ba101 engineering mathematic chapter 1 basic algebra

BASIC ALGEBRA BA101/CHAPTER 1

1 Prepared By : Lim Yeong Chyng

Chapter 1: Basic Algebra

Introduction

Algebra is the branch of mathematics that uses letters in place of some unknown numbers.

Literal numbers (the letters used in algebra) can either stand for variables (where the value

of the letter can change, such as the area of a rectangle and the area of a square) or constants

(where the value does not change), for example e (which has a constant value of 2.781828...).

Algebra is a powerful tool for problem solving in science, engineering, economics, finance,

architecture, ship-building and many other day-to-day tasks. If we did not use letters in place

of numbers (and used words instead), we would be writing many pages for each problem

and it would be much more confusing.

The 'Father of Algebra'

Abu Ja'far Muhammad ibn Musa al-Khwarizmi

Al-Khwarizmi

Al-Khwarizmi lived in Baghdad, 780 to 850 AD. He was one of the first to write about

algebra (using words, not letters).

Around 825 he wrote Al-jabr w’al muq abala, from which we get the word algebra (meaning

'restoration of broken parts'). This book included many word problems, especially to do with

inheritance.

He was also influential in the establishment of Hindu-Arabic numbers (1, 2, 3, ...) which

replaced Roman numerals (I, II, III, IV,...). The Hindu-Arabic system was much easier to

use when performing mathematical operations, since it is a base-10 system.

1.1 Simplifying Algebraic Fractions

Algebraic fraction has the same properties as numerical fraction. The only difference being

that the numerator (top) and denominator (bottom) are both algebraic expressions. Fractional



algebra is a rational number usually stated in the form of q

p , where p and q are integers.

Integer 'p' is known as the numerator and integer 'q' as denominator.

Meanwhile, an ordinary fraction is usually use to represent a part of an object or figure. For

example: a cake is cut into 6 equal parts. One part can be represented with fractional

expression of 1/6, similarly if expressed in the fractional algebra. There are some important

terms to be familiarize with before solving fractional algebra.

i. Equivalent fractions are fractions having same value.

2

1 =

4

2 =

8

4

ii. Single fraction is any fractional phrase.

iii. Lowest fractional form is a fraction that cannot be simplified further, or, its

numerator and denominator does not have a common factor.

q

p

Example 1

Complete the equivalent fractions below :

a) 20

6

105

2

b) 20

312

c) AB

CABBC

A

2

d)

XYZ

XYZ

XZ

XYZ

2

Numerator

Denominator



Solution:

a) 20

8

15

6

10

4

5

2

b) 10

20

3

6

1

22

c) CB

AB

CAB

BA

BC

A22

2

d)

XYZ

ZXY

YZX

XYZ

XZ

XYZ 2

22

2

Example 2

Simplify each of the following fractions:

a) 27

2

b

b b)

2

2

6

3

x

xx c) mmm 4)2(6

Solution:

(a) bbb

b

b

b

7

2

7

2

7

22

(b) x

x

xx

xx

x

xx

6

3

)6(

)3(

6

32

2

NOTE: The cancellation in (b) is allowed since x is a common factor of the numerator and

the denominator. Sometimes extra work is necessary before an algebraic fraction can be

reduced to a simpler form.

(c) mmmmmm 4264)2(6

m12

Example 3

Simplify the algebraic fraction: 32

122

2

xx

xx

Solution

In this case, the numerator and denominator can be factorised into two terms. Thus,

22 )1(12 xxx and )3)(1(322 xxxx



So,

)1)(3(

)1)(1(

32

122

2

xx

xx

xx

xx

3

1

x

x

Exercise 1

Simplify each of the following algebraic fractions.

(a) 45

36

14

7

ba

ba

(b) 25

1072

2

y

yy

Solution

(a) The fraction is 45

36

14

7

ba

ba. Instead of expanding the factors, it is easier to use the rule of

indices (powers) :

nm

n

m

aa

a , Thus,

4

3

5

6

45

36

14

7

14

7

b

b

a

a

ba

ba

4356

2

1 ba

4356

2

1 ba

b

a

2

(b) In this case, initial factorization is needed. So,

)2)(5(1072 yyyy and

)5)(5(252 yyy

Thus,



)5)(5(

)2)(5(

25

1072

2

yy

yy

y

yy

5

2

y

y

Exercise 2

Which of the following is a simplified version of 23

432

2

tt

tt?

2

4)(

t

ta

2

4)(

t

tb

2

4)(

t

tc

2

4)(

t

td

Solution

Factorize the numerator and denominator is respectively

)4)(1(432 tttt and )2)(1(232 tttt

so that,

)2)(1(

)4)(1(

23

432

2

tt

tt

tt

tt

2

4

t

t

So far, simplification has been achieved by cancelling common factors from the numerator

and denominator. There are fractions which can be simplified by multiplying the numerator

and denominator by an appropriate common factor, thus obtaining an equivalent, simpler

expression.

Exercise 3

Simplify the following fractions:

a)

2

14

1y

b)

2

13

xx



Solution

(a) In this case, multiplying both the numerator and the denominator by 4 gives:

2

41

)2

1(4

)4

1(4

2

14

1

yyy

(b) To simplify this expression, multiply the numerator and denominator

by 2 x. Thus

x

x

x

xxx

xx

2

13

2

)1

3(

2

13 2

Exercise 4

Simplify each of the following algebra fractions.

(a) 2

2

34 y

(b)

2

13

1

z

z

Solution

(a) The fraction is simplified by multiplying both the numerator and the denominator by

2.

4

38

22

)2

34(2

2

2

34

y

yy

(b) In this case, since the numerator contains the fraction 1/3 and the denominator

contains the fraction 1/2, the common factor needed is 2 × 3 = 6. Thus



36

26

)2

1(6

)3

1(6

2

13

1

z

z

z

z

z

z

Exercise 5:

Which of the following is a simplified version of1

1

1

x

xx

?

1

1)(

2

2

xx

xxa

1

1)(

2

2

x

xxb

1

1)(

2

2

xx

xc

1

1)(

2

2

x

xd

Solution:

For 1

1

1

x

xx

, the common multiplier is ( x + 1). Multiplying the numerator and the

denominator by this gives:

)1)(1(

)1

1)(1(

1

1

1

xx

xxx

x

xx

)1(

)1

1)(1()1(

2

x

xxxx

)1(

)1(2

2

x

xx

Exercise 7:

1. Simplified the following.

a) tsrrs 3232

b) )12

5(4 232 xqpqp



c) abcab 624 2

d) hgf

hgf22

23

9

36

e) npq

npqqp

10

153 22

f) xzxyzzxy 181226 32

g) 22222 3)4(7 rssrsr

h) 45326 xymnxymn

i) )32(5 byab

j) 3)12156( 2 cabmk

k) )42()35( 22 zxyzxy

l) )3()23( 22 kpmnrsmnkp

m) )48(2

19 nm

n) 2

1210)35(2

xx

Answer for Exercise 7:

1. a) tsr 436

b)

xqp 35

3

5

c) bc4

d) f4

e) 22

2

9qp

f) 339xyz

g) 222 311 rssr

h) 473 xymn

i) xyab 1510

j) 2452 cabmk

k) 27 zxy

l) rsmnkp 24

m) nm 249

n) 1215 x

2. 12



2. Given that a=1, b=-2, x=-1 and y=3,

find the value of



3. axxyba 222



4.



1.2 Solving Algebraic Fractions Using Addition, Subtraction, Multiplication and Division

It is make revenue process of finding increase and revenue push of two or more fraction algebra. In

settling these operations there were 3 moves that must be followed.

ADDITION

Example 1: Calculate

5

1

5

3

.

Solution: Since the denominators are the same, the denominator of the answer will be 5. Adding the

numerators , 3 + 1 = 4. The result will be:

5

4

5

13

5

1

5

3

Example 2:

Calculate

3

1

2

1

Solution: The denominators are different, so you must build each fraction to a form where both have

the same denominator. Since 6 is the common factor for 2 and 3, build both fractions to a

denominator of 6.

6

3

3

3

2

11

2

1

and,

6

2

2

2

3

11

3

1

Then

3

1

2

1

6

5

6

23

6

2

6

3

SUBTRACTION



Example 1:

Calculate 5

1

5

3

Solution: The denominators are the same, so you can skip step 1. The denominator of the answer will

be 5. Subtract the numerators for the numerator in the answer. 3 - 1 = 2. The answer is

5

2

5

13

5

1

5

3

Example 2: Calculate

3

1

2

1

Solution:

Since the denominators are not the same, you must build each fraction to a form where both

have the same denominator. As 6 is the common factor for 2 and 3, use 6 as the

denominator.

6

3

3

3

2

11

2

1

and,

6

2

2

2

3

11

3

1

Then,

3

1

2

1

6

1

6

23

6

2

6

3

MULTIPLICATION

Example 1:

9

8

4

3

Solution:

Multiply the numerators and the denominators, and simplify the result. Thus,



3

2

36

24

94

83

Example 2:

15

6

7

2

Solution:

Multiply the numerators and the denominators, and simplify the result. Thus,

35

4

105

12

157

62

DIVISION

Example 1:

16

9

4

3

Solution:

Change the division sign to multiplication and invert the fraction to the right of the sign.

9

16

4

3

Multiply the numerators and the denominators, and simplify the result.

3

4

36

48

94

163

Example 2:

8

9

40

3

Solution:

Change the division sign to multiplication and invert the fraction to the right of the sign.

9

8

40

3

Multiply the numerators and the denominators, and simplify the result.



15

1

360

24

940

83

9

8

40

3

Exercise 1: Simplify 23

32

10

)8)(5(

ts

tsst

Solution: 2

23

43

23

32

410

40

10

)8)(5(t

ts

ts

ts

tsst

Exercise 2: Simplify 22

2

)3(

12

ab

ba

Solution: 342

2

22

2

3

4

9

12

)3(

12

bba

ba

ab

ba

Exercise 3: Simplify m

mnnm

2

62 2

Solution: Form the given fraction into 2 fractions,

both with denominator 2m. Thus,

nmnm

mn

m

nm

m

mnnm3

2

6

2

2

2

62 22

Exercise 4: Simplify

45

13

x

x

Solution 1 - Multiplying by the reciprocal

Change the top expression into a single fraction with denominator x.

x

x

x

1313

Change the below expression into a single fraction with denominator x.

x

x

x

454

5

Thus, the question has become:



x

xx

x

x

x45

13

45

13

Look at the right hand side expression as a division process. Thus,

x

x

x

x 4513

Instead of division, we can carry out the multiplication by a reciprocal method.

x

x

x

x

x

x

45

13

45

13

The x's cancelled out, and we have our final answer, which is in its simplest form.

Solution 2 - Multiplying top and bottom

Multiply "x" to both the numerator and denominator. So, by just multiplying

the top and bottom part by x, everything will be simplified.

x

x

x

x

x

x

45

13

)45

(

)1

3(

1.3 Conversion of Formulas.

ADDITION

Example 1:

Given a = b + c, make c as the subject.

Solution

Since: a = b + c,

Subtract b from both sides of the equation,

a – b = b + c – b

or a – b = b – b + c

then, a – b = c

Therefore: c = a – b

SUBTRACTION



Example 1:

Given a = c – d, make d as the subject.

Solution

Since: a = c – d,

Add d to both sides of the equation.

Then, a + d = c – d + d

a + d = c + d – d

a + d = c

Deduct a from both sides of the equation.

We have: a + d = c

Then, a + d – a = c – a

a – a + d = c – a

d = c – a

Example 2:

Solve the equation

x − 6 = 10

Solution:

Add 6 to both sides of the equations.

We have: x − 6 = 10

Then, x − 6 + 6 = 10 + 6

So, x = 16

Thus, the value of x needs to be 16 to make the equation true.

Check with the given question:

16 − 6 = 10. It checks out to be correct.

MULTIPLICATION

Example 1:

If V = IR, make I as the subject.

Solution

Given V = IR.



Divide both sides of the equation with R,

Then: R

IR

R

V

IR

V or

R

VI

Example 2:

Solve: 5x = 35

This time we are asking:

5 × ? = 35

Setting x as the subject, divide both sides by 5:

355 x

5

35

5

5

x

7x

Check: 5 × 7 = 35. It checks out to be correct.

DIVISION

Example 1:

Make t as the subject for formula t

sv .

Solution

Given t

sv . Find t.

Multiply both equations with t,

tt

svt , then svt .

Divide both sides of the equation with v,

Since: svt

Then, v

s

v

vt or

v

st



Combination of Arithmatic Operations

Example 1:

Make x as subject for formula cmxy .

Solution

Given cmxy .

Subtract c from both sides of the equation,

y – c = mx + c – c

thus, y – c = mx.

Divided both sides with m,

y – c = mx,

then m

mx

m

cy

xm

cy

Thus, m

cyx

Example 2

Given V = E – Ir, make r as the subject.

Solution:

V = E – Ir ,

then V + Ir = E

Ir = E – V.

Divide both sides of the equations with I,

I

VE

I

Ir

So, I

VEr

Example 3

Solve 5 − (x + 2) = 5x



Solution

First, we open the bracket:

5 − (x + 2) = 5x

5 − x − 2 = 5x

3 − x = 5x

By adding x to both sides of the equation,

We have: 3 = 6x

Now divide both sides by 6 and swap sides:

Thus: x = 0.5

Example 4

Solve 5x − 2(x − 5) = 4x

Solution:

Expanding the bracket:

5x − 2(x − 5) = 4x

5x − 2x + 10 = 4x

3x + 10 = 4x

Subtracting 3x from both sides gives: x = 10

ARITHMETIC OPERATION WITH PARENTHESIS

Example 1:

Make C as the subject given that C

DCBA

Solution

Given: C

DCBA

,

Multiply both sides of the equation with C,

C

C

DCBAC

= BC – BD



Move BC to the left side of the equation,

AC – BC = – BD

Factorization of C,

C (A – B) = – BD

BA

BDC

BA

BD

BA

BDC

or

AB

BDC

Exercises

Example 1:

Make d as the subject for 4

2hdV

,

Solution:

Given4

2hdV

,

Multiply both sides of the equation with 4,

44

42hd

V

4V = d2h

Divide both sides with h,

h

hd

h

V

24

24d

h

V

Get the square root of both sides,

24d

h

V



dh

V

4

Example 2:

The area of a circle is given by A = π r2 . Find the value of r.

Solution

Making r as the subject, we have,

π r2 = A

Divide both sides with π:

Ar 2

So,

Ar

Problem Solving

1. A copper wire with a length of = 2 cm, resistance R= 4Ω and resistivity

ρ= 17.2 x 10-6

cm. Calculate the cross-section area A, of the wire, by using

the resistivity formula, A

R

Solution

AR

A

)2)(102.17(4

6

26106.8 cmA

Exercise 1:

1. Simplify the following:

(a) bc

a

a 32

3

(b) mmm

2

)1(

1



(c) xy

y

x 15

452

(d) yx

5

2

3

(e) 22

2

2 qp

ppqp

(f) yxy

2

9

4

(g) 2

3

3

5

10 t

w

z

t

(h) yx

yx

xy

yx23

22 22

3

2. If F Newton's force, m kg mass and a ms-2

acceleration are connected with the formula

maF , determine the acceleration if a force of 2 kN was imposed on a mass of 1000 kg.

3. Given the kinetic energy as 2

2

1mvK , where m is the mass in kg and v is the velocity in

ms-1

. Determine the mass if it is being hurled vertically up with a velocity of 20 ms-1

and a

kinetic energy of 1000 Joule.

4. Three resistor R1, R2 and R3 are connected in parallel in an electrical circuit. Determine the

total resistance RT, using the formula below:

321

1111

RRRRT

(a) Calculate RT, if R1 = 5 Ω, R2 = 10 Ω and R3 = 30 Ω.

(b) Calculate R2, if RT = 1 Ω, R1 = 2 Ω and R3 = 6 Ω.

Answer for Exercise 1:

1. a) abc

abc

6

29 2

b)

)1(

12

mm

m

c) xy

y

3

12

d) xy2

15

e) )(2

2

qp

p

f) x9

2

g) t

w

6

2

h)

26

)(

y

yxx

2. m 028.0

3. 22 msa

4. kgm 5



5. a) 3 Ω

b) 3 Ω

1.4 Solve the Quadratic Equations Using Factorization, Formulas And Completing The

Squares.

QUADRATIC EQUATION

A Quadratic equation in a unknown (variable) is one equation have one unknown only and his

unknown's supreme power is 2.

ROOT OF A QUADRATIC EQUATION

Root of a quadratic equation can be found by using three methods:

a) Factorization

b) Quadratic formula

c) Completing the square

a. Factorization

Example:

Solve the quadratic equations below, using factorization method.

a) 2x2 + 13x + 10 = 0

b) 6x2 – 20 = 2x

c) 3x = - 6x2

Solution:

a) Given: 010132 2 xx

Then: 0)5)(32( xx

03 x or 05 x

3x or 5x

b) Given: xx 2206 2

Then: 02026 2 xx

0)42)(53( xx

053 x or 042 x

3

5x or 2x



c) Given: 263 xx

Then: 036 2 xx

0)36( xx

0x or 3x

b. Quadratic formula

If quadratic equation given by form 02 cbxax , then value x can find through quadratic

formula, a

acbbx

2

42

Example:

Solve quadratic equation using quadratic formula method

a. 018219 2 xx

b. 35172 2 xx

c. 0245 2 xx

Solution:

a. 018219 2 xx

Thus: a = 9, b = 21 and c = -18

18

64844121 x

18

108921x

18

3321x

18

3321x or

18

3321

18

12x or

18

54

92

18942121 2 x



3

2x or 3

b. 35172 2 xx then: 035172 2 xx

So, a=2, b=-17, and c=35

)2(2

)35)(2(417)17( 2 x

4

28028917 x

4

917 x

4

317 x

4

317 x or

4

317

5x or 2

7

c. 0245 2 xx

52

2544)4( 2 x

10

40164 x

10

564 x

10

48.74 x



10

48.74 x or

10

48.74

10

48.11x or

10

48.3

148.1x or 0.348

c) Completing the square

Example

Solve the quadratic equation using completing the square

a) 4x2 + 5 = -9x

b) x2 = 4x + 4

c) 2x2 +4x – 8=0

Solution

a) Given 4x2 + 5 = -9x

Step 1: Arrange to the form into cbxax 2

594 2 xx

Step 2: Divide with 4 to make the coefficient of x2 equal to 1

4

5

4

9

4

4 2

xx

4

5

4

92 x

x

Step 3: Add 22 )8

9()

2

1

4

9( to both sides

222 )8

9(

4

5)

8

9(

4

9

xx

22 )8

9(

4

5)

8

9( x

64

8180

64

1)

8

9( 2 x



64

1)

8

9( x

8

1

8

1

8

9x or

8

1

8

9x

8

8x or

8

10x

1x or 4

5x

b) Given 4x2 = 4x + 4


4x2 – 4x = 4

Step 2: Divided by 4 to make the coefficient of x2 equal to 1

4

4

4

4

4

4 2

xx

x2 – x = 1

Step 3: Add 22 )2

1()

2

11( to both sides

222 )2

1(1)

2

1( xx

4

11)

2

1( 2 x

4

5

4

5

2

1x



4

5

2

1x

118.15.0 x

118.15.0 x or 118.15.0 x

618.1x or 618.0x

c) Given 2x2+4x-8 =0


2x2 + 4x = 8

Step 2: Divided by 2 to make the coefficient of x2 equal to 1

2

8

2

4

2

2 2

xx

422 xx

Step 3: Add 22 )1()2

12( to both sides

222 1412 xx

5)1( 2 x

51 x

51x

236.21x

236.21x or 236.21x

236.1x or 236.3x

Exercise:

1. Solve quadratic equation

i. by factoring

(a) 0862 xx

(b) 05143 2 xx

(c) 04133 2 xx

ii. by quadratic formula



(a) 0162 2 xx

(b) 0935 2 xx

(c) 1092

1 2 xx

iii. by completing the square

(a) 0762 xx

(b) 062 2 xx

(c) 01263 2 xx

Answer:

1. i. By factoring

(a) 4,2 xx

(b) 5,3

1 xx

(c) 4,3

1 xx

ii. By quadratic formula

(a) 177.0,823.2 xx

(b) 675.1,075.1 xx

(c) 05.19,05.1 xx

iii. By completing the square

(a) 1,7 xx

(b) 0,3 xx

(c) 236.3,236.1 xx



1.5 To solve Simultaneous Linear Equation with two variables using the Elimination and

Subtitution Methods.

A linear equation is an equation in one or more variables where each term's degree is not

more than 1. That means a variable x may appear, but neither any higher power of x, such as

x2, nor any product of variables, such as xy, may appear. It has to be a pretty simple equation

such as ax + by + cz = d ( Example : 3x + 2y - 5z = 8 ). Any other equation in other

degree is known as a non-linear equation.

A system is just a collection of such linear equations, and to solve a system, look for the values

of the variables which make all the equations true simultaneously. For instance, if x and y are

the variables, then an example system of linear equations is

5x - 2y = 4

x + 2y = 8

There are various ways of solving this system, and they lead to the unique solution where x = 2

and y = 3. We'll look next at two common algorithm for solving systems of simultaneous

equations called substitution and elimination.

Substitution method

This method is used when one of the variables is the subject of the equation and the

coefficient of the variables is 1.

Example:

a) nm 2 and 9 nm

b) 32 yx and 732 yx

Steps for the substitution method:

Step 1: Express one variable in terms of the other variable by rearranging one of the

equations.

Step 2: Substitute the expression into the other equation.

Step 3: Solve the value of the chosen variable from the resulting equation.

Step 4: Find the value of the other variable.



Example 1

Given a linear equation of : 3x – 4y = 24

Use y as subject.

Solution: 3x – 4y = 24

– 4y = 24 – 3x

y = 4

324

x

Example 2

Solve the linear equations below, using the substitution method.

Solution:

Step 1 :

Choose equation (1), and use y as the subject.

2x – y = 7

– y = 7 – 2x

y = – 7 + 2x ………… (3)

Step 2 :

Substitute equation (3) into equation (2).

3x + 2y = 14

3x + 2 (-7 + 2x) = 14

3x – 14 + 4x = 14

7x = 14 + 14

7x = 28

x = 7

28

So, x = 4

Step 3 :

Substitute the value x = 4 from step (2), into equation (3).

y = –7 + 2(4)

= –7 + 8

y = 1

Thus, x = 4 and y = 1.

2x – y = 7 (1)

3x + 2y = 14 (2) Let the equations to be equation (1) and (2) for easy reference.



Example 3

Solve the simultaneous equations below, using the substitution method.

523 yx

1252 yx

Solution:

Let, 523 yx ……………..(1)

1252 yx ……………(2)

Let us choose equation (2) and use variable y as the subject.

2x + 5y = 12

5y = 12 – 2x

5

212 xy

………..(3)

Substitute 5

212 xy

into equation (1) and solve the value of x.

3x – 2 55

212

x

2542415 xx

19 x = 25 + 24

= 49

x = 19

49

Then substitute the value of x =19

49 into equation (3)

y = 5

212 x

= 5

19

49212

= 5

19

9812

= 5

1998228

Multiplies 5 to both sides of the equation to eliminate the

denominator 5.



= 5

1

19

130

= 95

130

y = 19

26

Therefore, value of x = 19

49 and value of y = 19

26

Elimination method

This method is suitable when the substitution method involves an awkward fraction. To

eliminate, remove one of the variables to give an equation with only one variable.

Example:

5

52

yx

and 82 yx

Steps involved in elimination method of solution:

Step 1: If necessary, multiply either one or both equations by a suitable non-zero number to

make the coefficients of one of the variable equal.

Step 2: Add or subtract the equations to eliminate one of the variables.

Step 3: Solve the resulting equation in one unknown.

Step 4: Find the value of the other variable by substitution.

Example 1

Solve the linear equations below:

x – y = 2

x + y = 6

Solution

x – y = 2 ………….(1)

x + y = 6 ………….(2)

Step 1 :

Choose the variable to be eliminated. Multiply the equation with any suitable number, so that

the variable’s coefficient to be eliminated is the same.



For this example, variable x is selected to be elimunated. Therefore equation (1) – (2):

Equation (1) – (2): x – y = 2

(–) x + y = 6

_____________

– 2 y = – 4

y = 2

4

y = 2

Step 2 :

Then substitute the value of y = 2 into the equation (1) to obtain of value x.

x – 2 = 2

= 2 + 2

x = 4

Therefore, value x = 4 and y = 2

Example 2

Using the Elimination Method, solve the linear equations below:

2x – y = 7

3x + 2y = 14

Solution:

2x – y = 7……………..(1)

3x + 2y = 14…………..(2)

Choose the variable y for elimination. Therefore, equation (1) needs to multiply

with 2.

Equation (1) 2 : 4x – 2y = 14………..(3)

3x + 2y = 14……….(2)



Add equation (3) to equation (2) for the elimination of variable y.

Equation (3) + (2): 7x = 28

x = 7

28

So, x = 4

Substitute x = 4 into equation (2), to obtain the value for y :

Equation (2): 3x + 2y = 14

3(4) + 2y = 14

12 + 2y = 14

2y = 14 – 12

y =2

2

y = 1

Exercise

1. Solve the simultaneous linear equations below, using substitution method.

3 yx and 132 yx

2. Solve the simultaneous linear equations below, using elimination method.

732 nm and 952 nm

3. Solve the simultaneous linear equations below:

(a) 12 yx

532 xy

(b) 325 xy

432 xy

(c) 52 yx

62 yx

(d) 92 yx

43 yx

(e) 452 yx

73 yx

(f) 432 ed

125 ed



Answer:

1. 1,2 yx

2. 1,2 nm

3.

(a) 1,1 yx

(b) 19

17,

19

14 yx

(c) 3

4,

3

7 yx

(d) 7,1 yx

(e) 2,3 yx

(f) 2,1 ed

ba101 engineering mathematic chapter 1 basic algebra

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