business math chapter 3

35
CHAPTER 3 : SYSTEM OF EQUATION 3.1 LINEAR EQUATIONS 3.2 QUADRATIC EQUATIONS 3.3 SYSTEM OF EQUATIONS 3.4 INEQUALITIES

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Page 1: Business Math Chapter 3

CHAPTER 3 : SYSTEM OF EQUATION

3.1 LINEAR EQUATIONS

3.2 QUADRATIC EQUATIONS

3.3 SYSTEM OF EQUATIONS

3.4 INEQUALITIES

Page 2: Business Math Chapter 3

3.1 LINEAR EQUATION

An equation is simply a statement that 2 quantities are the same.

E.g.

A statement can be true or false is said to be a conditional (or open) equation as it is true for some values of the variable & not true for others.

For example: 2x + 3 = 7 is true for x = 2 but is false for x = 8 or other values.

Any number that makes the equation true is called a solution or root of the equation.

RHSLHS

Equation =

Think of an equation as a pair of perfectly balanced old – fashioned scales

RHSLHS

954

Introduction

Page 3: Business Math Chapter 3

3.1 LINEAR EQUATION

A statement such as is called an identity as it is true for all real numbers x.

While solving equation, there may be no solution, one solution or there may be more than one solution.

For example:

112222 xx

Equation Solution Comment

x + 9 = 12 x = 3 Only one solution

x2 = 81 x = 9 or – 9 Two solutions

x = 5 + x No solution No solution

x2 – 16 = (x – 4)(x + 4) All values of x Equation is true for all values of x.

Introduction

Page 4: Business Math Chapter 3

3.1 LINEAR EQUATION

To solve an equation, we find the values of the unknown that satisfy it.

How: Isolate the unknown / variable & solve its value. A linear equation is an equation that can be written in

the form;

where a & b are real numbers. Linear equations have only one root/ solution.

0 bax 0 bax

Solving Linear equation with 1 variable

Page 5: Business Math Chapter 3

15

31233

123

x

x

x

9

31233

123

x

x

x

43

12

3

3

123

x

x

x

36

1233

3

123

x

x

x

Add 3 to both sides Subtract 3 from both sides

Multiply both sides by 3 Divide both sides by 3

Perform the same

operation on both

sides

Always check in the

original equation

3.1 LINEAR EQUATIONSolving Linear equation with 1 variable

Page 6: Business Math Chapter 3

Example 1 .8637 equation the Solve xxx

874 x

154 x

4

15x

837 xx

8637 xxx Parentheses removed

x – terms combined

Both sides divided by 4

7 added to both sides

3x added to both sides

Step Working

Page 7: Business Math Chapter 3

PRACTICE 1

4

37532 e)

242137 d)

643.05.1 c)

43 b)

3546 a)

equation. given the Solve 2.

VV

pp

xx

nn

LL

5

2

1

32d)

5

36

510

3

2c)

124

3

2b)

1035 a)

:equation following the Solve 1.

x

x

xx

xx

x

Page 8: Business Math Chapter 3

A quadratic equation contains terms of the first and second degree of unknown is given by;

where a, b, c are real numbers and a ≠ 0. A quadratic equation that contains terms of the 2nd

degree only of the unknown is called a pure quadratic equation, can be expressed as ;

A value of the unknown that will satisfy the equation is called a solution or a root of the equation.

02 cbxax 02 cbxax

02 cax 02 cax

3.2 QUADRATIC EQUATIONIntroduction

Standard form

Standard form

Page 9: Business Math Chapter 3

3.2 QUADRATIC EQUATIONSolving quadratic equation

All quadratic equations have two roots . In pure quadratic equation , the roots are equal but of

opposite sign. It can solve by extraction of the root.There are two methods of solving complete quadratic

equation; By factorization By using quadratic formula

Page 10: Business Math Chapter 3

219.45

89

5

89

895

0895 c)

416

16

016 b)

39

9 a)

:equation the Solve

2

2

2

2

2

2

R

R

R

R

x

x

x

x

x

3552 e)

6

361

36

36

036 d)

2

2

2

x

ix

x

x

x

x

Where Where

i 1

Example 2

Page 11: Business Math Chapter 3

The expression is written as the product of two factors.

If the product of 2 factors is zero, then one of those factors must be zero. Thus if then either

Only be used if the quadratic expression can be factorized completely.

cbxax 2

0mn.0 or 0 nm

3.2 QUADRATIC EQUATIONSolving quadratic equation - Factorization

Page 12: Business Math Chapter 3

34

0304

034

012 c)

5

2

0250

025

025 b)

6

060

06

06 a)

:equationquadratic the Solve

2

2

2

xx

xorx

xx

xx

x

xorx

xx

xx

x

xorx

xx

xx

x

x

4

-3

= 4x

= -3x

= xx2

(x-3)

(x+4)

-12

Example 3

Page 13: Business Math Chapter 3

32

0302

032

06

61 f)

71

0701

071

076 e)

34

0304

034

0127 d)

:equationquadratic the Solve

2

2

2

xx

xorx

xx

xx

x x

xx

xorx

xx

xx

xx

xorx

xx

xxx

x

4

3

= 4x

= 3x

= 7xx2

(x+3)

(x+4)

12

Example 3

Page 14: Business Math Chapter 3

PRACTICE 2

1642 e)

035116 d)

0253c)

8103 b)

0145 a)

:equationquadratic the Solve

2

2

2

2

xx

xx

xx

xx

xx

Page 15: Business Math Chapter 3

For any quadratic equation ax2 + bx + c = 0, the solutions for x can be found by using the quadratic formula:

The quantity b2 – 4ac is called discriminant of the quadratic equation that characterize the solution of the quadratic equations: If b2 – 4ac = 0 one real solution If b2 – 4ac > 0 2 real and unequal solutions If b2 – 4ac < 0 no real but 2 imaginary solutions

(complex number)

a

acbbx

2

42 a

acbbx

2

42

3.2 QUADRATIC EQUATIONSolving quadratic equation – Quadratic Formula

Page 16: Business Math Chapter 3

formulaquadratic the usingby 0582 equation the Solve 1. 2 kk

4

1048

22

52488

2

4

2

2

a

acbbk

5495.44

10482

k

5495.04

10481

k

Step 1

Step 2

Step 3

Identify a, b & c, by comparing with ax2 + bx + c = 0.

a = 2, b = 8, c = -5

Substitute into the formula:

The solutions are: b2 - 4ac > 0,2 real & unequal

roots

b2 - 4ac > 0,2 real & unequal

roots

Example 4

Page 17: Business Math Chapter 3

Example 4formulaquadratic the usingby 09124equation the Solve 2. 2 pp

8

012

42

9441212

2

4

2

2

a

acbbp

2

3

8

12p

Step 1

Step 2

Step 3

Identify a, b & c, by comparing with ax2 + bx + c = 0.

a = 4, b = -12, c = 9

Substitute into the formula:

The solution is: b2 - 4ac = 0,1 real root

b2 - 4ac = 0,1 real root

Page 18: Business Math Chapter 3

Example 4formulaquadratic the usingby 0453 equation the Solve 3. 2 kk

6

235

6

2315

6

235

32

43455

2

4

2

2

i

a

acbbk

808306

2352

.i.

i

k

8.083.06

2351

i

ik

Step 1

Step 2

Step 3

Identify a, b & c, by comparing with ax2 + bx + c = 0.

a = 3, b = -5, c = 4

Substitute into the formula:

The solutions are:

b2 - 4ac < 0,no real, 2

imaginary roots

b2 - 4ac < 0,no real, 2

imaginary roots

Page 19: Business Math Chapter 3

PRACTICE 3

sss

RR

xx

xx

xx

219 e)

25

3 d)

0373 c)

42499 b)

087 a)

:formulaquadratic using

by below equationquadratic the Solve

2

2

2

2

Page 20: Business Math Chapter 3

A system of equations contains 2 or more equations. Each equation contains 1 or more variables.

A solution of the system is any pair of values (x , y)

that satisfies the both equations.2 methods that are usually used:

Substitution method

Elimination method

222

111

cybxa

cybxa

222

111

cybxa

cybxa

Where a & b are

coefficients of x & y, c is a constant.

Where a & b are coefficients of x & y,

c is a constant.

3.3 SYSTEMS OF EQUATIONIntroduction

Page 21: Business Math Chapter 3

Step for solving by substitution: Pick 1 of the equation & solve for 1 of the

variables in terms of the remaining variables. Substitute the result in the remaining

equations. If 1 equation in 1 variable results, solve the

equation. Otherwise, repeat Step 1 – 3 again. Find the values of the remaining variables by

back – substitution. Check the solution found.

3.3 SYSTEMS OF EQUATIONSubstitution Method

Page 22: Business Math Chapter 3

2332

163

equations of system following the Solve

Eqyx

Eqyx

1

99

33612

33362

y

y

yy

yy

yx 36 From Eq 1, make x as subject:

Substitute x in Eq 2 :

Solve for y :

Substitute y in Eq 1, to solve for x: 3136 x

Solution:

Step 1

Step 2

Step 3

Step 4

Step 5 Check the solution : Eq 1, 3 – 3(-1) = 6

Eq 2, 2(3) + 3(-1) = 3

Example 5 a

Page 23: Business Math Chapter 3

Example 5 b

2143

182

equations of system following the Solve

Eqxy

Eqyx

yx 28 From Eq 1, make x as subject:

Substitute x in Eq 2 :

Solve for y :

Substitute y in Eq 1, to solve for x: 2328 x

Solution:

Step 1

Step 2

Step 3

Step 4

Step 5 Check the solution :

Eq 1; 2 + 2(3) = 8

Eq 2; 3(3) – 4(2) = 1

The solution is the intersection

point (2,3)

The solution is the intersection

point (2,3)

3

3311

18323

12843

y

y

yy

yy

Page 24: Business Math Chapter 3

Example 5 c

252

1824

equations of system following the Solve

Eqyx

Eqyx

xy 25 From Eq 2, make y as subject:

Substitute y in Eq 1 :

Solution:

Step 1

Step 2

Step 3 The result is a false statement.

Therefore the system is inconsistent. It means that there is no solutions for this system of equations

The lines are parallel & no intersection

The lines are parallel & no intersection

810

84104

82524

xx

xx

Page 25: Business Math Chapter 3

21236

142

equations of system following the Solve

Eqyx

Eqyx

1212

126126

122436

xx

xx

xy 24 From Eq 1, make y as subject:

Substitute y in Eq 2 :

Solution:

Step 1

Step 2

Step 3 The result is a true statement.

Therefore the system is dependent. It means that there is no unique solutions can be determined.

The lines is coincide, the systems has

infinitely many solution

The lines is coincide, the systems has

infinitely many solution

Example 5 d

Page 26: Business Math Chapter 3

Step for solving by elimination: Write x’s, y’s & numbers in the same order in both

equations. Compare x’s & y’s in the 2 equations & decide which

unknown is easier to eliminate. Make sure the number of this unknown is the same in both equations.

Eliminate the equal terms by adding or subtracting the 2 equations

Solve the resulting equation & substitute in the simpler original equation in order to find the value of the other unknown.

Check the solution found.

Elimination Method

3.3 SYSTEMS OF EQUATION

Page 27: Business Math Chapter 3

21923

1467

equations of system following the Solve

Eqyx

Eqxy

2

5829

7618821

y

y

yy

54

20

2764

x

x

Rearrange the eq:

To eliminate x, Eq1 multiply by 3, Eq 2 multiply by 4

Substitute y in Eq 1, to solve for x:

21923

1674

Eqyx

Eqyx

476812

3182112

Eqyx

Eqyx

Eq 3 – Eq 4, solve for y :

Multiply by a constants to

get same coefficient of x

Multiply by a constants to

get same coefficient of x

Solution:

Step 1

Step 2

Step 3

Step 4

Step 5Check the solution :

Example 6 a

Page 28: Business Math Chapter 3

223

1423

equations of system following the Solve

Eqyx

Eqyx

11

2

211

6492

y

y

yy

11

16

22611

211

23

x

x

x

To eliminate x, Eq2 multiply by 3;

Substitute y in Eq 2, to solve for x:

3693

1423

Eqyx

Eqyx

Eq 1 – Eq 3, solve for y :

Solution:

Step 1

Step 2

Step 3

Step 4Check the solution :

Example 6 b

Page 29: Business Math Chapter 3

PRACTICE 4

2

4)4

3610

425 3)

5186

362 2)

526

52 1)

A.

usingby system following the Solve

22

yx

yx

LK

LK

nm

nm

yx

yx

method onSubstituti

83

22)4

1236

42 3)

462

1269 2)

16

522 1)

B.

2

xy

xxy

SR

SR

nm

nm

yx

yx

method nEliminatio

Page 30: Business Math Chapter 3

An inequality is a statement involving 2 expressions separated by any of the inequality symbols below.

3.4 INEQUALITYIntroduction

Inequality Meaning

a > b a is greater than b

a < b a is less than b

a ≥ b a is greater than or equal to b

a ≤ b a is less than or equal to b

A real number is a solution of an inequality involving a variable if a true statement is obtained when the variable is replaced by the number.

The set of all real numbers satisfy the inequality is called the solution set.

Page 31: Business Math Chapter 3

3.4 INEQUALITYInequality, Interval Notation & Line Graph Inequalit

yNotation

Interval Notation

Graph

a ≤ x ≤ b [ a, b ]

a < x ≤ b ( a, b ]

a ≤ x < b [ a, b )

a < x < b ( a, b )

x ≥ b [ b, ∞ )

x > b ( b, ∞ )

x ≤ a ( -∞, a ]

x < a ( -∞, a )

a b

Page 32: Business Math Chapter 3

3.4 INEQUALITYProperties of Inequality

PropertiesLet a, b & c be any real

number

Example

If a < b and b < c, then a < c If 2 < 3 & 3 < 8 so 2 < 8

If a < b, then a ± c < b ± c 5 < 7, 5 + 3 < 7 + 3, 8 < 10

5 < 7, 5 - 3 < 7 - 3, 2 < 4

If a < b, c > 0 then ac < bc 5 < 7, 5(2) < 7(2), 10 < 14

If a < b, c < 0 then ac > bc 5 < 7, 5(-2) > 7(-2), -10 > -14

If a < b, c > 0 then a/c < b/c 5 < 7, 5/2 < 7/2, 2.5 < 3.5

If a < b, c < 0 then a/c > b/c 5 < 7, 5/(-2) > 7/(-2), -2.5 > -3.5Similar properties hold if < between a and b is replaced

by ≥ , > , or ≤ .

Page 33: Business Math Chapter 3

3

12

5

46 d)21273c)

17

17

10743

9254 b)74103 a)

:esinequaliti following the of each Solve

xxxx

x

x

xx

xxxx

Divide by –ve no , reverse

sign

Divide by –ve no , reverse

sign

Example 7

Page 34: Business Math Chapter 3

1,0,1,2or23Therefore

23

6339

423and2311

42311 a)

:esinequaliti following the of each Solve

xx

xx

xx

xx

x

Solution set

Solution set

. is thereTherefore5

17

6175

1372and72310

1372310 b)

solution no

x

xx

xxxx

xxx

Example 8

Impossible to have x that satisfies both inequalities

Impossible to have x that satisfies both inequalities

Page 35: Business Math Chapter 3

PRACTICE 5

7521b)

715123a)

:inequality the Solve .3

.41

satisfying of values of range the Determine .2

.513 for set solution the find integer, an is If .1

x

x xx

xx

xx

2400425006

that given dollars) (in profit maximum the Find 5.

5.275.1255

that given dollars) (in cost minimum the Find 4.

PP

P

CC

C