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MATHEMATICS

Grade 8

THE NUMBER SYSTEM

(Natural and whole numbers)

Module 3

ALGEBRA

ALGEBRA

CLASS ASSIGNMENT1

  • Discover ALGEBRA step by step...
  • In Algebra, we make use of letters in the place of unknowns (numbers that we do not know).
  • Letters represent variables (values that may vary) and numbers are the constants (the values remain the same).

Look at the polynomial, for example

Figure 1
Figure 1 (Picture 63.png)

From the above, you will be able to recognise the following:

  • The number of terms (terms are separated by + and - signs): 3 terms
  • Coefficient of xx size 12{x} {}² (the number immediately before xx size 12{x} {}²): 3
  • Coefficient of xx size 12{x} {} (the number immediately before xx size 12{x} {}): - 1414 size 12{ { { size 8{1} } over { size 8{4} } } } {}
  • Constant: 5
  • The degree of expression (highest power of xx size 12{x} {}): 2
  • The expression is arranged in descending powers of xx size 12{x} {}.
  • 3 xx size 12{x} {}² means 3 x xx size 12{x} {}² (3 multiplied by xx size 12{x} {}²)
  • xx size 12{x} {}² means ( xx size 12{x} {}) x ( xx size 12{x} {}) ( xx size 12{x} {} multiplied by xx size 12{x} {})
  • What happens to (+)and (-) signs during multiplication and division?

Here you have it:

  • (+) x of ÷ (+) = (+)
  • (-) x of ÷ (-) = (+)
  • (+) x of ÷ (-) = (-)

1.Study the following in your groups and supply the answers:

( 1 4 x 2 x ) 4 + 6 ( 1 4 x 2 x ) 4 + 6 size 12{ { { \( { { size 8{1} } over { size 8{4} } } x rSup { size 8{2} } ` - `x \) } over {4} } `+`6} {}

  • Indicate the following:

1.1 number of terms

1.2 coefficient of xx size 12{x} {}

1.3 constant

1.4 degree of the expression

2. Now we can use variables to define the following with the magical language of mathematics --- i.e. algebraic expressions.

See if you can define these in the form of algebraic expressions:

Given Algebraic Expression

2.1 The sum of a number and 9

2.2 A number multiplied by 7

2.3 The difference between a and b

2.4 6 less than a number reduced by 7

2.5 The product of a number and b

2.6 Quotient of a number and 7

2.7 Square of a

2.8 Square root of a

2.9 Subtract the difference between a and b from their product

3. The following are referred to as flow diagrams – They consist ofa) inputb) formula in which the input number is substitutedc) output

Complete (a), (b) and (c)

Figure 2
Figure 2 (Picture 67.png)

4. See if you can determine a formula for the following and complete the table.

Table 1
x x size 12{x} {} 2 5 8 10 15 47
y 7 11 17      

formula: y =

HOMEWORK ASSIGNMENT1

1. Determine a formula for each of the following and complete the table.

1.1 formula: y = ……………………………………………………

Table 2
x x size 12{x} {} 2 5 8 9 12 20
y 10 16 22      

1.2 formula: y = ……………………………………………………

Table 3
x x size 12{x} {} 3 7 10 9 12 20
y 12 32 47      

1.3 formula: y = ……………………………………………………

Table 4
x x size 12{x} {} 1 3 4 9 12 20
y 1 9 16      

1.4 formula: y = ……………………………………………………

Table 5
x x size 12{x} {} 1 2 3 6 7 10
y 1 8 27      

1.5 formula: y = ……………………………………………………

Table 6
x x size 12{x} {} 1 2 4 9 12 20
y 2 5 17      

2. The sketch shows matches arranged to form squares and combinations of squares.

Figure 3
Figure 3 (Picture 69.png)

2.1 Make a sketch to show four squares and indicate how many matches were used.

Matches? …………………………

2.2 Can you determine a formula that will provide a quick way for determining how many matches you will need to form ( xx size 12{x} {}) number of squares?

y = ………………………………… (with y representing the number of matches)

2.3 Now make use of your formula to determine how many matches you will need to form 110 squares.

2.4 Determine how many squares you will be able to form with 2 005 matches.

3. Examine the following expression and answer the questions that follow:

1 4 a + a 2 5 + 7 + 3a 3 1 4 a + a 2 5 + 7 + 3a 3 size 12{ - { {1} over {4} } a``+`` { {a rSup { size 8{2} } } over {5`} } ``+`7`+3a rSup { size 8{3} } } {}

3.1 Arrange the expression in ascending powers of a.

3.2 Determine:

3.2.1 number of terms

3.2.2 coefficient of a²

3.2.3 degree of the expression

3.2.4 constant term

3.2.5 the value of the expression if a = -2

4. Write an algebraic expression for each of the following.

4.1 the product of a and p, multiplied by the sum of a and p.

4.2 the sum of a and p, multiplied by 3

4.3 the quotient of a and p multiplied by 3

4.4 the cost of a bus trip is p rand per km. Calculate the cost of the entire, trip if the distance travelled is 45 km.

4.5 5 is added to the product of 3 and a, and the answer is reduced by the sum of 9 and b

5. You rent a car at Cape Town International airport at R 125,50 per day.

5.1 Compile a table to indicate how much it will cost you in hire for the following periods: 6; 7; ..... 12 days.

5.2 Determine a formula for representing the data with y (total cost) and xx size 12{x} {} (number of days).

5.3 What will the total hiring costs for 2½ months come to?

6. How many terms in each of the following expressions?

6.1 ab + m/n - 2(a + b)

6.2 (p + q + r)3 - 4r²

6.3 m/n + 7m² ÷ 5 x p - q x r

6.4 (6 x q) ÷ (r x 7)

6.5 mn - pr - a5mn - pr - a5 size 12{ { { bold "mn - pr - a"} over {5} } } {}

Assessment

Table 7
Assessment of myself:   by myself:   Assessment by Teacher:
I can…     1 2 3 4   Critical Outcomes 1 2 3 4
distinguish between the terms of a polynomial; (Lo 2.4; 2.8.2; 2.9)                     Critical and creative thinking        
identify the coefficient of an unknown; (Lo 2.4; 2.9)                     Collaborating        
identify the constant in a polynomial; (Lo 2.4; 2.9)                     Organising en managing        
determine the degree of an expression; (Lo 2.4; 2.9; 2.8.1)                     Processing of information        
arrange the expression in a descending order; (Lo 2.4; 2.9)                     Communication        
accurately multiply and divide the signs (+ / -); (Lo 2.4; 2.8.4)                     Problem solving        
write algebraic expressions; (Lo 2.4; 2.2; 2.8.4)                     Independence        
determine the formulas for flow diagrams and tables. (Lo 2.1; 2.3; 2.4; 2.7)                              

good average not so good

Table 8
Comments by the learner:     My plan of action:     My marks:
I am very satisfied with the standard of my work.     <   Date:      
I am satisfied with the steady progress I have made.         Out of:      
I have worked hard, but my achievement is not satisfactory.         Learner:      
I did not give my best.     >          
Table 9
Comments by parents:   Comments by teacher:
     
     
     
Parent signature: Date:   Signature: Date:

Assessment

Table 10
Learning outcomes(LOs)
 
LO 2
Patterns Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems, using algebraic language and skills.
We know this when the learner :
2.1 investigates and extends numerical and geometrical patterns to find relationships and rules, including patterns that:2.1.1 are presented in physical or diagrammatic form;2.1.2 are not limited to series with constant difference or ratio;2.1.3 occur in natural and cultural contexts; 2.1.4 are created by the learner him/herself;2.1.5 are presented in tables;2.1.6 are presented algebraically;
2.2 describes, explains and justifies observed relationships or rules in own words or in algebra;
2.3 represents and uses relationships between variables to determine input an output values in a variety of ways by making use of:2.3.1 verbal descriptions;2.3.2 flow diagrams;2.3.3 tables;2.3.4 formulas and equations;
2.4 builds mathematical models that represent, describe and provide solutions to problem situations, thereby revealing responsibility towards the environment and the health of other people (including problems in the contexts of human rights, social, economic, cultural and environmental issues);
2.7 is able to determine, analyse and interpret the equivalence of different descriptions of the same relationship or rule which can be represented:2.7.1 verbally;2.7.2 by means of flow diagrams;2.7.3 in tables;2.7.4 by means of equations or expressions to thereby select the most practical representation of a given situation;
2.8 is able to use conventions of algebraic notation and the variable, reconcilable and distributive laws to:2.8.1 classify terms like even and odd and to account for the classification;2.8.2 assemble equal terms;2.8.3 multiply or divide an algebraic expression with one, two, or three terms by a monomial;
2.8.4 simplify algebraic expressions in bracketed notation using one or two sets of brackets and two types of operation;2.8.5 compare different versions of algebraic expressions having one or two operations, select those that are equivalent and motivate the selected examples;2.8.6 rewrite algebraic expressions, formulas or equations in context in simpler or more usable form;
2.9 is able to interpret and use the following algebraic ideas in context: term, expression, coefficient, exponent (or index), basis, constant, variable, equation, formula (or rule).

Memorandum

CLASS ASSIGNMENT 1

  • 2
  • 1 4 1 4 size 12{ - { {1} over {4} } } {}
    (1)
  • 6
  • 2

2.1 x + 7

2.2 x + 7

2.3 a – b

  • (x + 7) – 6

= x – 13

  • x x bb size 12{b} {} = xbb size 12{b} {}
  • x 7 x 7 size 12{ { {x} over {7} } } {}
    (2)
  • a a size 12{a} {} 2
  • a a size 12{ sqrt {a} } {}
    (3)

2.9 abab size 12{ ital "ab"} {} – ( aa size 12{a} {}bb size 12{b} {})

3.1 ac

7 -4

3.2 ac

9 1212 size 12{ - { {1} over {2} } } {}

4. 21; 31; 95; y=2x+1y=2x+1 size 12{y=2x+1} {}

HOMEWORK ASSIGNMENT 1

  • y = 2x + 6
  • y = 5x – 3
  • y = x2
  • y = x3
  • y = x2 + 1
  • Sketch: (3 x 4) + 1 = 13
  • y = 3x + 1
  • y = 3(110) + 1 = 331
  • (2 005 – 1) ÷ 3 = 668
  • 7 – 14a14a size 12{ { {1} over {4} } a} {} + a25a25 size 12{ { {a rSup { size 8{2} } } over {5} } } {} + 3 a3a3 size 12{a rSup { size 8{3} } } {}
  • 4
  • 1 5 1 5 size 12{ { {1} over {5} } } {}
    (4)
  • 3
  • 7
  • 14211421 size 12{ - { {1} over {4} } left ( - { {2} over {1} } right )} {} + 252252 size 12{ left ( { { - 2} over {5} } right ) rSup { size 8{2} } } {} + 7 3(-2)3

= 1212 size 12{ { {1} over {2} } } {} + 4545 size 12{ { {4} over {5} } } {} + 7 – 24

= 5+8+70240105+8+7024010 size 12{ { {5+8+"70" - "240"} over {"10"} } } {}

= 1571015710 size 12{ { { - "157"} over {"10"} } } {} = -15,7

  • ap + (a + p)
  • 3(a + p)
  • apap size 12{ { {a} over {p} } } {} + 3
  • 45p
  • (3a + 5) – (9 + b)

5.1

Table 11
Days 6 7 8 9 10 11 12
R 753 878,50 1 004 1 629,50 1 255 1 380,50 1 506
  • y = 125,5x
  • 2 1212 size 12{ { {1} over {2} } } {} months (2 x 30 ) + 15 75 x R125,50 = R9 412,50

or (30 + 31 + 15) 76 x R125,50 = R9 538,00

  • 3
  • 2
  • 3
  • 1
  • 1

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