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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 45 36 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? 6a 2 b 3 ab size 12{ { {6a rSup { size 8{2} } b} over {3 bold "ab"} } } {} = .........................

Yes, it is actually like this: 6 × a × a × b 3 × a × b size 12{ { {6 times a times a times b} over {3 times `a` times `b} } } {} = 2 × a × 1 × 1 1 × 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: m 5 m 7 = 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 m ² what remains below the line.

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

1. Now simplify the following:

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

1.2 15 p 4 y 7 3p 7 y 3 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 ( 4m 2 ) 2 2m 6 size 12{ { { \( 4m rSup { size 8{2} } \) rSup { size 8{2} } } over {2m rSup { size 8{6} } } } } {} ..................................................

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

1.6 25 a 2 b 3 c 4 . 4 ab 3 c 2 15 a 3 b 4 c 5 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: 1 3 b ( 9a ) size 12{ { {1} over {3} } b \( 9a \) } {} = 1 × b 3 × 9a 1 size 12{ { {1 times b} over {3} } times { {9a} over {1} } } {} = b × 9a 3 = 9 ab 3 size 12{ { {b times 9a} over {3} } `=` { {9 bold "ab"} over {3} } } {} = 3 ab

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

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

2.1 Write each of the following in simplified form.

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

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

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

2.1.4: 6a 2 b 2 c 3 - 15 a 4 b 6 c 7 + 27 b 9 c 10 3a 2 b 4 c 3 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: 7m 2 pq 9 - 49 m 6 n 7 - 35 p 6 q 12 7 mn 3 q 4 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 - 56 p 7 q 7 -8 mn 3 q 4 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 3a 2 bc 4 - 36 a 4 b 7 + 24 a 3 b 2 -3 ab 2 c 2 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 (5 a ² - 25 b )

1.4 ( a 2 b 2 ) 3 . ( ab 2 ) 4 a 2 b 3 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 ( 4 kp 4 ) 2 2k 3 p 2 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 = 3 ab ² + 6 a ² and Q = 2 ab , calculate:

2.1 2 P - 3 Q

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

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

3. Supposing that 5 a 3 b ² books cost (-5 ab + 15 a 4 b 7 ) rands, calculate the price of one book.

Assessment

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

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. >
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 x size 12{x} {} 3 is 2 x size 12{x} {} .

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

3. -10 ½<-10,499

4. -2² = -4

5. The first prime number is 1.

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Source:  OpenStax, Algebra. OpenStax CNX. Aug 17, 2012 Download for free at http://cnx.org/content/col11445/1.1
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