Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th 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 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Recap Page No. 66 and 67

Question 1.
Write the numbers of terms in the following expressions.
(i) x + y + z – xyz
Solution:
4 terms

(ii) m2n2c
Solution:
1 term

(iii) a2b2c – ab2c2 + a2bc2 + 3abc
Solution:
4 terms

(iv) 8x2 – 4xy + 7xy2
Solution:
3 terms
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
Question 1.
2x2 – 5xy + 6y2 + 7x – 10y + 9
Solution:
Numerical co efficient in 2x2 is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y2 is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is – 10
Numerical co-efficient in 9 is 9

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 1
Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 3.
Pick out the like terms from the following.
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 6
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 7

Question 4.
Add : 2x, 6y, 9x – 2y
Solution:
2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y

Question 5.
Simplify : (5x3 y3 – 3x2 y2 + xy + 7) + (2xy + x3y3 – 5 + 2x2y2)
Solution:
(5x3y3 – 3x2y2 + xy + 7) + (2xy + x3y3 – 5 + 2x2y2)
= 5x3y3 + x3y3 – 3x2y2 + 2x2y2 + xy + 2xy + 7 – 5
= (5 + 1)x3y3 + (-3 + 2)x2y2 +(1 +2)xy + 2
= 6x3y3 – x2y2 + 3xy + 2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 6.
The sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
Solution:
Perimeter of the triangle = Sum of three sides
= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
= 4x – y + 12
∴ Perimeter of the triangle = 4x – y + 12 units.

Question 7.
Subtract -2mn from 6mn.
Solution:
6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 8.
Subtract 6a2 – 5ab + 3b2 from 4a2 – 3ab + b2.
Solution:
(4a2 – 3ab+ b2) – (6a2– 5ab + 3b2)
= (4a2 – 6a2) + (- 3ab -(-5 ab)] + (b2– 3b2)
= (4 – 6) a2 + [-3ab + (+ 5ab)] + (1 – 3) b2
= [4 + (- 6)] a2 + (-3 + 5) ab + [1+ (-3)]b2
= -2a2 + 2ab – 2b2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 9.
The length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
Solution:
Length of the log = 3a + 4b – 2
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 70
Length of the piece removed = 2a – b
Remaining length of the log = (3a + 4b – 2) – (2a – b)
= (3a – 2a) + [4b – (-b)] – 2
= (3 – 2)a + (4 + 1)b – 2
= a + 5b – 2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 10.
A tin had ‘x’ litre oil. Another tin had (3x2 + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x2 + 6) litres of oil from the second tin. How much oil was left In the second tin?
Solution:
Quantity of oil in the second tin = 3x2 + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x2 + 6x – 5) + (x + 7) litres
= 3x2 + (6x + x) + (-5 + 7)
= 3x2 + (6 + 1)x + 2
= 3x2 + 7x + 2 litres
Quantity of oil sold = x + 6 litres
∴ Quantity of oil left in the second tin = (3x2 + 7x + 2) – (x2 + 6)(3x2 – x2 ) + 7x + (2 – 6)
= (3 – 1)x2 + 7x + (-4) = 2x2 + 7x – 4
Quantity of oil left = 2x2 + 7x – 4 litres

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Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Solution:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y2 + 5y-1 – 3 is a an algebraic expression. But not a polynomial.

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Question 2.
-(5y2 + 2y – 6) Is this correct? If not, correct the mistake.
Solution:
Taking -(5y2 + 2y – 6) = 5y2 + [(-)(+) 2y] + [(-) × (-)6]
= -5y2 – 2y + 6
≠ -5y2 – 2y + 6
∴ Correct answer is -5y2 + 2y – 6 = -(5y2 + 2y + 6)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

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(i) 3ab2, -2a2b3
(ii) 4xy, 5y2x, (-x2)
(iii) 2m, -5n, -3p
Solution:
(i) (3ab2) × (-2a2b2) = (+) × (-) × (3 × 2) × (a × a2) × (b2 × b3) = -6a3 b5

(ii) (4xy) × (5y2x) × (-x2)
= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x2) × (y × y2)
= -20x4y3

(iii) (2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30mnp = 30 mnp

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Question 1.
Why 3 + (4x – 7y) ≠ 12x – 21y?
Solution:
Addition and multiplication are different
3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

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Question 1.
Which is corrcet? (3a)2 is equal to
(i) 3a2
(ii) 32a
(iii) 6a2
(iv) 9a2
Solution:
(3a) =32a2 = 9a2
(iv) 9a2 is the correct answer

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Question 1.
Multiply
(i) (5x2 + 7x – 3) by 4x2
Solution:
(5x2 + 7x – 3) × 4x2
= 4x2(5x2 + 7x – 3) Multiplication is commutative
= 4x2 (5x2 + 4x2 (7x) + 4x2 (-3)
= (4 × 5)(x2 × x2) + (4 × 7)(x2 × x) + (4 × -3)(x2)
= 20x4 + 28x3 – 12x2

(ii) (10x – 7y + 5z) by 6xyz
Solution:
(10x – 7y + 5z) by 6xyz
(10x – 7y + 5z) × 6xyz = 6xyz (10x – 7y + 5z) [∵ Multiplication is commutative]
= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × -7) + (x × y × y × z) + (6 × 5)(x × y × z × z)
= 60x2yz + (-42xy2z) + 30xyz2
= 60x2yz – 42x2z + 30xyz2

(iii) (ab + 3bc – 5ca) by – 3abc
Solution:
(ab + 3bc – 5ca) × (- 3abc) = (-3abc) (ab + 3bc – 5ca)
[∵ Multiplication is commutativel
= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
= (-3)(a × a × b × b × c) + (- 3 × 3) + (a × b × b × c × c)
= -3a2b2c – 9ab2c2 – 30a2bc2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

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Question 1.
Multiply
(i) (a – 5) and (a + 4)
Solution:
(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
= (a × a) + (a × 4) + (-5 × a) + (-5 × 4)
= a2 + 4a – 5a – 20 = a2 – a – 20

(ii) (a + b) and (a – b)
Solution:
(a + b) (a – b) = a(a – b) + b (a – b)
= (a × a) + (a × -b)+(b × a) + b(-b)
= a2 – ab + ab – b2 = a2 – b2

(iii) (m4 + n4) and (m – n)
Solution:
(m4 + n4)(m – n) = m4(m – n) + n4(m – n)
= (m4 × m) + (m4 × (-n)) + (n4 × m (n4 × (-n))
= m5 – m4n + mn4 – n5

(iv) (2x + 3)(x – 4)
Solution:
(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
= (2x2 × x) – (2x × 4) + (3 × x) – (3 × 4)
= 2x2 – 8x + 3x – 12 = 2x2 – 5x – 12

(v) (x – 5)(3x + 7)
Solution:
(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
= (x × 3x) + (x × 7) + (-5 × 3x) + (-5 × 7)
= 3x2 + 7x – 15x – 35
= 3x2 – 8x – 35

(vi) (x – 2)(6x – 3)
Solution:
(x – 2)(6x – 3) × (6x – 3) – 2(6x – 3)
= (x × 6x)+(x × (-3) × (2 × 6x) – (2 × 3)
= 6x2 – 3x – 12x + 6
= 6x2 – 15x + 6

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Question 2.
3x2 (x4 – 7x3 + 2), what is the highest power in the expression.
Solution:
3x2(x4 – 7x3 + 2) = (3x2) (x4) + 3x2 (-7x3)+ (3x2)2
= 3x6 – 21x5 + 6x2
Highest power is 6 in x6.

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.2

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Question 1.
Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\)
(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
(iii) When a monomial is divided by itself, we will get I?
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 50

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Question 1.
Divide
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 61
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 625

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Question 1.
Are the following divisions correct ?
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 51
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 52

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Question 1.
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 600
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 53
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 54

Exercise 3.3

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Question 1.
1. (p + 2)2 = …….
2. (3 – a)2 = …….
3. (62 – x2) = ………
4. (a + b)2 – (a – b)2 = …….
= a2 + 2ab + b2 – a2 – 2ab – b2
= (1 – 1)a2 + (2 + 2)ab + (+1 – 1 )b2 = 4ab
5. (a + b)2 = (a + b) × (a + b)
6. (m + n)( m – n) = m2 – n2
7. (m + 7)2 = m2 + 14m + 49
8. (k2 – 36) ≡ k2 – 62 = (k + 6)(k – 6)
9. m2 – 6m + 9 = (m – 3)2
10. (m – 10)(m + 5) = m2 + (-10 + 5)m + (-10)(5) = m2 – 5m – 50
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 90

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

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Question 1.
Expand using appropriate identities
Question 1.
(3p + 2q)2
Solution:
(3p + 2q)2
Comparing (3p + 2q)2 with (a + b)2, we get a = 3p and b = 2q.
(a + b)2 = a2 + 2ab + b2
(3p + 2q)2 = (3p)2+ 2(3p) (2q) + (2q)2
= 9p2 + 12pq + 4q2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 2.
(105)2
Solution:
(105)2 = (100 + 5)2
Comparing (100 + 5)2 with (a + b)2, we get a = 100 and b = 5.
(a + b)2 = a2 + 2ab + b2
(100 + 5)2 = (100)2 + 2(100)(5) + 52 = 1oooo + 1000 + 25
1052 = 11,025

Question 3.
( 2x – 5d)2
Solution:
(2x – 5d)2
Comparing with (a – b)2, we get a = 2x b = 5d.
(a – b)2 = a2 – 2ab + b2
(2x – 5d)2 = (2x)2 – 2(2x)(5d) + (5d)2
= 2x2 – 20 xd + 52d2 = 4x2 – 20 xd + 25d2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 4.
(98)2
Solution:
(98)2 = (100 – 2)2
Comparing (100 – 2)2 with (a – b)2 we get
a = 100, b = 2
(a – b)2 = a2 – 2ab + b2
(100 – 2)2 = 1002 – 2(100)(2) + 22
= 10000 – 400 + 4 = 9600 + 4 = 9604

Question 5.
(y – 5)(y + 5)
Solution:
(y – 5)(y + 5)
Comparing (y – 5) (y + 5) with (a – b) (a + b) we get
a = y; b = 5
(a – b)(a + b) = a2 – b2
(y – 5)(y + 5) = y2 – 52 = y2 – 25

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 6.
(3x)2 – 52
Solution:
(3x)2 – 52
Comparing (3x)2 – 52 with a2 – b2 we have
a = 3x; b = 5
(a2 – b2) = (a + b)(a – b)
(3x)2 – 52 = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
= 9x2 – 15x + 15x – 25 = 9x2 – 25

Question 7.
(2m + n)(2m +p)
Solution:
(2m + n) (2m + p)
Comparing (2m + n) (2m + p) with (x + a) (x + b) we have
x = 2n; a = n ;b = p
(x – a)(x + b) = x2 + (a + b)x + ab
(2m +n) (2m +p) = (2m2) + (n +p)(2m) + (n) (p)
= 22m2 + n(2m) + p(2m) + np
= 4m2 + 2mn + 2mp + np

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 8.
203 × 197
Solution:
203 × 197 = (200 + 3)(200 – 3)
Comparing (a + b) (a – b) we have
a = 200, b = 3
(a + b)(a – b) = a2 – b2
(200 + 3)(200 – 3) = 2002 – 32
203 × 197 = 40000 – 9
203 × 197 = 39991

Question 9.
Find the area of the square whose side is (x – 2)
Solution:
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
= x – 2x – 2x + 4x2 – 4x + 4

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 10.
Find the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
Solution:
Length of the rectangle = y+ 4
breadth of the rectangle = y – 3
Area of the rectangle = length × breadth
= (y + 4)(y – 3) = y2 + (4 +(-3))y + (4)(-3)
= y2 + y – 12

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Question 1.
Expand :
Question 1.
(x + 4)3
Solution:
Comparing (x + 4)3 with (a + b)3, we have a = x and b = 4.
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(x + 4)3 = x3 + 3x2(4) + 3(x)(4)2 + 43
= x3 + 12x2 + 48x + 64

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 2.
( y – 2)2
Solution:
Comparing (y – 2) with (a – b)3 we have a = y b = z
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(y – 2)2 = y3 – 3y(2) + 3y(2)2 + 23
= y3 – 6y2 + 12y + 8

Question 3.
(x + 1)(x + 3)(x + 5)
Solution:
Comparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
a = 1
b = 3
and c = 5
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 63

Exercise 3.4

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Question 1.
Factorize the following:
Question 1.
3y + 6
Solution:
3y + 6
3y + 6 = 3 × y + 2 × 3
Taking out the common factor 3 from each term we get 3 (y + 2)
∴ 3y + 6 = 3(y + 2)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 2.
10x2 + 15y
Solution:
10x2 + 15y2
10x2 + 15y2 = (2 × 5 × x × x) + (3 × 5 × y × y)
Taking out the common factor 5 we have
10x2 + 15y2 = 5(2x2 + 3y2)

Question 3.
7m(m – 5) + 1(5 – m)
Solution:
7m(m – 5) + 1(5 – m)
7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
= 7m(m – 5) – 1 (m – 5)
Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Question 4.
64 – x2
Solution:
64 – x2
64 – x2 = 82 – x2
This is of the form a2 – b2
Comparing with a2 – b2 we have a = 8, b = x
a2 – b2 = (a + b)(a – b)
64 – x2 = (8 + x)(8 – x)

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