Students can Download Maths Chapter 3 Algebra Ex 3.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks.
1. (p – q)2 = _______
2. The product of (x + 5) and (x – 5) is _______
3. The factors of x2 – 4x + 4 are _______
4. Express 24ab2c2 as product of its factors is _______
Answers:
1. p2 – 2pq + q2
2. x2 – 25
3. (x – 2) and (x – 2)
4. 2 × 2 × 2 × 3 × a × b × b × c × c

Question 2.
Say whether the following statements are True or False.
(i) (7x + 3) (7x – 4) = 49 x2 – 7x – 12
(ii) (a – 1)2 = a2 – 1.
(iii) (x2 + y2)(y2 + x2) = (x2 + y2)2
(iv) 2p is the factor of 8pq.
Answers:
(i) True
(ii) False
(iii) True
(iv) True

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 3.
Express the following as the product of its factors.
(i) 24ab2c2
(ii) 36 x3y2z
(iii) 56 mn2p2
Solution:
(i) 24ab2c2 = 2 × 2 × 2 × 3 × a × b × b × c × c
(ii) 36 x3y2z = 2 × 2 × 3 × 3 × x × x × x × y × y × z
(iii) 56 mn2p2 = 2 × 2 × 2 × 7 × m × n × n × p × p

Question 4.
Using the identity (x + a)(x + b) – x2 + x(a + b) + ab, find the following product.
(i) (x + 3) (x + 7)
(ii) (6a + 9) (6a – 5)
(iii) (4x + 3y) (4x + 5y)
(iv) (8 + pq) (pq + 7)
Solution:
(i) (x + 3) (x + 7)
Let a = 3; b = 7, then
(x + 3) (x + 7) is of the form x2 + x (a + b) + ab
(x + 3) (x + 7) = x2 + x (3 + 7) + (3 × 7) = x2 + 10x + 21

(ii) (6a + 9) (6a – 5)
Substituting x = 6a ; a = 9 and b = -5
In (x + a) (x + b) = x2 + x (a + b) + ab, we get
(6a + 9)(6a – 5) = (6a)2 + 6a (9 + (-5)) + (9 × (-5))
62 a2 + 6a (4) + (-45) = 36a2 + 24a – 45
(6a + 9) (6a – 5) = 36a2 + 24a – 45

(iii) (4x + 3y) (4x + 5y)
Substituting x = 4x ; a = 3y and b = 5y in
(x + a) (x + b) = x2 + x (a + b) + ab, we get
(4x + 3y) (4x – 5y) = (4x)2 + 4x (3y + 5y) + (3y) (5y)
= 42 x2 + 4x (8y) + 15y2 = 16x2 + 32xy + 15y2
(4x + 3y) (4x + 5y) = 16x2 + 32xy + 15y2

(iv) (8 + pq) (pq + 7)
Substituting x = pq ; a = 8 and b = 7 in
(x + a) (x + b) = x2 + x (a + b) + ab, we get
(pq + 8) (pq + 7) = (pq)2 + pq (8 + 7) + (8) (7)
= p2 q2 + pq (15) + 56
(8 + pq) (pq + 7) = p2 q2 + 15pq + 56

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 5.
Expand the following squares, using suitable identities.
(i) (2x + 5)2
(ii) (b – 7)2
(iii) (mn + 3p)2
(iv) (xyz – 1)2
Solution:
(i) (2x + 5)2
Comparing (2x + 5)2 with (a + b)2 we have a = 2x and b = 5
a = 2x and b = 5,
(a + b)2 = a2 + 2ab + b2
(2x + 5)2 = (2x)2 + 2(2x) (5) + 52 = 22 x2 + 20x + 25
= 22 x2 + 20x + 25
(2x + 5)2 = 4x2 + 20x + 25

(ii) (b – 7)2
Comparing (b – 7)2 with (a – b)2 we have a = b and b = 7
(a – b)2 = a2 – 2ab + b2
(b – 7)2 = b2 – 2(b) (7) + 72
(b – 7)2 = b2 – 14b + 49

(iii) (mn + 3p)2
Comparing (mn + 3p)2 with (a + b)2 we have
(a + b)2 = a2 + 2ab + b2
(mn + 3p)2 = (mn)2 + 2(mn) (3p) + (3p)2
(mn + 3p)2 = m2 n2 + 6mnp + 9p2

(iv) (xyz – 1)2
Comparing (xyz – 1)2 with (a – b)2 we have = a + xyz and b = 1
a = xyz and b = 1
(a – b)2 = a2 – 2ab + b2
(xyz – 1)2 = (xyz)2 – 2 (xyz) (1) + 12
(xyz -1)2 = x2 y2 z2 – 2 xyz + 1

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 6.
Using the identity (a + b)(a – b) = a2 – b2, find the following product.
(i) (p + 2) (p – 2)
(ii) (1 + 3b) (3b – 1)
(iii) (4 – mn) (mn + 4)
(iv) (6x + 7y) (6x – 7y)
Solution:
(i) (p + 2) (p – 2)
Substituting a = p ; b = 2 in the identity (a + b) (a – b) = a2 – b2, we get
(p + 2) (p – 2) = p2 – 22

(ii) (1 + 3b)(3b – 1)
(1 + 3b) (3b -1) can be written as (3b + 1) (3b – 1)
Substituting a = 36 and b = 1 in the identity
(a + b) (a – b) = a2 – b2, we get
(3b + 1)(3b – 1) = (3b)2 – 12 = 32 × b2 – 12
(3b + 1) (3b – 1) = 9b2 – 12

(iii) (4 – mn) (mn + 4)
(4 – mn) (mn + 4) can be written as (4 – mn) (4 + mn) = (4 + mn) (4 – mn)
Substituting a = 4 and b = mn is
(a + b) (a – b) = a2 – b2, we get
(4 + mn) (4 – mn) = 42 – (mn)2 = 16 – m2 n2

(iv) (6x + 7y) (6x – 7y)
Substituting a = 6x and b = 7y in
(a + b) (a – b) = a2 – b2, We get
(6x + 7y) (6x – 7y) = (6x)2 – (7y)2 = 62x2 – 72y2
(6x + 7y) (6x – 7y) = (6x)2 – (7y)2 = 62x2 – 72y2
(6x + 7y) (6x – 7y) = 36x2 – 49y2

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 7.
Evaluate the following, using suitable identity.
(i) 512
(ii) 1032
(iii) 9982
(iv) 472
(v) 297 × 303
(vi) 990 × 1010
(vii) 51 × 52
Solution:
512
= (50 + 1)2
Taking a = 50 and b = 1 we get
(a + b)2 = a2 + 2ab + b2
(50 + 1)2 = 502 + 2 (50) (1) + 12 = 2500 + 100 + 1
512 = 2601

(ii) 1032
1032 = (100 + 3)2
Taking a = 100 and b = 3
(a + b)2 = a2 + 2ab + b2 becomes
(100 + 3)2 = 1002 + 2 (100) (3) + 32 = 10000 + 600 + 9
1032 = 10609

(iii) 9982
9982 = (1000 – 2)2
Taking a = 1000 and b = 2
(a – b)2 = a2 + 2ab + b2 becomes
(1000 – 2)2 = 10002 – 2 (1000) (2) + 22
= 1000000 – 4000 + 4
9982 = 10,04,004

(iv) 472
472 = (50 – 3)2
Taking a = 50 and b = 3
(a – b)2 = a2 – 2ab + b2 becomes
(50 – 3)2 = 502 – 2 (50) (3) + 32
= 2500 – 300 + 9 = 2200 + 9
472 = 2209

(v) 297 × 303
297 × 303 = (300 – 3) (300 + 3)
Taking a = 300 and b = 3, then
(a + b) (a – b) = a2 – b2 becomes
(300 + 3) (300 – 3) = 3002 – 32
303 × 297 = 90000 – 9
297 × 303 = 89,991

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

(vi) 990 × 1010
990 × 1010 = (1000 – 10) (1000 + 10)
Taking a = 1000 and b = 10, then
(a – b) (a + b) = a2 – b2 becomes
(1000 – 10) (1000 + 10) = 10002 – 102
990 × 1010 = 1000000 – 100
990 × 1010 = 999900

(vii) 51 × 52
= (50 + 1) (50 + 1)
Taking x = 50, a = 1 and b = 2
then (x + a) (x + b) = x2 + (a + b) x + ab becomes
(50 + 1) (50 + 2) = 502 + (1 + 2) 50 + (1 × 2)
2500 + (3) 50 + 2 = 2500 + 150 + 2
51 × 52 = 2652

Question 8.
Simplify: (a + b)2 – 4ab
Solution:
(a + b)2 – 4ab = a2 + b2 + 2ab – 4ab = a2 + b2 – 2ab = (a – b)2

Question 9.
Show that (m – n)2 + (m + n)2 = 2(m2 + n2)
Solution:
Taking the LHS = (m – n)2 + (m + n)2
Samacheer-Kalvi-7th-Maths-Solutions-Term-3-Chapter-3-Algebra-Ex-3.1-1

Question 10.
If a + b = 10 , and ab = 18, find the value of a2 + b2.
Solution:
We have (a + b)2 = a2 + 2ab + b2
(a + b)2 = a2 + b2 + 2ab
given a + b = 0 and ab = 18
102 = = a2 + b2 + 2(18)
100 = = a2 + b2 + 36
100 – 36 = a2 + b2
a2 + b2 = 64

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 11.
Factorise the following algebraic expressions by using the identity a2 – b2 = (a + b)(a – b).
(i) z2 – 16
(ii) 9 – 4y2
(iii) 25a2 – 49b2
(iv) x4 – y4
Solution:
(i) z2 – 16
z2 – 16 = z2 – 42
We have a2 – b2 = (a + b) (a – b)
let a = z and b = 4,
z2 – 42 = (z + 4) (z – 4)

(ii) 9 – 4y2
9 – 4y2 = 32 – 22 y2 = 32 – (2y)2
let a = 3 and b = 2y, then
a2 – b2 = (a + b) (a – b)
∴ 32 – (2y)2 = (3 + 2y) (3 – 2y)
9 – 4y2 = (3 + 2y) (3 – 2y)

(iii) 25a2 – 49b2
25a2 – 49b2 = 52 – a2 – 72 = (5a)2 – (7b)2
let A = 5a and B = 7b
AB2
(5a)2 – (7b)2 = (5a + 7b) (5a – 7b)

(iv) x4 – y4
Let x4 – y4 = (x2)2 – (y2)2
We have a2 – b2 = (a + b) (a – b)
(x2)2 – (y2)2 = (x2 + y2) (x2 – y2)
x4 – y4 = (x2 + y2) (x2 – y2)
Again we have x2 – y2 = (x + y) (x – y)
∴ x4 – y4 = (x2 + y2) (x + y) (x – y)

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 12.
Factorise the following using suitable identity.
(i) x2 – 8x + 16
(ii) y2 + 20y + 100
(iii) 36m2 + 60m + 25
(iv) 64x2 – 112xy + 49y2
(v) a2 + 6ab + 9b2 – c2
Solution:
(i) x2 – 8x + 16
x2 – 8x + 16 = x2 – (2 × 4 × x) + 42
This expression is in the form of identity
a2 – 2ab + b2 = (a – b)2
x2 – 2 × 4 × x + 42 = (x – 4)2
∴ x2 – 8x + 16 = (x – 4) (x – 4)

(ii) y2 + 20y + 100
y2 + 20y + 100 = y2 + (2 × (10)) y + (10 × 10)
= y2 + (2 × 10 × y) + 102
This is of the form of identity
a2 + 2 ab + b2 = (a + b)2
y2 + (2 × 10 × y) + 102 = (y + 10)2
y2 + 20y + 100 = (y + 10)2
y2 + 20y + 100 = (y + 10) (y + 10)

(iii) 36m2 + 60m + 25
36m2 + 60m + 25 = 62 m2 + 2 × 6m × 5 + 52
This expression is of the form of identity
a2 + 2ab + b2 = {a + b)2
(6m)2 + (2 × 6m × 5) + 52
= (6m + 5)2
36m2 + 60m + 25 = (6m + 5) (6m + 5)

(iv) 64x2 – 112xy + 49y2
64x2 – 112xy + 49y2 = 82 x2 – (2 × 8x × 7y) + 72y2
This expression is of the form of identity
a2 – 2ab + b2 = (a- b)2
(8x)2 – (2 × 8x × 7y) + (7y)2 = (8x – 7y)2
64x2 – 112xy + 49y2 = (8x – 7y) (8x – 7y)

(v) a2 + 6ab + 9b2 – c2
a2 + 6ab + 9b2 – c2 = a2 + 2 × a × 3b + 32 b2 – c2
= a2 + (2 × a × 3b) + (3b)2 – c2
This expression is of the form of identity
[a2 + 2ab + b2] – c2 = (a + b)2 – c2
a2 + (2 × a × 36) + (3b)2 – c2 = (a + 3b)2 – c2
Again this RHS is of the form of identity
a2 – b2 = (a + b) (a – b)
(a + 3b)2 – c2 = [(a + 3b) + c] [(a + 3b) – c]
a2 + 6ab + 9b2 – c2 = (a + 3b + c) (a + 3b – c)

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Objective Type Questions

Question 1.
If a + b = 5 and a2 + b2 = 13, then ab = ?
(i) 12
(ii) 6
(iii) 5
(iv) 13
Answer:
(ii) 6
Hint: (a + b)2 = 25
13 + 2ab = 25
2ab = 12
ab = 6

Question 2.
(5 + 20)(-20 – 5) = ?
(i) -425
(ii) 375
(iii) -625
(iv) 0
Answer:
(iii) -625
Hint: (50 + 20) (-20 – 5) = -(5 + 20)2 = – (25)2 = – 625

Question 3.
The factors of x2 – 6x + 9 are
(i) (x – 3)(x – 3)
(ii) (x – 3)(x + 3)
(iii) (x + 3)(x + 3)
(iv) (x – 6)(x + 9)
Answer:
(i) (x – 3)(x – 3)
Hint: x2 – 6x + 9 = x2 – 2(x) (3) + 32
a2 – 2ab + b2 – (a- b)2 = (x – 3)2 = (x – 3) (x – 3)

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.1

Question 4.
The common factors of the algebraic expression ax2y, bxy2 and cxyz is
(i) x2y
(ii) xy2
(iii) xyz
(iv) x
Ans :
(iv) xy
Hint: ax2y = a × x × x × y
bxy2 = b × x × y × y
cxyz = C × x × y × z
Common factor = xy

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