Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.2
Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.2
Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.2
Question 1.
Fill in the blanks
Question 5.
Divide:
(i) 32y2 – 8yz by 2y
(ii) (4m2n3 + 16m4 n2 – mn) by 2mn
(iii) 5xy2 – 18x2y3 + 6xy by 6xy
(iv) 81(p4 q2 r3 + 2p3q3 r2 – 5p2q2r2) by (3pqr)2
Answer:
(i) 32y2 – 8yz by 2y
(ii) (4m2n3 + 16m4 n2 – mn) by 2mn
(iii) 5xy2 – 18x2y3 + 6xy by 6xy
(iv) 81(p4 q2 r3 + 2p3q3 r2 – 5p2q2r2) by (3pqr)2
= 81/9(p2q2r2)1-1 (p2r + 2pq – 5)
= 9(p2r + 2pq – 5) = 9 p2r + 18pq – 45
Question 6.
Identify the errors and correct them.
(i) 7y2 – y2 + 3y2 = 10y2
Answer:
7y2 – y2 + 3y2 = 10y2 = (7 – 1 + 3)y2
= (6 + 3)y2
= 9y2
(ii) 6xy + 3xy = 9x2y2
Answer:
6xy + 3xy = (6 + 3) xy
= 9 xy
(iii) m(4m – 3) = 4m2 – 3
Answer:
m(4m – 3) = m(4m) + m(-3)
= 4m2 – 3m
(iv) (4n)2 – 2n + 3 = 4n2 – 2n + 3
Answer:
(4n)2 – 2n + 3 = 16n2 – 2n + 3
(v) (x – 2)(x + 3) = x2 – 6
Answer:
(x – 2)(x + 3) = x(x + 3) – 2 (x + 3)
= x(x) + (x) × 3 + (-2) (x) + (-2) (3)
= x2 + 3x – 2x – 6
= x2 + x – 6
(vi) -3p2 + 4p – 7 = -(3p2 + 4p – 7)
Answer:
-3p2 + 4p – 7 = -(3p2 – 4p + 7)
Question 7.
Statement A: If 24p2q is divided by 3pq, then the quotient is 8p.
Statement B: Simplification of (5x+5)/5 is 5x.
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer:
(ii) A is true but B is false
Hint:
Question 8.
Statement A: 4x2 + 3x – 2 = 2(2x2 + 3x/2 – 1)
Statement B: (2m – 5) – (5 – 2m) = (2m – 5) + (2m – 5)
(i) Both A and B are true
(ii) A is true but B is false
(iii) A is false but B is true
(iv) Both A and B are false
Answer:
(i) Both A and B are true
Hint:
(2m – 5) – (5 – 2m) = 2m – 5 – 5 + 2m = 4m – 10
(2m – 5) + (2m – 5) = 4m – 10
The Complete Educational Website