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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