LEAVING CERT ALGEBRALEAVING CERT ALGEBRA

1. SIMPLIFY

Simplifying Expressions

Substitution

Squaring Rule

Division in Algebra

SUMMARY OF THE SECTIONS IN L.C. ALGEBRA

Surds

2. FACTORS

Common (grouping)

Quadratic

Difference of two squares

Sum and difference of two cubes

3. FUNCTIONS AND EQUATIONS

Linear

Quadratic

Cubic

Simultaneous

2 unknowns

Non-linear

Express in terms of

5. INEQUALITIES

Linear

Quadratic from Graphs

Single

Double

6. INDICES

SECTION 1 SIMPLIFYING

2Example 1. Simplify ( 3)( 2 )x x x

3 2 22 3 6x x x x

2 2( 2 ) 3( 2 )x x x x x

2( 3)( 2 )x x x

3 25 6x x x

Simplifying Expressions

2Example. 2 Simplify (3 -1)(1- 2 ) -5( - 1) x x x x

2 23 - 6 -1 2 -5 5 -5x x x x x 2-11 10 - 6x x

2(3 -1)(1- 2 ) --5( 1)x x x x

x 3 4Example 3. Simplify (Express as a single fraction)

2 5

x

3 4

2 5

x x

When dealing with fractions always get a common denominator.

5 3 2( 4)

(2)(5)

(x ) x

5 15 2 8

10

x x

3 23

10

x

2

3 5Example 4. Simplify (Express as a single fraction)

2 2x x

2

3 5

2 2x x

When dealing with fractions always get a common denominator.

2

3 5

2 2x x

2

2

3 2 5( 2)

( - 2)( 2)

(x ) x

x x

2

2

3 6 5 10

( - 2)( 2)

x x

x x

2

2

3 5 16 ( - 2)( 2)

x x

x x

Substitution2 2 -3 1

Example 1. Find the value of when 3 4 2

x xx

1 Substitute in for every .

2( ) 2 2( ) 3

3 4

x

12

12

122 1 3

3 4

2.5 2

3 4

2.5 2

3 4

0.833 0.5

1.333

Calculator

1 1Example 2. Find the value of when and 2

1 2

a ba b

a b

1

Substitute for every and 2 for every 2

( ) ( ) 1

( ) ( ) 1

a b

1

2 21

2 2

12

12

2 1

2 1

12

12

1 2

3

12

12

3

1 7

2 2

1 2

2 7

1

7

Calculator

Squaring2Example 1. Simplify ( )x y

2( )x y

( )( )x y x y

2x xy yx2y

2 22x xy y

x ( x + y ) + y ( x + y )

Simplifying SurdsWhen simplifying surds we use the following :

2224248 Example:

.

baab 1.

b

a

b

a2.

5

3

25

3

25

3Example:

bbb 3. 555 Example:

Only like surds can be added or subtracted.4. 3232 Example:

3 7 7 2 7 Example:

5. Multiplying surds 2 3 5 2 10 6 Example:

Example: (3 2)(3 2) 9 3 2 3 2 2 2

= 9 2

= 7

6. Irrational Denominator Example: 2

Simplify 3

Irrational Denominator

2 3 2 3

33 3 Rational Denominator

Example: 3

Simplify 1 3

Irrational Denominator

3 1 3

1 3 1 3

3 3 3

1 3

3 3 3

2

Rational Denominator

2Example 1. Simplify

1x

2 1

1 1

x

x x

2 1

1

x

x

1Example 2. Simplify

1

x

x

1 1

1 1

x x

x x

1

1

x x x x

x

SECTION 2 FACTORSType 1 Common Factor (Grouping)

2Example 1. Factorise 3 3x x xy y

2 3 3 Group like termsx x xy y 2 ( 3 ) ( 3 )x x xy y

( 3) ( 3)x x y x

( 3)( )x x y

Type 2 Quadratic Factors

2Example 1. Factorise 3 2x x

Method 1 Brackets Method 2 Big X Method 3 Guide Number

2 3 2x x

( )( )x x2 1

( 2)( 1)x x

2 3 2x x x

x

2

1

( 2)( 1)x x

21 3 2x x

Guide Number ( )( )1 2 = 2Factors for Guide Number

2 1 21 3 2x x

21 2 1 2x x x 2(1 2 ) (1 2)x x x ( 2) 1( 2)x x x ( 2)( 1)x x

2Example 2. Factorise 12 7 10x x

Method 1 Brackets

212 7 10x x

( )( )4x 3x5 2

(4 5)(3 2)x x

4x

3x

5

2

Guide Number ( )( ) =12 10 120

Factors for Guide Number

120 1, 60 2, 40 3

30 4, 24 5, 2

15 8

0 6

, 12 10.

2(12 15 ) (8 10)x x x 3 (4 5) 2(4 5)x x x

( 2)( 1)x x

Method 2 Big X

212 7 10x x

(4 5)(3 2)x x

Method 3 Guide Number

2 712 10x x

212 15 8 10x x x

212 7 10x x

Type 3 Difference of Two Squares2 2 = ( )( )x y x y x y

2Example 1. Factorise 16 25x

216 25x 2 2(4 ) (5)x

(4 5)(4 5)x x

2Example 2. Factorise 1 16a

21 16a2 2(1) (4 )a

(1 4 )(1 4 )a a

Quadratic EquationsQuadratic equations have two solutions (roots).

2Example 1. Solve for if 3 2 0x x x Method 1 Using Factors

2 3 2 0x x ( 2)( 1) 0x x 2 0 or 1 0x x

2 or 1x x

Method 2 Using Quadratic Formula2 3 2 0x x

2 4

2

b b acx

a

1a 3b 2c 2( ) ( ) 4( )( )

2( )x

13 23

1

3 9 8 3 1

2 2x

3 1 3 1 or

2 2x x

1 or 2x x

2Example 2. Solve for if 4 0x x x

Method 1 Using Factors2 4 0x x ( 4) 0x x

0 or 4 0x x

0 or 4x x

Method 2 Using Quadratic Formula2 4 0x x

2 4

2

b b acx

a

1a 4b 0c 2( ) ( ) 4( )( )

2( )x

14 04

1

4 16 0 4 4

2 2x

4 4 4 4 or

2 2x x

4 or 0x x

2Example 2. Solve for if 4 0x x

Method 1 Using Factors2 4 0x

( 2( 2) 0x x 2 0 or 2 0x x

2 or 2x x

Method 3 Using Quadratic Formula2 4 0x

2 4

2

b b acx

a

1a 0b 4c 2( ) ( ) 4( )( )

2( )x

10 40

1

0 0 16 0 4

2 2x

0 4 0 4 or

2 2x x

2 or 2x x

Method 2 Using 2 4 0x 2 4x

4x 2x

2 or 2x x

Example 1. Solve 2 5 3

3 - 2 7

x y

x y

72y- 3x

35y 2x

5310y-15x

610y4x

1941

x

41 x19

195-

y

Method 1

3 52 3 5

2

3 2 7

yx y x

y

Method 2

Simultaneous Equations

Type 1 Two Linear.

3 3 5 4 149 15 4 14 19 5

5

19

53 5

19

2

41

19

y yy y

y

y

x

x

3 5

2

y

2 2Example 1 Solve 2 - 14

1

x y

x y

Type 2 One Linear, One Non Linear.

2 22 - 14

1

x y

x y

1x y 2 22( ) - 14y 1y

2 22( +1+2 ) 14y y y

2 22 +2 + 4 14y y y 2 + 4 12 0y y

( 6)( 2) 0y y

6 0 or 2 0y y

6 or 2y y 5x 3x

This rearranging is often called “changing the subject of the formula” or “express in terms of ”.

Express in terms of.

Example 1 Make the subject of the formula 3 y ax y z

3 ax y z

3 ax ax y z ax

3 y z ax

3

3 3 3

y z ax

3 3

z axy

2Example 2 If 2 express in terms of , and . v u as a s u v

2 2v u as

22 2( ) 2v u as

2 2 2v u as

2 2 2 2 2v u u u as 2 2 2v u as

2 2 2

2 2 2

v u as

s s s

2 2

2 2

v ua

s s

SECTION 5 INEQUALITIES

-1x

-33x

sign. a also is symbol inequality The

signs

33

693

963

Change

x

x

x

Single.

Example 1 Show on a numberline the solution set of 3 6 9, .x x R

1 0- 2 - 1- 3- 4 2 3 4 5

Double.

Example 1 Show on a numberline the solution set of 5 3 1 7, .x x R

3 1 1 7 1x 3 1 7x

3 6x

2x

bits. twointo upSplit

5 5 3 1 5x 0 3 6x

6 3x 2 x

2x

5 3 1x

1 0- 2 - 1- 3- 4 2 3 4 5

5 3 1 7x

5 3 1 7x

5 3 1 7x

’04, LCO, Paper 1

3( -8 -5)

3(-13)

Value is -39

3(2(-4) – 5)

2. (a) Find the value of 3(2p – q) when p = -4 and q = 52. (a) Find the value of 3(2p – q) when p = -4 and q = 52. (a) Find the value of 3(2p – q) when p = -4 and q = 5

Method: Get a common denominator

1

1

x

x

( ) 1 1

(1)

x x

x

(1) xcommon denominator =

1x

x

Method: Use previous answer and cancel

2 1

1

xx

x x

2 1

1

x x

x x

2

Method: Use previous answer and solve

2 3x

2 3 x

5 x

Method: Isolate x

ax b c

Step 1: Take b from both sides

ax c b

c bx

a

Step 2: Divide both sides by a

Method: Solve the inequality and then select all appropriate integers for the set

3 2 4x

3 4 2x 3 6x

2x

............, 3, 2, 1,0,1,2

1 2 3 4 5 6 70-1-2-3-4-5 8-6-7-8

Remember the set of

integers Z contains all

positive and negative whole

numbers and zero.

A =

1 35

2

x

1 3 5 2x 1 3 10x

3 10 1x 3 9x

3x

3x

B 2, 1,0,1,2,3,4,............

1 2 3 4 5 6 70-1-2-3-4-5 8-6-7-8

Multiply both sides by 2

Take 1 from both sides

Divide both sides by 3

Multiply both sides by -1

Remember this will change the direction of the inequality

List the solution set

Or show the solution set on the number line

............, 3, 2, 1,0,1,2 2, 1,0,1,2,3,............

A B= 2, 1,0,1,2

A B=

1 2 3 4 5 6 70-1-2-3-4-5 8-6-7-8

6 12x 5 5x

6 12 5 5x x

6 5 12 5x x

7x

3(2 4)x 5( 1)x

2006 Paper 1: Question 2

Solve simultaneously between Equation 1 and Equation 2 to find the values of a and b

2 15 Equation 1 a b

12 Equation 2 a b

3 27a

( 1592 ) b

18 15b

15 18b

3b

9a

2006 Paper 1: Question 3

2

ab c 3a 2

3b 1c

23( )( )3 (

2

1)

2 1

2

1

2

2 2

2 10

20

x y

x y

10 2x y

2 2( 20 2 )1 0y y 2 2100 4 4 200 yy y

2 80 05 40y y 2 08 16y y

4) 0( 4)(y y

4 0 or 4 0y y

One solution 4y

Therefore line is a tangent.

10 2 )4(x

2x

3 2Simplify by multiplying both sides by ( 2) ( 2) ( 2)

2

xx x x x

x

2 2 3 2x x x 2 4 3 0x x

2 4

2

b b acx

a

1a 4b 3c

24 4( ) ( ) 4( )( )

2( )

1 3

1x

4 16 12

2x

4 28

2

4 2 7

2

x

x

2 7x