1.1 Algebra. Note 1: Adding and Subtracting Terms Algebraic terms have numbers and variables that...

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


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


Examples of LIKE TERMS:

2x, 9x, 0.74x, ¾ x, 300x

2x2y, -6x2y, 15x2y, 0.9x2y

3b, -2b, ½ b, 100b


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


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


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


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


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



QUADRATIC EXPANSION


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


-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


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


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


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’.


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


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



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


372 2 xx

362 2 xxx

)12(3)12( xxx

)3)(12( xx

372

362)3)(12(2

2

xx







1 x 6

632


2032 2 xx

20852 2 xxx

)52(4)52( xxx

)4)(52( xx

2032

20852)4)(52(2

2

xx







5 x -8

40202


1816 2 xx

14416 2 xxx

)14(1)14(4 xxx

)14)(14( xx

1816

14416)14)(14(2

2

xx


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

1.1 Algebra. Note 1: Adding and Subtracting Terms Algebraic terms have numbers and variables that...

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