0lil ~ 0•b.( I I I I I ) d.( I I I I I )-2 -1 0 2 3 -2 -1 0 2 3

12. The solution set ofthe system oflinear inequalities 2x -y <2 and x + 2y <2 is:

a.

b.

c.

d.

31

13. The perimeter ofarectangular lot must not be greater that 30 meters (m). If the lengthis 8 m, what is the set of possible values for the width?

a. w=7m c. w 2: 7m

d. w~7mb. O~w ~7

Compare your answers with those in the Answer Key on page 39. If you got a score of:

9-13 Congratulations! You learned a lot from this module. Just review the parts of themodule that you did not understand very well.

6-8 Good! Review the parts you did not understand very well.

0-5 You should study the whole module again.

32

"Answer Key

A. Let's See What You Already Know (pages 2-5)

1. b 8. b

2. b 9. b

3. a 10. c

4. b 11. d

5. a 12. b

6. c 13. a

7. c 14. a

B. Lesson 1

Let s Review (page 9)

1. multiplication property by a negative number

2. addition property


4. multiplication property by a positive number


Let s Study and Analyze (page 9)

1. No. It is also possible that x can be a number less than 2, suchas-3, whichalso satisfies the inequality x2 >4.

Let s See What You Have Learned (page 9)



3. multiplication property by a negative number



33

C. Lesson 2

Let s Try This (page 13)

• The addition property can be used to solve the inequality.

7r-52:16

7r-5+52: 16+5

7r 2: 21

r2:3

• The solution to the inequality is r 2: 3.




o 3

Let s Study and Analyze (page 16)

1. Yes. Adding two bigger quantities will definitely result in a sum that's bigger thanthe sum of two lesser quantities.

2. Yes. Multiplying two bigger quantities will result in a product bigger than theproduct of the two lesser quantities.

Let s See What You Have Learned (page 16)

A. 1. 15 S 3xx2:5

• III52x>4x>2

0 .,-4 -3 -2 -1 0 1 2 3 4 5 6 7

5. 5x + 6 <x- 24x<-8x<-2

... 0

 402x + 12> 402x> 28x> 14

The possible numbers are x > 14.

D. Lesson 3


• -2 < 3x-4 < 2

• The number 4 should be added to all parts of the inequality.

Thus, -2 + 4 < 3x - 4 + 4 < 2 + 42 < 3x < 6

2• The solution set of the absolute inequalities is "3 < x < 2.

• The graph ofthe inequality is:

< I I Io I 23 3

2 3I)

35

Lets Try This (page 22)

• -10 < 5 - 3x ::s 10

• The number (-5) must be added to all the parts of the inequality. Thus,-1 0 - 5 ::s5 - 3x - 5 ::s10 - 5-15 ::s- 3x ::s5

-5• The solution set ofthe absolute inequality is x::s 3or x > 5.


• •< I I_____.

I I I I I )-3 -2 -1 0 2 345 6

Lets Study and Analyze (page 22)

No. There is no number that is less than -1 but is at the same time greater than 1.

Lets See What You Have Learned (page 23)

1. Ix + 41 < 8

-8 <x+ 4 < 8

-12<x<4

( I I I )-12 0 4

2. ~ + 21::s3-3 ::sx + 2 ::s3-5 ::Sx::S1

( I I >-5 0 1

3. 15x- 61< 1-1 < 5x- 6 < 15 < 5x < 7

71 <x<"5

0-0< II" II" " I" " III" I" " I )-1 0 1 2. 2 3 4

5

36

4. ~ -31> 2

x - 3 > 2 or x - 3 <-2

x> 5 or x < 1

( I-1 0

5. -10 2: 2 - 3x 2: 10

-122:-3x 2: 8

84>x>--- - 3

o •

1 >5

< 111111111111111111111111111111111111 >-3-8 -2 -1 0 1 2 3 4

3

E. Lesson 4


x + y -2 < 0 and

x-y-1>O

• Slope-intercept forms:

y=x+2

y=x-1

• The region described by x +Y - 2 < 0 is to the left of the boundary line. This isbecause the relation is "less than."

• The region described by x - y - 1 > 0 is to the right of the boundary line. Thisis because the relation is "greater than."

• Points in the region of intersection include (1.5, -2), (-2,-3) and (-1, --4).

37

Lets See What You Have Learned (page 28)

1. x +y:s 5 and x - y;::: 4 2. 2x +y:s -3 and 5x + 1 ;:::2y

3. 4y:s 2x -10 andy -x> 6

38

4. 2x- 3y ~-1 and x ~y

5. x + 2y ~ 7 and 2x + y ~-2

F. What Have You Learned? (pages 29-30)

1. e 8. e

2. d 9. a

3. d 10. d

4. a 11. e

5. b 12. d

6. a 13. d

7. d

39

References

Capitulo, F. M. Algebra: A Simplified Approach. Manila: National Bookstore, 1989.

Sia, Lucy 0., et al. 21st Century Mathematics, Second Year.Quezon City: PhoenixPublishing House, Inc. 2000.

40

Powersof 10

10-' = _!_10

10-2 =_1_100

10-3 =_1_1,000

106 = 1,000,000105 = 100,000

104 =10,000

103 =1,000

102 = 100

10' = 10 10-4 =_1_10,000

110-5

100,000

10-6 = __ 1__1,000,000

Pythagorean Theorem

In a right angle, if a and b are the lenghtsof the perpendicular sides and c is thelengthof the hypotenuse, then

b

Point-Slope Form

If a straight line has slope m and passes through thepoirit(Xl' y,), then an equation of the line isy- Yl = m(x -xJ

Slope-Intercept Form

If a linear equation is written as y = mx + b, then m istheslope of the line and b is the y-intercept,

DistanceFormula

Thedistance between 2 points PI(XI'Y2) and P2(X2'Y2)is given by

Midpoint Formula

If M(x,y) is the midpoint of the line segment fromPI(~"Y.) to P2(X2'Y2) then

z

Arithmetic Sequences

nth term: a = al + (n - l)dn

Geometric Sequences

nth term: an = al",-I

Arithmetic Series

Sum of n terms:

Geometric Series

al (1- rn)Sum of n terms: Sn = 1- r

Permutations

The number of arrangements of n things taken r at a

. . p( ) n!time IS n, r = -(--).n-r .

Combinations

The number of ways to chose r elements from a group

(n) n!of n elements is ( _ ). Ir n r .r.

Quadratic Formula

Trigonometric Identities

oppositehypotenuse

. adjacentcosme =--=---

hypotenuse

1cscx=--

sin x1 cosxcotx=--=--

tan x sin xcos 2 X + sin 2 X = 1

tan 2 X + 1= sec 2 x

cot 2x + 1= esc 2 X

sine

tangent oppositeadjacent

sin x = cos x tan x

sin x 'cosx=--

tan xsin x

tanx=--cosx

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the Year 20100 Uil.HESCO gn~emaltiolr'!al NOMA li~eracy Prize for its

Nonfolrmal Education Accreditation and Equivalency

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but also tothe entire Filipino nation. "

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~ I

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