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Division in algebra

Module by: Siyavula Uploaders. E-mail the author

MATHEMATICS

Grade 8

THE NUMBER SYSTEM

(Natural and whole numbers)

Module 6

DIVISION IN ALGEBRA

CLASS ASSIGNMENT 1

  • Discover more and more about division in ALGEBRA!
  • Write the following fraction in its simplest form in 45364536 size 12{ { {"45"} over {"36"} } } {} = ........................
  • Like common fractions, you can also write algebraic fractions in the simplest form.

What would the following be in its simplest form? 6a2b3ab6a2b3ab size 12{ { {6a rSup { size 8{2} } b} over {3 bold "ab"} } } {} = .........................

Yes, it is actually like this: 6×a×a×b3×a×b6×a×a×b3×a×b size 12{ { {6 times a times a times b} over {3 times `a` times `b} } } {} = 2×a×1×11×1×12×a×1×11×1×1 size 12{ { {2 times a times 1 times 1} over {1 times `1` times `1} } } {}

(Now you may cancel all the like terms above and below)(What remains above and below? Just write down the answer)

  • There is a shortcut for terms with exponents: m5m7=m5m7= size 12{ { {m rSup { size 8{5} } } over {m rSup { size 8{7} } } } `={}} {} .........................

Are you able to identify the shortcut? Yes, 7 - 5 = 2. Therefore m² what remains below the line.

Answer: 1m21m2 size 12{ { {1} over {m rSup { size 8{2} } } } } {}

1. Now simplify the following:

1.1 18m59m218m59m2 size 12{ { {"18"m rSup { size 8{5} } } over {9m rSup { size 8{2} } } } } {} ..................................................

1.2 15p4y73p7y315p4y73p7y3 size 12{ { {"15"p rSup { size 8{4} } y rSup { size 8{7} } } over {3p rSup { size 8{7} } y rSup { size 8{3} } } } } {} ..................................................

1.3 (4m2)22m6(4m2)22m6 size 12{ { { \( 4m rSup { size 8{2} } \) rSup { size 8{2} } } over {2m rSup { size 8{6} } } } } {} ..................................................

1.4 4(2m2)52(m2)24(2m2)52(m2)2 size 12{ { {4 \( 2m rSup { size 8{2} } \) rSup { size 8{5} } } over {2 \( m rSup { size 8{2} } \) rSup { size 8{2} } } } } {} ..................................................

1.5 8a2(a2)28a28a2(a2)28a2 size 12{ { {8a rSup { size 8{2} } \( a rSup { size 8{2} } \) rSup { size 8{2} } } over {8a rSup { size 8{2} } } } } {} ..................................................

1.6 25a2b3c4.4ab3c215a3b4c525a2b3c4.4ab3c215a3b4c5 size 12{ { {"25"a rSup { size 8{2} } b rSup { size 8{3} } c rSup { size 8{4} } "." 4 bold "ab" rSup { size 8{3} } c rSup { size 8{2} } } over {"15"a rSup { size 8{3} } b rSup { size 8{4} } c rSup { size 8{5} } } } } {} ...........................................

2.Remember: 13b(9a)13b(9a) size 12{ { {1} over {3} } b \( 9a \) } {} = 1×b3×9a11×b3×9a1 size 12{ { {1 times b} over {3} } times { {9a} over {1} } } {} = b×9a3=9ab3b×9a3=9ab3 size 12{ { {b times 9a} over {3} } `=` { {9 bold "ab"} over {3} } } {} = 3ab

Therefore: 1313 size 12{ { {1} over {3} } } {} means: x 1 ÷ 3

Now try to simplify the following: 1/3 ( 4a - 6b)

2.1 Write each of the following in simplified form.

2.1.1: 1515 size 12{ { {1} over {5} } } {} x a ..................................................

2.1.2: 1515 size 12{ { {1} over {5} } } {}(2a² - 15) ..................................................

2.1.3: (2a - 8b + 12c) ÷ 2 ..................................................

2.1.4: 6a2b2c3 - 15a4b6c7+ 27b9c103a2b4c36a2b2c3 - 15a4b6c7+ 27b9c103a2b4c3 size 12{ { {6a rSup { size 8{2} } b rSup { size 8{2} } c rSup { size 8{3} } " - 15"a rSup { size 8{4} } b rSup { size 8{6} } c rSup { size 8{7} } " "+" 27"b rSup { size 8{9} } c rSup { size 8{"10"} } } over {3a rSup { size 8{2} } b rSup { size 8{4} } c rSup { size 8{3} } } } } {}

2.1.5: 7m2pq9 - 49m6n7 - 35p6q12−7mn3q47m2pq9 - 49m6n7 - 35p6q12−7mn3q4 size 12{ { {7m rSup { size 8{2} } bold "pq" rSup { size 8{9} } " - 49"m rSup { size 8{6} } n rSup { size 8{7} } " - 35"p rSup { size 8{6} } q rSup { size 8{"12"} } } over { - 7 bold "mn" rSup { size 8{3} } q rSup { size 8{4} } } } } {}

HOMEWORK ASSIGNMENT 1

1. Simplify:

1.1 - 56p7q7-8mn3q4- 56p7q7-8mn3q4 size 12{ { {"- 56"p rSup { size 8{7} } q rSup { size 8{7} } } over {"-8" bold "mn" rSup { size 8{3} } q rSup { size 8{4} } } } } {}

1.2 3a2bc4 - 36a4b7+ 24a3b2-3ab2c23a2bc4 - 36a4b7+ 24a3b2-3ab2c2 size 12{ { {3a rSup { size 8{2} } bold "bc" rSup { size 8{4} } " - 36"a rSup { size 8{4} } b rSup { size 8{7} } " "+" 24"a rSup { size 8{3} } b rSup { size 8{2} } } over {"-3" bold "ab" rSup { size 8{2} } c rSup { size 8{2} } } } } {}

1.3 ½ (5a² - 25b)

1.4 (a2b2)3.(ab2)4a2b3(a2b2)3.(ab2)4a2b3 size 12{ { { \( a rSup { size 8{2} } b rSup { size 8{2} } \) rSup { size 8{3} } "." \( bold "ab" rSup { size 8{2} } \) rSup { size 8{4} } } over {a rSup { size 8{2} } b rSup { size 8{3} } } } } {}

1.5 3(4kp4)22k3p23(4kp4)22k3p2 size 12{ { {3 \( 4 bold "kp" rSup { size 8{4} } \) rSup { size 8{2} } } over {2k rSup { size 8{3} } p rSup { size 8{2} } } } } {}

2. If P = 3ab² + 6a² and Q = 2ab, calculate:

2.1 2P - 3Q

2.2 PQPQ size 12{ { {P} over {Q} } } {}

2.3 P+Q2QP+Q2Q size 12{ { {P`+`Q} over {2Q} } } {}

3. Supposing that 5a3b² books cost (-5ab + 15a4b7) rands, calculate the price of one book.

Assessment

Table 1
Assessment of myself:   by myself:   Assessment by Teacher:
I can…     1 2 3 4   Critical Outcomes 1 2 3 4
express fractions in their simplest form; (Lo 2.2; 1.6.2)                     Critical and creative thinking        
express algebraic fractions in their simplest form; (Lo 2.2; 1.6.2; 2.8.3; 2.8.4; 2.8.5; 2.8.6)                     Collaborating        
calculate the shortest path for terms with exponents. (Lo 2.2; 1.6.2; 1.6.3)                     Organising en managing        
                      Processing of information        
                      Communication        
                      Problem solving        
                      Independence        
                               

good average not so good

Table 2
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 3
Comments by parents:   Comments by teacher:
     
     
     
Signature: Date:   Signature: Date:

Tutorial 2: (Algebra)

Total: 70

Question 1

A. Indicate whether the following statements are TRUE or FALSE and improve the wrong statements.

1. The base of 2 xx size 12{x} {}3 is 2 xx size 12{x} {}.

2. The number of terms in the expression are THREE:3(5a + 2) + 5b - 6

3. -10 ½ < -10,499

4. -2² = -4

5. The first prime number is 1.

6. 0 ÷ 6 = 0

7. The 5 in 53 is known as the power.

8. -3 > -6

9. The square of van 8 is 64.

10. The coefficient of xx size 12{x} {} in 4 xx size 12{x} {}y² is 4.

11. 7( xx size 12{x} {} - 2y) = 7 xx size 12{x} {} - 2y

12. 2 xx size 12{x} {}² + 3 xx size 12{x} {}² = 5 xx size 12{x} {}4

13. 5 - 2 + (-2) = 1

14. a6b123a6b123 size 12{ nroot { size 8{3} } {a rSup { size 8{6} } b rSup { size 8{"12"} } } } {}= a2b4

[14]

B. Complete the following:

1. The coefficient of a in 3ab is …

2. The exponent 5a + b is ....

[2]

[16]

Question 2

Here matches are arranged to form squares.

Figure 1
Figure 1 (Picture 109.png)

1. Complete the following table. [4]

Table 4
Number of squares 2 3 5 9
Number of matches        

2. How many matches are required to build 150 squares? [2]

3. How many squares can be built with 1 000 matches? [2]

[8]

Question 3

A. Supposing a = 3 ; b = -2 and c = 6, determine the value of:

1. 2a(3b + 5c) [3]

2. a−b−cb−ca−b−cb−c size 12{ { {a` - `b` - `c} over {b` - `c} } } {} [4]

B. Addition and subtraction:

1. Add: 3a + 4b - 6c ; -6b + 3c + 2a ; 5c - 3b ; a - 2c [3]

2. Subtract 8 xx size 12{x} {} + 2y + 5 from 10 xx size 12{x} {} - 2y + 4 [3]

[13]

Question 4

Simplify each of the following:

1. 7a²bc - 2ab²c - 3ba²c + acb² [2]

2. 5a² x 3a3 - 3a4 x 2a [3]

3. 3(2a²b3)² . (-ab)3 [4]

4. -5ab² (3a² - 4b) [2]

5. 2a (7a + 4) - 3( a + 3a²) - ( -4a² - a ) [4]

6. (- 3ab)2(2a)34(ab)2(- 3ab)2(2a)34(ab)2 size 12{ { { \( "- 3" bold "ab" \) rSup { size 8{2} } \( 2a \) rSup { size 8{3} } } over {4 \( bold "ab" \) rSup { size 8{2} } } } } {} [4]

7. 14a4b36c2×12c4−7a3b214a4b36c2×12c4−7a3b2 size 12{ { {"14"a rSup { size 8{4} } b rSup { size 8{3} } } over {6c rSup { size 8{2} } } } ` times ` { {"12"c rSup { size 8{4} } } over { - 7a rSup { size 8{3} } b rSup { size 8{2} } } } } {} [3]

[22]

Question 5

Supposing A = a²b + 5ab² - 6ab and B = ab, determine the following:

5.1 ABAB size 12{ { {A`} over {B} } } {} [3]

5.2 3AB3AB size 12{ { {3A`} over {B} } } {} [4]

5.3 A+BBA+BB size 12{ { {A`+`B`} over {B} } } {} [4]

[11]

TOTAL: 70

AlgebraTutorial

Table 5
I demonstrate knowledge and understanding of: Learning outcomes   0000 000 00 0
1. different terms that can be distinguished in a polynomial; 2.4; 2.8.2; 2.9          
2. identifying the coefficient of an unknown; 2.4; 2.9          
3. identifying the constant in a polynomial; 2.4; 2.9          
4. determining the degree of an expression; 2.4; 2.8.1; 2.9          
5. arranging the expression in a descending order; 2.4; 2.9          
6. the correct usage of signs ( + / - ) in an expression; 2.4; 2.8.4 & .6          
7. algebraic expression; 2.2; 2.4; 2.8.4          
8. determining the formulas of flow diagrams and tables; 2.1 & .3; .4; .7          
9. adding and subtracting of numbers; 2.4; 2.8.2 & .4          
10. writing expressions in exponent form; 1.6.3; 2.4          
11. solving expressions that have brackets; 2.4; 2.8.3 & .5          
12. the correct usage of the order of calculations; 2.4; 2.8.5          
13. substitution of unknowns with constant values; 2.4; 2.8.4;1.6.2 & .3          
14. expressing fractions in their simplest form; 2.2; 2.4; 1.6.2          
15. expressing algebraic fractions in their simplest form. 2.4; 1.6.2; 2.8.3          
16.              
17.              
Table 6
The learner’s … 1 2 3 4
work is… Not done. Partially done. Mostly complete. Complete.
layout of the work is… Not understandable. Difficult to follow. Sometimes easy to follow. Easy to follow.
accuracy of calculations… Are mathematically incorrect. Contain major errors. Contain minor errors. Are correct.
Table 7
  My BEST marks:   Comments by teacher:
Date:          
Out of:          
Learner:          
           
          Signature: Date:

Parent Signature: Date:

Test 2: (Algebra)

Total: 60

1. Simplify:

1.1: 17a²bc - 21ab²c - 3ba²c + 4acb² [2]

1.2: 5a² x 3a3 - 3a4 x 5a [3]

1.3: 7ab - (-4ab) [2]

1.4: 3(2a²b3)3 . (-a²b²)3 [3]

1.5: -7ab²(3a² - 5b) [2]

1.6: 5a(3a + 5) - 2(2a + 4a²) - 2a(-3a) [4]

1.7: (-2ab)2(3a)34(ab2)2(-2ab)2(3a)34(ab2)2 size 12{ { { \( "-2" bold "ab" \) rSup { size 8{2} } \( 3a \) rSup { size 8{3} } } over {4 \( bold "ab" rSup { size 8{2} } \) rSup { size 8{2} } } } } {} [4]

[20]

2. If a = 3 ; b = -2 and c = 6, determine the values for:

2.1 3a(2b + 4c) [3]

2.2 a - b - cb+ca - b - cb+c size 12{ { {a" - " bold "b "" - "c} over {b`+`c} } } {} [4]

2.3 3ca.b.c3ca.b.c size 12{ { {3c} over {a "." b "." c} } } {} [3]

2.4 9a5-3a39a5-3a3 size 12{ { {9a rSup { size 8{5} } } over {"-3"a rSup { size 8{3} } } } } {} [3]

[13]

3. Given: 5a² - (m + n)a - 5p

3.1 How many terms are there in the above expression? ......................... [1]

3.2 Write down the coefficient of p. ......................... [1]

3.3 Write down the exponent of p. ......................... [1]

3.4 What is the reciprocal of -5? ......................... [1]

[4]

4.

4.1 Determine by how much 7a + 5b + 9c is bigger than 2c + 3a - 7c . [3]

4.2 Determine by how much n + 3m + 3k is smaller than -3k - 7m + 2n [3]

4.3 Subtract xx size 12{x} {}² - 2 xx size 12{x} {} + 4 from 5c² + 6 xx size 12{x} {} - 9 [3]

[9]

5. Write down an algebraic expression for each of the following.

5.1: 23 m reduced by the square of n ........................... [2]

5.2 Subtract the product of m and n from the difference between m and n. [2]

5.3 You have twenty coins. y of them are fifty-cent coins and the rest are ten-cent coins.

5.3.1 how many ten-cent coins do you have (in terms of y) [1]

5.3.2 What is the total value of your money? [2]

[7]

6. Complete the following tables and give a formula for each in the form y = ......

6.1 y = .................................................. [3]

Table 8
x x size 12{x} {} 3 5 7 9 10 15  
y 9 25          

6.2 y = .................................................. [4]

Table 9
x x size 12{x} {} 3 5 7 9  
y 11   23   38

[7]

7. Bonus question

Study the following pattern:

1 = 1 x 1

1 + 3 = 2 x 2

1 + 3 + 5 = 3 x 3

1 + 3 + 5 + 7 = 4 x 4

Now determine:

1 + 3 + 5 + 7 + 9 + ..... + 21 + 23 + 25 + 27 + 29

2]

Enrichment exercise for the quick learner

1. If pq means 3 p + q2, (3  4)  5 will be equal to...

a) 60 b) 100 c) 87 d) 72 e) 91

2. In the multiplication shown, S and T are different digits between 1 and 9. The value of S + T is ...

S 6

x 2 T

2 1 5 0

a) 13 b) 14 c) 15 d) 16 e) 17

3. How many digits are there in 5 8 ?

a) 2 b) 5 c) 6 d) 8 e) 40

4. The average of the numbers 0,1 ; 0,11 and 0,111 is ...

a) 0,041 b) 0,107 c) 0,11 d) 0,1111 e) 0,17

5. In a certain class 1/3 of the pupils are girls. If 6 boys represent one quarter of the number of boys in the class, how many pupils are there in the class?

a) 18 b) 24 c) 27 d) 32 e) 36

6. If ½ - x = x - 1/3 , x will be equal to...

a) 1/12 b) 1/6 c) 5/12d) 5/6 e) 3/12

7. Calculate the value of x if .....

4 + x 3 = 2 4 + x 3 = 2 size 12{ nroot { size 8{3} } {4+ sqrt {x} } =2} {}
(1)

a) 4 b) 2

Memorandum

CLASSWORK ASSIGNMENT 1

  • 2m3
  • 5y 4 p 3 5y 4 p 3 size 12{ { {5y rSup { size 8{4} } } over {p rSup { size 8{3} } } } } {}
    (2)
  • 8 m 2 8 m 2 size 12{ { {8} over {m rSup { size 8{2} } } } } {}
    (3)
  • 432m102m4432m102m4 size 12{ { {4 left ("32"m rSup { size 8{"10"} } right )} over {2m rSup { size 8{4} } } } } {} = 64m6

1.5 a4

1.6 20b2c320b2c3 size 12{ { {"20"b rSup { size 8{2} } c} over {3} } } {}

2. 4a34a3 size 12{ { {4a} over {3} } } {} – 2b

2.1.1 1a51a5 size 12{ { {1a} over {5} } } {}

2.1.2 2a252a25 size 12{ { {2a rSup { size 8{2} } } over {5} } } {} – 3

  • a – 4b + 6c
  • 2b22b2 size 12{ { {2} over {b rSup { size 8{2} } } } } {} – 5a2b2c4 + 9b5c7a29b5c7a2 size 12{ { {9b rSup { size 8{5} } c rSup { size 8{7} } } over {a rSup { size 8{2} } } } } {}
  • −mpq5n3−mpq5n3 size 12{ { { - ital "mpq" rSup { size 8{5} } } over {n rSup { size 8{3} } } } } {} + 7m5n4q47m5n4q4 size 12{ { {7m rSup { size 8{5} } n rSup { size 8{4} } } over {q rSup { size 8{4} } } } } {} + 5p6q8mn35p6q8mn3 size 12{ { {5p rSup { size 8{6} } q rSup { size 8{8} } } over { ital "mn" rSup { size 8{3} } } } } {}

HOMEWORK ASSIGNMENT 1

1.1 7p7q3mn37p7q3mn3 size 12{ { {7p rSup { size 8{7} } q rSup { size 8{3} } } over { ital "mn" rSup { size 8{3} } } } } {}

1.2 −ac2b−ac2b size 12{ { { - ital "ac" rSup { size 8{2} } } over {b} } } {} + 12a3b5c212a3b5c2 size 12{ { {"12"a rSup { size 8{3} } b rSup { size 8{5} } } over {c rSup { size 8{2} } } } } {} – 8a2c28a2c2 size 12{ { {8a rSup { size 8{2} } } over {c rSup { size 8{2} } } } } {}

1.3 5a225a22 size 12{ { {5a rSup { size 8{2} } } over {2} } } {} – 25b225b2 size 12{ { {"25"b} over {2} } } {}

= 5a2−25b25a2−25b2 size 12{ left ( { {5a rSup { size 8{2} } - "25"b} over {2} } right )} {}

1.4 a6b6.a4b8a2b3a6b6.a4b8a2b3 size 12{ { {a rSup { size 8{6} } b rSup { size 8{6} } "." a rSup { size 8{4} } b rSup { size 8{8} } } over {a rSup { size 8{2} } b rSup { size 8{3} } } } } {}

= a8b11a8b11 size 12{a rSup { size 8{8} } b rSup { size 8{"11"} } } {}

1.5 316k2p82k3p2316k2p82k3p2 size 12{ { {3 left ("16"k rSup { size 8{2} } p rSup { size 8{8} } right )} over {2k rSup { size 8{3} } p rSup { size 8{2} } } } } {}

= 24p6k24p6k size 12{ { {"24"p rSup { size 8{6} } } over {k} } } {}

  • 6ab2 + 12a2 – (6ab)
  • 3b23b2 size 12{ { {3b} over {2} } } {} + 3ab3ab size 12{ { {3a} over {b} } } {}
  • 3 ab 2 + 6a 2 + 2 ab 4 ab 3 ab 2 + 6a 2 + 2 ab 4 ab size 12{ { {3 ital "ab" rSup { size 8{2} } +6a rSup { size 8{2} } +2 ital "ab"} over {4 ital "ab"} } } {}

= 3b43b4 size 12{ { {3b} over {4} } } {} + 3a2b3a2b size 12{ { {3a} over {2b} } } {} + 1212 size 12{ { {1} over {2} } } {}

3. −5ab+15a4b75a3b2−5ab+15a4b75a3b2 size 12{ { { - 5 ital "ab"+"15"a rSup { size 8{4} } b rSup { size 8{7} } } over {5a rSup { size 8{3} } b rSup { size 8{2} } } } } {} = – 1a2b1a2b size 12{ { {1} over {a rSup { size 8{2} } b} } } {} + 3ab5

TUTORIAL

QUESTION 1

A.1. False: x

  1. True
  2. True
  3. True
  4. False: 2
  5. True
  6. False: 5 base
  7. False
  8. False
  9. False: 4y2
  10. False: 7x – 14y
  11. False: 5x2
  12. True
  13. True

B.1. 3b

2. a+b

QUESTION 2

1. 7; 10; 16; 28

  1. (150 x 3) + 1 = 451
  2. (1 000 – 1) ÷ 3 = 333

QUESTION 3

A.1. 2(3)[3(–2) + 5(6)]

= 6[–6 + 30]

= 6[+24] = 144

2. 3−(−2)−(6)(−2)−(6)3−(−2)−(6)(−2)−(6) size 12{ { {3 - \( - 2 \) - \( 6 \) } over { \( - 2 \) - \( 6 \) } } } {} or 3+2−6−2−63+2−6−2−6 size 12{ { {3+2 - 6} over { - 2 - 6} } } {}

= −1−8−1−8 size 12{ { { - 1} over { - 8} } } {} = 1818 size 12{ { {1} over {8} } } {}

B.1. 6a – 5b

2. 2x – 4y – 1

QUESTION 4

  1. 4a2bc – ab2c
  2. 15a5 – 6a5 = 9a5
  3. 12a4b6(–a3b3)

= –12a7b9

  1. –15a3b2 + 20ab3
  2. 14a2 + 8a – 3a – 9a2 + 4a2 + a

= 9a2 + 6a

6. 9a2b2.8a34a2b29a2b2.8a34a2b2 size 12{ { {9a rSup { size 8{2} } b rSup { size 8{2} } "." 8a rSup { size 8{3} } } over {4a rSup { size 8{2} } b rSup { size 8{2} } } } } {} = 18a3

7. 142aa4bb3bc1142aa4bb3bc1 size 12{ { { { {1}} { {4}} rSup { size 8{2} } { {a}} rSup { size 8{a { {4}}} } { {b}} rSup { size 8{b { {3}}} } } over { { {b}} { {c}} rSub { size 8{1} } } } } {} x 122c4c2−7a3b2122c4c2−7a3b2 size 12{ { { { {1}} { {2}} rSup { size 8{2} } { {c}} rSup { size 8{ { {4}}c2} } } over { - { {7}} { {a}} rSup { size 8{ { {3}}} } { {b}} rSup { size 8{ { {2}}} } } } } {}

= -4abc2

QUESTION 5

5.1 a2b+5ab2−6ababa2b+5ab2−6abab size 12{ { {a rSup { size 8{2} } b+5 ital "ab" rSup { size 8{2} } - 6 ital "ab"} over { ital "ab"} } } {}

= a + 5b – 6

5.2 3a2b+15ab2−18abab3a2b+15ab2−18abab size 12{ { {3a rSup { size 8{2} } b+"15" ital "ab" rSup { size 8{2} } - "18" ital "ab"} over { ital "ab"} } } {}

= 3a + 15b – 18

5.3 a2b+5ab2−5ababa2b+5ab2−5abab size 12{ { {a rSup { size 8{2} } b+5 ital "ab" rSup { size 8{2} } - 5 ital "ab"} over { ital "ab"} } } {}

= a + 5b – 5

TEST

  • 4a2bc – 17ab2c
  • 15a5 – 15a5 = 0
  • 7ab + 4ab = 11ab
  • 3(8a6b9) (–a6b6)

= –24a12b15

  • –7a3b2 + 35ab3
  • 15a2 + 25a – 4a –8a2 + 6a2

=13a2 + 21a

  • 3(3)[2(–2) + 4(6)]

= 9 [–4 + 24]

= 180

2.2 3−(−2)−(6)(−2)+(6)3−(−2)−(6)(−2)+(6) size 12{ { {3 - \( - 2 \) - \( 6 \) } over { \( - 2 \) + \( 6 \) } } } {}

= 3+2−6−2+63+2−6−2+6 size 12{ { {3+2 - 6} over { - 2+6} } } {}

= 0404 size 12{ { {0} over {4} } } {} = 0

2.3 3(6)(3)(−2)(6)3(6)(3)(−2)(6) size 12{ { {3 \( 6 \) } over { \( 3 \) \( - 2 \) \( 6 \) } } } {}

= 18−3618−36 size 12{ { {"18"} over { - "36"} } } {}

= – 1212 size 12{ { {1} over {2} } } {}

2.4 9(3)532−3(3)39(3)532−3(3)3 size 12{ { {9 \( { {3}} \) rSup { size 8{ { {5}}} rSup { size 8{ {} rSup { size 6{3 rSup {2} } } } } } } over { - 3 \( { {3}} \) rSup { { {3}}} } } } {}

= 9(9)3−39(9)3−3 size 12{ { {9 \( { { {9}} \) } cSup { size 8{3} } } over { - { {3}}} } } {}

= –27

  • 3
  • –5
  • 1
  • − 1 5 − 1 5 size 12{ - { {1} over {5} } } {}
    (4)
  • 7a + 5b + 9c – (2c + 3a – 7c)

= 7a + 5b + 9c – 2c + 3a – 7c

= 4a + 5b + 14c

  • –3k – 7m + 2n – (n + 3m 3k)

= –3k – 7m + 2n – n + 3m 3k

= –6k – 10m + n

  • 5x2 + 6x – 9 – (x2 – 2x + 4)

= 5x2 + 6x – 9 – x2 – 2x + 4

= 4c2 + 8x – 13

  • 23m – n2
  • (m – n) – mn
  • 20 – y
  • 50y – 10(20 – y)

= 50y – 200 + 10y

= 60y – 200

  • y = x2

49; 81; 100; 225

  • y = 3x + 2

17; 29; 18

7. 152 = 225

ENRICHMENT EXERCISE

  1. b
  2. a
  3. c
  4. b
  5. e
  6. c
  7. d

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