Work out the following divisions.
(i) (10x - 25) ÷ 5 (ii) (10x - 25) ÷ (2x - 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7) (iv) 9x2y2(3z - 24) ÷ 27xy(z - 8)
(v) 96abc(3a -12)(5b - 30) ÷ 144(a - 4)(b - 6)
Solution:
We will find the factors of the algebraic expression then cancel out common factors of the numerator and denominator.
(i) (10x - 25) ÷ 5
Factorising (10x - 25) , we get
(10x - 25) = 5× 2 × x - 5 × 5
= 5(2x - 5)
(10x - 25) ÷ 5 = 5(2x - 5) / 5
= 2x - 5
(ii) (10x - 25) ÷ (2x - 5)
Factorising (10x - 25) , we get
(10x - 25) = 5 × 2 × x - 5× 5
= 5(2x - 5)
(10x - 25) ÷ (2x - 5) = 5(2x - 5) / (2x - 5)
= 5
(iii) 10y(6y + 21) ÷ 5(2y + 7)
Factorising 10y(6y + 21) , we get
10y(6y + 21) = 5 × 2 × y ×(2 × 3 × y + 3 × 7)
= 5 × 2 × y × 3(2 × y + 7)
= 30y(2y + 7)
10y(6y + 21) ÷ 5(2y + 7) = 30y(2y + 7) / 5(2y + 7)
= 6y
(iv) 9x2y2(3z - 24) ÷ 27xy(z - 8)
Factorising 9x2y2(3z - 24) , we get
9x2y2(3z - 24) = 3 × 3 × x × x × y × y ×(3 × z - 2 × 2 × 2 × 3)
= 3 × 3 × x × x × y × y × 3( z - 2 × 2 × 2)
= 27x2y2(z - 8)
9x2y2(3z - 24) ÷ 27xy(z - 8) = 27x2y2(z - 8) / 27xy(z - 8)
= xy
(v) 96abc(3a - 12)(5b - 30) ÷ 144(a - 4)(b - 6)
Factorising 96abc(3a - 12)(5b - 30) , we get
96abc(3a - 12)(5b - 30) = 96abc ×(3 × a - 2 × 2 × 3) × (5 × b - 5 × 2 × 3)
= 96abc × 3(a - 2 × 2)× 5(b - 2 × 3)
= 1440abc(a - 4)(b - 6)
Thus, 96abc(3a - 12)(5b - 30) ÷ 144(a - 4)(b - 6) = 1440abc(a - 4)(b - 6) / 144(a - 4)(b - 6)
= 10abc
☛ Check: NCERT Solutions for Class 8 Maths Chapter 14
Video Solution:
Work out the following divisions. (i) (10x - 25) ÷ 5 (ii) (10x - 25) ÷ (2x - 5) (iii) 10y(6y + 21) ÷ 5(2y + 7) (iv) 9x²y²(3z - 24) ÷ 27xy(z - 8) (v) 96abc(3a -12)(5b - 30) ÷ 144(a - 4)(b - 6)
Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.3 Question 3
Summary:
The following divisions have been performed (i) (10x - 25) ÷ 5 (ii) (10x - 25) ÷ (2x - 5) (iii) 10y(6y + 21) ÷ 5(2y + 7) (iv) 9x2y2(3z - 24) ÷ 27xy(z - 8) (v) 96abc(3a -12)(5b - 30) ÷ 144(a - 4)(b - 6) and the results are (i) 2x - 5 (ii) 5 (iii) 6y (iv) xy (v) 10abc
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