Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Students can Download Maths Chapter 3 Algebra Ex 3.3 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 2 Chapter 3 Algebra Ex 3.3

Question 1.
Fill in the blanks.

(i) The degree of the term a3b2c4d2 is _______
(ii) Degree of the constant term is _______
(iii) The coefficient of leading term of the expression 3z2y + 2x – 3 is _______
Answers:
(i) 11
(ii) 0
(iii) 3

Question 2.
Say True or False.

(i) The degree of m2 n and mn2 are equal.
(ii) 7a2b and -7ab2 are like terms.
(iii) The degree of the expression -4x2 yz is -4
(iv) Any integer can be the degree of the expression.
Answers:
(i) True
(ii) False
(iii) False
(iv) True

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 3.
Find the degree of the following terms.
(i) 5x2
(ii) -7 ab
(iii) 12pq2 r2
(iv) -125
(v) 3z
Solution:
(i) 5x2
In 5x2, the exponent is 2. Thus the degree of the expression is 2.

(ii) -7ab
In -7ab, the sum of powers of a and b is 2. (That is 1 + 1 = 2).
Thus the degree of the expression is 2.

(iii) 12pq2 r2
In 12pq2 r2, the sum of powers of p, q and r is 5. (That is 1 +2 + 2 = 5).
Thus the degree of the expression is 5.

(iv) -125
Here – 125 is the constant term. Degree of constant term is 0.
∴ Degree of -125 is 0.

(v) 3z
The exponent is 3z is 1.
Thus the degree of the expression is 1.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 4.
Find the degree of the following expressions.
(i) x3 – 1
(ii) 3x2 + 2x + 1
(iii) 3t4 – 5st2 + 7s2t2
(iv) 5 – 9y + 15y2 – 6y3
(v) u5 + u4v + u3v2 + u2v3 + uv4
Solution:
(i) x3 – 1
The terms of the given expression are x3, -1
Degree of each of the terms: 3,0
Terms with highest degree: x3.
Therefore, degree of the expression is 3.

(ii) 3x2 + 2x + 1
The terms of the given expression are 3x2, 2x, 1
Degree of each of the terms: 2, 1, 0
Terms with highest degree: 3x2
Therefore, degree of the expression is 2.

(iii) 3t4 – 5st2 + 7s2t2
The terms of the given expression are 3t4, – 5st2, 7s3t2
Degree of each of the terms: 4, 3, 5
Terms with highest degree: 7s2t2
Therefore, degree of the expression is 5.

(iv) 5 – 9y + 15y2 – 6y3
The terms of the given expression are 5, – 9y , 15y2, – 6y3
Degree of each of the terms: 0, 1, 2, 3
Terms with highest degree: – 6y3
Therefore, degree of the expression is 3.

(v) u5 + u4v + u3v2 + u2v3 + uv4
The terms of the given expression are u5, u4v , u3v2, u2v3, uv4
Degree of each of the terms: 5, 5, 5, 5, 5
Terms with highest degree: u5, u4v , u3v2, u2v3, uv4
Therefore, degree of the expression is 5.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 5.
Identify the like terms : 12x3y2z, – y3x2z, 4z3y2x, 6x3z2y, -5y3x2z
Solution:
-y3 x2z and -5y3x2z are like terms.

Question 6.
Add and find the degree of the following expressions.
(i) (9x + 3y) and (10x – 9y)
(ii) (k2 – 25k + 46) and (23 – 2k2 + 21 k)
(iii) (3m2n + 4pq2) and (5nm2 – 2q2p)
Solution:
(i) (9x + 3y) and (10x – 9y)
This can be written as (9x + 3y) + (10x – 9y)
Grouping the like terms, we get
(9x + 10x) + (3y – 9y) = x(9 + 10) + y(3 – 0) = 19x + y(-6) = 19x – 6y
Thus degree of the expression is 1.

(ii) (k2 – 25k + 46) and (23 – 2k2 + 21k)
This can be written as (k2 – 25k + 46) + (23 – 2k2 + 21k)
Grouping the like terms, we get
(k2 – 2k2) + (-25 k + 21 k) + (46 + 23)
= k2 (1 – 2) + k(-25 + 21) + 69 = – 1k2 – 4k + 69
Thus degree of the expression is 2.

(iii) (3m2n + 4pq2) and (5nm2 – 2q2p)
This can be written as (3m2n + 4pq2) + (5nm2 – 2q2p)
Grouping the like terms, we get
(3m2n + 5m2n) + (4pq2 – 2pq2)
= m2n(3 + 5) + pq2(4 – 2) = 8m2n + 2pq2
Thus degree of the expression is 3.

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 7.
Simplify and find the degree of the following expressions.
(i) 10x2 – 3xy + 9y2 – (3x2 – 6xy – 3y2)
(ii) 9a4 – 6a3 – 6a4 – 3a2 + 7a3 + 5a2
(iii) 4x2 – 3x – [8x – (5x2 – 8)]
Solution:
(i) 10x2 – 3xy + 9y2 – (3x2 – 6xy – 3y2)
= 10x2 – 3xy + 9y2 + (-3x2 + 6xy + 3y2)
= 10x2 – 3xy + 9y2 – 3x2 + 6xy + 3y2
= (10x2 – 3x2) + (- 3xy + 6xy) + (9y2 + 3y2)
= x2(10 – 3) + xy(-3 + 6) + y2(9 + 3)
= x2(7) + xy(3) + y2(12)
Hence, the degree of the expression is 2.

(ii) 9a4 – 6a3 – 6a4 – 3a2 + 7a3 + 5a2
= (9a4 – 6a4) + (- 6a3 + 7a3) + (- 3a2 + 5a2)
= a4(9-6) + a3 (- 6 + 7) + a2(-3 + 5)
= 3a4 + a3 + 2a2
Hence, the degree of the expression is 4.

(iii) 4x2 – 3x – [8x – (5x2 – 8)]
= 4x2 – 3x – [8x + -5x2 + 8)]
= 4x2 – 3x – [8x – 5x2 – 8]
= 4x2 – 3x – 8x + 5x2 – 8
(4x2 + 5x2) + (- 3x – 8x) – 8
= x2(4+ 5) + x(-3-8) – 8
= x2(9) + x(- 11) – 8
= 9x2 – 11x – 8
Hence, the degree of the expression is 2.

Objective Type Question

Question 8.
3p2 – 5pq + 2q2 + 6pq – q2 +pq is a
(i) Monomial
(ii) Binomial
(iii) Trinomial
(iv) Quadrinomial
Answer:
(iii) Trinomial

Question 9.
The degree of 6x7 – 7x3 + 4 is
(i) 7
(ii) 3
(iii) 6
(iv) 4
Answer:
(i) 7

Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 3 Algebra Ex 3.3

Question 10.
If p(x) and q(x) are two expressions of degree 3, then the degree of p(x) + q(x) is
(i) 6
(ii) 0
(iii) 3
(iv) Undefined
Answer:
(iii) 3

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