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1.1 Algebra
Note 1: Adding and Subtracting Terms
Algebraic terms have numbers and variables that are sometimes raised to an exponent
8x3
Coefficient is 8
Variable is x
Exponent is 3To add/subtract terms they must have the same variable and the same exponent
Note 1: Adding and Subtracting Terms
Terms with the same variable and same exponent are called Like Terms
8x3
Coefficient is 8
Variable is x
Exponent is 3To add/subtract terms they must have the same variable and the same exponent
Note 1: Adding and Subtracting Terms
Examples of LIKE TERMS:
2x, 9x, 0.74x, ¾ x, 300x
2x2y, -6x2y, 15x2y, 0.9x2y
3b, -2b, ½ b, 100b
Note 1: Adding and Subtracting Terms
Variables that have different exponents are not the same variable therefore variables can only be added if the variables are the same.
e.g. 3r + r2 = 3r + r2
e.g. 4r + r2 + r + 2r2
= 5r + 3r2
e.g. 6b + 3r2 + 4r – b + 6b2
= 6b – b + 6b2 + 3r2 + 4r = 5b + 6b2 + 3r2 + 4r
Start by identifying like terms
GAMMA Ex 1.03 Pg 7
Note 2: Exponent Rules
To multiply algebraic terms, we multiply the coefficients together and add the exponents.
Remember : x is the same as 1x y is the same as y1
am x an = am+n
e.g. 3 4r = 3 4 r = 12r e.g. b2 b2
= b4
e.g. 2x2y 4y2
= 8x2y3
1.) Index the number. 2.) Multiply each variable index by the index outside
the brackets.3.) If the bracket can be simplified, do this first.
Note 2: Exponent Rules
e.g. Simplify
(2x2)3
= 23 x23
= 8x6
(-4h2g6)2
= 16h4g12
22
3
12
x
x
= (4x)2
= 16x2
When simplifying expressions with square roots:1.) Square root the number.2.) Half the power of each variable.
Note 2: Exponents Rules
e.g. Simplify
12k25
= 5 k122
= 5 k6
GAMMA Ex 1.02 Pg 4-5Ex 1.04 Pg 10
681k
= 9 k62
= 9 k3
2625k
= 25 k22
= 25 k
Note 3: Algebraic FractionsWhen dividing algebraic terms:Simplify the numbers by using the fraction button on your calculator.Subtract the powers for each of the different base variables.
Remember : x is the same as x1
am = am-n
an
e.g. = 2 e.g.
= 2a2
e.g. =
3
6
4
6
4
8
a
a
12
6a
2
a
e.g. 6
4
12
6
s
s
ssssss
ssss
12
6
22
1
s
=
=
Note 3: Algebraic FractionsMultiply these fractions and write in simplest form
Remember : x is the same as x1
am = am-n
an
y
x3
7
2
4
8
b
a
7
32
12
16
ab
ba
43
4
b
a
e.g. xy
x5Multiply the numerators and denominators
= 2
215
y
x
e.g. xa
b
3
2 3
=
=
Note 3: Algebraic FractionsDivide these fractions and write in simplest form
Remember : x is the same as x1
am = am-n
an
y
x3
4
2
2
5
y
x
e.g. ÷x
y
5
6 3To divide; we multiply by the reciprocal
= 36
5
y
x
e.g. ÷
4
2
6
15
y
x=
=
y
x3 x
2
4
x 3
2x
= 2
4
x x
x2
3
= 32
12
x
= 3
6
x
GAMMA Ex 7.01 Pg 90Ex 7.02 Pg 91Ex 7.03 Pg 92
Starter3
1
2
1+
= +6
2
6
3
=6
5
5
3x− 7
2 y
= −57
37
x
75
25
y
= −35
21x
35
10y
= 35
1021 yx
2
4
x+ 22
3
x
= +22
8
x 22
3
x
= 22
11
x
Multiply the number on the outside by each of the numbers/variables on the inside of the brackets
Note 4: Expanding Brackets
e.g. Expand
7(2x – 3)= 7 2x + 7 -3
= 14x - 21
x(5x – 2)= x 5x + x -2
= 5x2 –2x
5(2a + 3) + 3(a – 4)
= 10a + 15 + 3a – 12= 10a + 3a + 15 – 12
= 13a + 3
QUADRATIC EXPANSION When we expand two brackets we use:
F – first (multiply the first variable or number from each bracket)O – outside (multiply the outside variables together)I – inside (multiply the two inside variables together)L – last (multiply the last variable in each bracket together)
Simplify, leaving your answer with the highest power first to the lowest power (or number) last.
Note 4: Expanding Two Brackets
e.g. (x + 4) (x – 2) F O I L
= x2 - 2x + 4x - 8
= x2 + 2x - 8
QUADRATIC EXPANSION
Note 4: Expanding Two Brackets
e.g. (x + 3) (x – 5)
= x2 + 3x - 5x - 15= x2 - 2x - 15
e.g. (x + 10) (x + 1)
= x2 + 10x + x + 10= x2 + 11x + 10
e.g. (x - 3) (x – 8)
= x2 - 3x - 8x + 24= x2 - 11x + 24
e.g. (x - 4) (x + 4)
= x2 - 4x + 4x - 16= x2 - 16
•Notice the middle term cancels out
DIFFERENCE OF SQUARES
QUADRATIC EXPANSION
Note 4: Expanding Two Brackets
e.g. (x + 7) (x – 9)
= x2 + 7x - 9x - 63= x2 - 2x - 63
e.g. (x – 5) (x + 4)
= x2 - 5x + 4x - 20= x2 - x - 20
e.g. (x - 2) (x – 6)
= x2 - 2x - 6x + 12= x2 - 8x + 12
e.g. (x - 9) (x + 9)
= x2 - 9x + 9x - 81= x2 - 81
•Notice the middle term cancels out
DIFFERENCE OF SQUARES
QUADRATIC EXPANSION
Note 4: Expanding Two Brackets
e.g. (2x + 3) (x – 5)
= 2x2 + 3x - 10x - 15= 2x2 - 7x - 15
(x + 5) (x + 5)
= x2 + 5x + 5x + 25= x2 + 10x + 25
e.g. 5(3x + 4) (2x – 1)
= 5[6x2 + 8x - 3x - 4]
= 30x2 + 25x - 20
e.g. (x + 5)2
= 5[6x2 + 5x - 4]
GAMMA (odd)Ex 2.01 Pg 12-13Ex 2.02 Pg 14Ex 2.03-2.04 Pg 15
NuLake Pg 14,23-25
Note 5: Solving Linear EquationsTo solve simple equations, do the opposite operation to what is happening to x
Solve
x + 5 = 7
e.g.
x = 2
= -3
5
42x =
2
x
x = 14
x = -6
x = 7 - 5 5x = 42 x - 7 = 7
x = 7 + 7
x = -3 x 2
Note 5: Solving Linear EquationsTo solve 2 stage simple equations, do the opposite operation to what is happening to x (aim to get x on the LHS)
Solve
6x + 5 = 17
e.g.
6x = 12
5
x
= 12
2
3x
= 9 + 3
x = 60
6x = 17 - 5
x = 12 x 5
= 7
x + 3 = 7 x 2
- 3 = 9
x = 2
Check that your answer works in the ORIGINAL equation
x + 3 = 14 x = 14 - 3x = 11
5
x
5
x
Note 5: Solving Linear EquationsSolve equations with like terms. Collect x terms on the LHS and collect numbers on the RHS
Solve
6x + 4 = 4x - 6
e.g.
2x = -10 -4x – 5x = -23 – 22
-4x + 22 = 5x − 23 6x – 4x = -6 - 4
x = 5
5x − 1 = 7x + 9 5x – 7x = 9 + 1
25 – 9x – 3 + 5x = 7x – 23 -2x
x = -5
Check that your answer works in the ORIGINAL equation
-2x = 10 x = -5
-9x = -45
Note 5: Solving Linear EquationsWhen equations have brackets, expand the brackets first and then solve the equation
Solve
6(x – 1) = 12
e.g.
6x = 18 -14x = -44
56 – 8x = 6x + 12
GAMMA - odd onlyEx 4.01 Pg 44 Ex 4.02 Pg 45Ex 4.03 Pg 46 Ex 4.04 Pg 46
6x – 6 = 12
x =
6x -2 + 2(x+5) = 06x – 2 +2x + 10 = 0
8(7 – x) = 6(x + 2)
x = 3
Check that your answer works in the ORIGINAL equation
8x + 8 = 0 8x = -8
x = x = -1 14
44
7
22
Startera
1
ab
2+
= +ab
b
ab
2
=ab
b 2
4x + 2 – (x – 3) = 0
3x + 5 = 0
5x = 6x + 21 – 8x
4x + 2 – x + 3 = 0
Simplify Solve for x
3x = -5
x = 3
5
5x = -2x + 21
Simplify…..Simplify……& Simplify!
5x + 2x = 21
7x = 21
x = 3
Note 6: Solving equations with fractions
When solving equations involving fractions, multiply both sides of the equation by a suitable number to eliminate the fractions.
2
3x= 7
32 x * Multiply both sides by 14
2
)3( x7
)32( x=14 14
7(x+3) = 2(2x-3)
* It is possible to skip this step by cross-multiplying
* Expand and simplify
7x+21 = 4x - 63x = -27 x = -9
Solving Linear Equations w/ fractionsTry these!
Solve
= 10
e.g.
x – 5 = 4020x = -6
20x + 8 = 2
x – 5 = 10 x 4 4 x 5y = 7(2y+3) 4(5x+2) = 2
x = 45
Check that your answer works in the ORIGINAL equation
20y = 14y + 21
20y – 14y = 21x = 6y = 21
6
21
4
5x
Cross-Multiply!
7
5y = 4
32 y
y =
2
25 x = 4
1
20
6
10
3x =
1
Note 7: Solving Linear InequationsInequations are expressions that include:
< (less than), > (greater than) ≤ (less than or equal to), ≥ (greater than or equal to)
To solve: we need to isolate the unknown variable just like we did for solving linear equations.
e.g. Solve the inequality
One exception: if at any time we multiply or divide by a negative number, we must REVERSE the inequality sign
3x + 8 ≥ 293x ≥ 29 - 83x ≥ 21
x ≥ 7
x ≤ 2
Note 7: Solving Linear InequationsInequations are expressions that include:
< (less than), > (greater than) ≤ (less than or equal to), ≥ (greater than or equal to)
To solve: we need to isolate the unknown variable just like we did for solving linear equations.
e.g. Solve the inequality
One exception: if at any time we multiply or divide by a negative number, we must REVERSE the inequality sign
2x + 10 ≥ 4x + 62x – 4x ≥ 6 - 10
-2x ≥ -4
55x < -205
Note 7: Solving Linear Inequations
e.g.
Try These!
5x + 8 < -12 -3(w – 6) > 9 < 72
32 x
5x < -12 - 8
x < - 4
-3w + 18 > 9
-3w > 9 - 18
-3w > -9
w < 3
-3-3
2 – 3x < 7 x 2
2 – 3x < 14
-3x < 12
x > -4
-3-3
NuLake Pg 20-21
Substitution means replacing a variable with a number.
Note 8: Substitution & Rearranging Formulae
e.g. Calculate the value of these expressions: 7x – 1 when x = 2= 7 2 – 1= 14 – 1= 13
when f = 2 and g = 62gf
= 4
262
=
5x2 - 3x + 2 when x = -3 = 5×(-3)2 - 3 -3 + 2 = 45 - −9 + 2 = 45 + 9 + 2 = 56
Note 8: Substitution & Rearranging Formulae
Rearranging formulae and equations involves manipulating them to make a certain variable ‘the subject’ of the equation (all by itself). Just like how we do when we ‘solve for x’.
Note 8: Substitution & Rearranging Formulae
e.g. Make d the subjectv2 = u2 + 2ad
v2 – u2 = 2ad2a 2a
v2 – u2 = d2a
ad + bd = 5
d (a + b) = 5(a + b) (a + b)
d = 5(a + b)
√π
Note 8: Substitution & Rearranging Formulae
e.g. Make r the subject
A = πr2
A = πr2
π π
A = r2√
π√A = r GAMMA - odd onlyEx 3.01 Pg 25-28Ex 3.04 Pg 39
±NuLake Pg 26-28
QUADRATIC EXPANSION
Starter - Expand
e.g. (x + 7) (x + 2)
= x2 + 7x + 2x + 14= x2 + 9x + 14
e.g. (x – 5) (x + 9)
= x2 - 5x + 9x - 45= x2 + 4x - 45
e.g. (x – 2) (x – 3)
= x2 – 2x – 3x + 6= x2 – 5x + 6
e.g. (x - 8) (x + 8)
= x2 - 8x + 8x - 64= x2 - 64
•Notice the middle term cancels out
DIFFERENCE OF SQUARES
Note 9: FactorisingFactorising is the reverse procedure of expanding.
(x + 7) (x + 2) x2 + 9x + 14
Expanding
Factorising
Note 9: FactorisingLook for common factors first.
e.g. 15ab + 10b= 5b(3a + 2)
e.g. 4x2 + 16x + 8
= 4(x2 + 4x + 2)
e.g. 30a2 – 15a= 15a(2a – 1)
e.g. 3x + 6y – 9z = 3(x + 2y – 3z)
Note 9: FactorisingIf there is no common factor, try splitting the expression into two groups, and look for a common factor in each pair.
e.g. 2ac + 2bc + 3ad + 3bd= 2c(a + b) + (a + b) is the common factor 3d(a + b)= (a + b)(2c + 3d)
= (x + 2y)(3 + 4z)
e.g. 3x + 6y + 4xz + 8yz = 3(x + 2y) + 4z(x + 2y) (x + 2y) is the common factor
e.g. Factorise simple quadratics:
a.)
b.)
x2 + 6x + 8 Find 2 numbers which multiply to 8 and add to 6
(x+2) (x+4)
x2 - 5x - 24
Put these number into brackets4 and 2The factors are (x+2) and (x+4)
Find 2 numbers which multiply to -24 and add to -5-8 and 3 Put these number into brackets(x-8) (x+3)
The factors are (x-8) and (x+3)
c.) x2 + 5x - 24 Find 2 numbers which multiply to -24 and add to 58 and -3 Put these number into brackets(x+8) (x-3)
Note 9: Factorising
The factors are (x+8) and (x-3)
GAMMA - oddEx 2.06 Pg 18 Ex 2.07 Pg 19-20
Expand (x + y)(x – y)
a.)
b.)
a2-16b2
Remember this result
The factors are (a-4b) and (a+4b)
Check your solution by expanding
(a-4b) (a+4b)
= x2 – y2
(a)2-(4b)2
The middle term cancels out
36a3b - 4ab3
4ab(9a2 - b2)4ab[(3a)2 – b2]
4ab [(3a – b)(3a+b)]
There is a common factor of 4ab
Note 10: Factorising – Difference of squares
StarterFactorise
4x2 – 49y2 9x2 – 36y2 144x2 – 9y8
( ) ( )2x – 7y 2x + 7y 9( ) ( ) ( )( )x - 2y x + 2y 12x – 3y4 12x + 3y4
x2 + 6x + 9 x2 – 12x + 36( ) ( )x + 3 x + 3
( )2x + 3
( ) ( )x – 6 x – 6
( )2x – 6
Evaluate
b.)
Factorise
Check your solution
= (81-80) (81+80) a.) 812 – 802
12342-12352
= (1) (161)
= 161
= (1234-1235) (1234+1235) = (-1) (2469) = -2469
Note 10: Factorising – Difference of Squares
GAMMA - oddEx 2.07 Pg 19-20 Ex 2.08 Pg 21 Ex 2.09 Pg 21
Factorisation of quadratic expressions with a ≠ 1
12112 2 xx
12832 2 xxx
)32(4)32( xxx
)4)(32( xx
12112
12382)4)(32(2
2
xx
xxxxxCheck by expanding
Mulitply the coefficient of x2 and the constant.
Find 2 numbers that multiply to give this value and add to give the coefficient of x
Write the quadratic with the x-term split into two x-terms using these numbers
Factorise the pairs of terms
Factorise again, taking the bracket as the common factor
3 x 8
24122
Factorisation of quadratic expressions with a ≠ 1
372 2 xx
362 2 xxx
)12(3)12( xxx
)3)(12( xx
372
362)3)(12(2
2
xx
xxxxxCheck by expanding
Mulitply the coefficient of x2 and the constant.
Find 2 numbers that multiply to give this value and add to give the coefficient of x
Write the quadratic with the x-term split into two x-terms using these numbers
Factorise the pairs of terms
Factorise again, taking the bracket as the common factor
1 x 6
632
Factorisation of quadratic expressions with a ≠ 1
2032 2 xx
20852 2 xxx
)52(4)52( xxx
)4)(52( xx
2032
20852)4)(52(2
2
xx
xxxxxCheck by expanding
Mulitply the coefficient of x2 and the constant.
Find 2 numbers that multiply to give this value and add to give the coefficient of x
Write the quadratic with the x-term split into two x-terms using these numbers
Factorise the pairs of terms
Factorise again, taking the bracket as the common factor
5 x -8
40202
Factorisation of quadratic expressions with a ≠ 1
1816 2 xx
14416 2 xxx
)14(1)14(4 xxx
)14)(14( xx
1816
14416)14)(14(2
2
xx
xxxxxCheck by expanding
4 x 4
16116
2)14( x
GAMMA - oddEx 2.09 Pg 21 Ex 2.10 Pg 22 Ex 2.11 Pg 23
Note 11: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions
e.g.
factorise first!
Solve the equation x2 + x – 12 = 0
If a x b = 0, this means that either a or b (or both) must be zero.
(x-3)(x+4) = 0
either x - 3 = 0 or x + 4 = 0 x = 3 x = -4 2 solutions
Note 11: Quadratic Equations Quadratic equations always have an x2 term and generally have 2 different solutions
e.g.split the x term
Solve the equation 12x2 + 17x – 14 = 0
12x (x+2) -7 (x+2) = 0
either 12x - 7 = 0 or x + 2 = 0
x = x = -2
factorise pairs
12 x -14 = -168
24, -7
12x2 + 24x – 7x – 14 = 0
(12x-7) (x+2) = 0
12
7
Note 11: Quadratic Equations
SOLVE THE EQUATIONS
e.g.factorise
x2 + 9x = 0
either x = 0 or x + 9 = 0 x = -9
x (x+9) = 0
GAMMA - odd Ex 5.01 Pg 65Ex 5.02 Pg 66Ex 5.03 Pg 68
e.g. 9x2 -16 = 0
9x2 = 16
x2 = 9
16
x = 9
16
x = ± 3
4
Use the quadratic formula to solve 2x2 + 7x + 5 = 0
a = b = c = 2 7 5
7 72 (2)(5)
(2)
x = -7 ± √94
x = -7 ± 34
x = -52
or x = -1Two SolutionsIGCSE Ex 28 Pg 73 oddEx 30 Pg 76 odd
StarterA wallet containing $40 has three times as many $1 notes as $5 notes. Find the number of each kind.
Let x be the number of $1 notes and y be the number of $5 notes
1x + 5y = 40
3y = x
x + 5y = 40
x – 3y = 0
(1)(2)
8y = 40 (1) – (2) y = 5
3(5) = x x = 15
There are 15 $1 notes and 5 $5 notes
Note 12: Simultaneous EquationsSubstitution
In order to solve for the value of two unknowns in a problem, you must have two different equations that relate to the unknownsSubstitution is often used when one of the equations contains a single unit quantity of an unknown.
e.g. 2x – 3y = 345x + y = 0
Label the equations(1)(2)
Rearrange for the single unit quantity (y here) y = -5x
Substitute this expression for y into (1) & solve for x 2x – 3(-5x) = 3417x = 34
x = 2 Solve for y by substituting x = 2 into (2)
5(2) + y = 0 y = -10
Note 12: Simultaneous EquationsSubstitutionSubstitution is often used when one of the
equations contains a single unit quantity of an unknown.
e.g. 2x – y = 26x = 4y – 1 Label the equations
(1)(2)
Substitute this expression for x into (1) & solve for y 2(4y – 1) – y = 26
8y − 2 – y = 26
y = 4Solve for x by substituting y = 4 into (2)x = 4(4) – 1
x = 15
GAMMA Ex 6.04 Pg 82Ex 6.05 Pg 83
7y = 28
Note 13: Simultaneous EquationsElimination
This method is often preferred and can be used if substitution is not suitable
e.g. x – y = 110x + y = 21
Label the equations(1)
(2)
We can add/subtract the equations to eliminate either x or y
* notice we have eliminated the y term
11x = 22
2 – y = 1y = 1
Solve for y by substituting x = 2 into either (1) or (2)
x = 2
add to eliminate y in this case
Note 13: Simultaneous EquationsElimination
This method is often preferred and can be used if substitution is not suitable
e.g. x + 2y = 102x + 3y = 14
Label the equations(1)
(2)
Multiply (1) by an appropriate # to eliminate either x or y 2x + 4y = 20
Subtract (2) from (1) – notice we have eliminated the x termy = 6
x + 2(6) = 10x + 12 = 10
Solve for x by substituting y = 6 into either (1) or (2)
x = -2
2x + 3y = 14
x 2 to eliminate x in this case
*try graphing this
x + 2y = 10
2x + 3y = 14Using the Cover-up method
GAMMA Ex 6.01 Pg 78Ex 6.02 Pg 78Ex 6.03 Pg 80
Note 14: Problems Solved using Simultaneous Equations
e.g. A motorist buys 24 L of petrol & 5 L of oil for $10.70, while another motorist buys 18 L of petrol and 10 L of oil for $12.40. Find the cost of 1 L of petrol and 1 L of oil at this garage.
24x + 5y = 107018x + 10y = 1240
Let x be the cost of 1 L of petrol, let y be the cost of 1 L of oil(in cents)
(2)
(1)
*multiply (1) by 248x + 10y = 214018x + 10y = 1240
30x = 900x = 30 *subst into (2)
18(30) + 10y = 1240 y = 70
The first OPEC oil shock occurred in 1973, gas prices doubled from 15 cents /L to 30 cents /L
GAMMA Ex 6.08 Pg 86Ex 6.09 Pg 87