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JSS3: MATHEMATICS - 1ST TERM

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  1. Binary Number System I | Week 1
    5Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1Topic
    |
    1 Quiz
  5. Factorisation I | Week 5
    4Topics
    |
    1 Quiz
  6. Factorisation II | Week 6
    3Topics
    |
    1 Quiz
  7. Factorisation III | Week 7
    3Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1Topic
    |
    1 Quiz
Lesson Progress
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To factorize an expression completely, take the H.C.F outside the bracket and then divide each term with the H.C.F.

Find the H.C.F of the following and then factorise; 

a. 6x and 3y

\( \scriptsize 6x = 2 \: \times \: 3 \: \times \: x \)

\( \scriptsize 3y = 3 \: \times \: y \)

H.C.F  =  3

Take the H.C.F outside the bracket and then divide each term with the HCF

∴\(\normalsize \frac{6x}{3} \scriptsize =2x \: and \: \normalsize \frac{3y}{3} \scriptsize = y \)

6x and 3y  =  3(2x and y)

b. 10xy and  20xy2

10xy = \( \scriptsize 2 \: \times \: 5 \: \times \: x \: \times \: y \)

20xy2  = \( \scriptsize 2 \: \times \: 2 \: \times \: 5 \: \times \: x \: \times \: y\: \times \: y \)

H.C.F = \( \scriptsize 2 \: \times \: 5 \: \times \: x \: \times \: y = 10xy\)

∴\(\normalsize \frac{10xy}{10xy} \scriptsize =1 \: and \: \normalsize \frac{20xy^2}{10xy} \scriptsize = 2y \)

10xy and  20xy2 = 10xy(1 and 2y)

c. 12y + 8y

Factors of 12y = 2 x 2 x 3 x y

Factors of 8y2 = 2 x 2 x 2 x y x y

H.C.F = 2 x 2 x y = 4y

∴\(\normalsize \frac{12y}{4y} \scriptsize = 3 \: and \: \normalsize \frac{8y^2}{4y} \scriptsize = 2y \)

= 4y  ( 3 + 2y)

d. 36ab  + 12ab2

36ab  =  2 x 2 x 9 x a x b

12ab2 = 2 x 2 x 3 x a x b x b

HCF =  2 x 2 x a x b = 4ab

∴\(\normalsize \frac{36ab}{4ab} \scriptsize = 9 \: and \: \normalsize \frac{12ab^2}{4ab} \scriptsize = 3b \)

= 4ab (9 + 3b)

e. 12p2 q – 9pq

12p2 q = (2 x 2 x 3 x p x p x q)

9pq = 3 x 3 x p x q

HCF  =  3 x p x q =  3pq

∴\( \normalsize \frac{12p^2q}{3pq} \scriptsize = 4p \: and \: \normalsize \frac{9pq}{3pq} \scriptsize = 3\)

= 3pq(4p – 3)

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