Chapter 1 Basic Algebra

8/3/2019 Chapter 1 Basic Algebra

1/37

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 forvariables (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 wroteAl-jabr wal 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, ...) whichreplaced 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


2/37



algebra is a rational number usually stated in the form ofq

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


3/37



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

bb)

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 andthe 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

12

2

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


4/37



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 is45

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

147

147

bb

aa

baba

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,


5/37



)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 of23

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

tt

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


6/37



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


7/37



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

xx

a 1

1

)( 2

2

x

xx

b

1

1)(

2

2

xx

xc

1

1)(

2

2

x

xd

Solution:

For1

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


8/37



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) tsr43

6

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


9/37



2. Given that a=1, b=-2, x=-1 and y=3,find the value of


10/37



3. axxyba 22

2


11/37



4.


12/37



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 havethe 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


13/37



Example 1:

Calculate5

1

5

3

Solution:

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


14/37



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.


15/37



15

1

360

24

940

83

9

8

40

3

Exercise 1: Simplify23

32

10

)8)(5(

ts

tsst

Solution: 223

43

23

32

410

40

10

)8)(5(t

ts

ts

ts

tsst

Exercise 2: Simplify22

2

)3(

12

ab

ba

Solution:342

2

22

2

3

4

9

12

)3(

12

bba

ba

ab

ba

Exercise 3: Simplifym

mnnm

2

622

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 denominatorx.

x

x

x

1313

Change the below expression into a single fraction with denominatorx.

x

x

x

454

5

Thus, the question has become:


16/37



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

Thex'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 byx, everything will be simplified.

x

x

x

x

x

x

45

13

)45

(

)1

3(

1.3Conversion 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


17/37



Example 1:

Given a = c d, make das 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 ofx needs to be 16 to make the equation true.

Checkwith the given question:

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

MULTIPLICATION

Example 1:

IfV = IR, makeIas the subject.

Solution

Given V = IR.


18/37



Divide both sides of the equation withR,

Then:R

IR

R

V

IR

V

or RV

I

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 tas the subject for formulat

sv .

Solution

Givent

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


19/37



Combination of Arithmatic Operations

Example 1:

Makexas subject for formula cmx .

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,

thenm

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 withI,

I

VE

I

Ir

So,I

VEr

Example 3

Solve 5 (x + 2) = 5x


20/37



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


21/37



Move BC to the left side of the equation,

AC BC =BD

Factorization ofC,

C (A B) =BD

BA

BDC

BABD

BA

BDC

or

AB

BDC

Exercises

Example 1:

Make das the subject for4

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

24 dh

V

Get the square root of both sides,

24 dh

V


22/37



dh

V

4

Example 2:

The area of a circle is given by A = r2

. Find the value of r.

Solution

Making ras the subject, we have,

r2 = A

Divide both sides with :

Ar 2

So,

Ar

Problem Solving

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

= 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


23/37



(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

wt

(h)yx

yx

xy

yx23

22 22

3

2. IfF 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

1

mvK , where m is the mass in kg and v is the velocity inms-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 thetotal resistance RT, using the formula below:

321

1111

RRRRT

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

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

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


24/37



5. a) 3 b) 3

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) Factorizationb) Quadratic formulac) Completing the square

a. Factorization

Example:

Solve the quadratic equations below, using factorization method.

a) 2x2 + 13x + 10 = 0b) 6x2 20 = 2xc) 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


25/37



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

108921 x

18

3321 x

18

3321 x or

18

3321

18

12x or

18

54

92

18942121 2 x


26/37



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 or2

7

c. 0245 2 xx

52

2544)4( 2 x

10

40164 x

10

564 x

10

48.74 x


27/37



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 = -9xb) x2 = 4x + 4c) 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


28/37



64

1)

8

9( x

8

1

8

1

8

9x or

8

1

8

9x

8

8x or

8

10x

1x or4

5x

b) Given 4x2 = 4x + 4

Step 1: Arrange to the form into cbxax 2 4x

2 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


29/37



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

Step 1: Arrange to the form into cbxax 2 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


30/37



(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


31/37



1.5To 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 notmore than 1. That means a variablexmay appear, but neither any higher power ofx, such as

x2

, nor any product of variables, such asxy, may appear. It has to be a pretty simple equationsuch as ax + by + cz = d ( Example : 3x+ 2y - 5z= 8 ). Any other equation in other

degree is known as a non-linear equation.

Asystem is just a collection of such linear equations, and to solve a system, look for the valuesof the variables which make all the equations true simultaneously. For instance, ifx 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 = 2and y = 3. We'll look next at two common algorithm for solving systems of simultaneousequations calledsubstitution 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 x

Steps for thesubstitution 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.


32/37



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 beequation (1) and (2) foreasy reference.


33/37



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)

Substitute5

212 xy

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

3x 2 55

212

x

2542415 xx

19x = 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 t o both sides of the

equation to eliminate the

denominator 5.


34/37



=5

1

19

130

=95

130

y =19

26

Therefore, value ofx =19

49 and value of y =19

26

Elimination method

This method is suitable when the substitution method involves an awkward fraction. Toeliminate, remove one of the variables to give an equation with only one variable.

Example:

5

52

yx

and 82 yx

Steps involved in eliminationmethodof solution:

Step 1: If necessary, multiply either one or both equations by a suitable non-zero number tomake 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 variables coefficient to be eliminated is the same.


35/37



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

Equation (1) (2): x y = 2() x +y = 6

_____________

2y = 4

y =2

4

y = 2

Step 2 :

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

x 2 = 2

= 2 + 2x = 4

Therefore, value x= 4 and y = 2

Example2

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 variabley for elimination. Therefore, equation (1) needs to multiply

with 2.

Equation (1) 2 : 4x 2y = 14..(3)3x + 2y = 14.(2)


36/37



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 = 1Exercise

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

3yx 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 x

(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


37/37


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

Chapter 1 Basic Algebra

Documents

Transcript of Chapter 1 Basic Algebra