LEAVING CERT ALGEBRA
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Transcript of LEAVING CERT ALGEBRA
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LEAVING CERT ALGEBRALEAVING CERT ALGEBRA
1. SIMPLIFY
Simplifying Expressions
Substitution
Squaring Rule
Division in Algebra
SUMMARY OF THE SECTIONS IN L.C. ALGEBRA
Surds
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2. FACTORS
Common (grouping)
Quadratic
Difference of two squares
Sum and difference of two cubes
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3. FUNCTIONS AND EQUATIONS
Linear
Quadratic
Cubic
Simultaneous
2 unknowns
Non-linear
Express in terms of
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5. INEQUALITIES
Linear
Quadratic from Graphs
Single
Double
6. INDICES
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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
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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
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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
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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
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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
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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 )
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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’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
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Method: Get a common denominator
1
1
x
x
( ) 1 1
(1)
x x
x
(1) xcommon denominator =
1x
x
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Method: Use previous answer and cancel
2 1
1
xx
x x
2 1
1
x x
x x
2
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Method: Use previous answer and solve
2 3x
2 3 x
5 x
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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
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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 =
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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
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............, 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
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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
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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
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2006 Paper 1: Question 3
2
ab c 3a 2
3b 1c
23( )( )3 (
2
1)
2 1
2
1
2
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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
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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