RBSE Maths Chapter 9 – Algebraic Expressions Mathematics Class 8 Important questions and solutions are available here. The additional, exercise important questions and solutions of Chapter 9, available at BYJU’S, contain detailed explanations. All these important questions are based on the new pattern designed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.
Chapter 9 of the RBSE Class 8 Maths will help the students to solve problems related to power of expression, like and unlike terms, addition and subtraction of algebraic expressions, multiplication of algebraic expressions, multiplying a monomial by a monomial, multiplying a binomial or a trinomial by a monomial, multiplying a binomial by a binomial, multiplying a binomial by trinomial, standard identities, application of identities.
RBSE Maths Chapter 9: Exercise 9.1 Textbook Important Questions and Solutions
Question 1: Find the product of the following pairs of monomials.
(i) 3, 5x
(ii) -5p, -2q
(iii) 7t2, -3n2
(iv) 6m, 3n
(v) -5x2, -2x
Solution:
[i] 3 * (5x)= 3 * 5 * (x)
= 15x
[ii] -5p, -2q= (-5) x p x (-2) x q
= (-5) x (-2) x pq
= 10pq
[iii] 7t2, -3n2= 7 x t2 x (-3) x n2
= 7 x (-3) x t2 x n2
= -21t2n2
[iv] 6m, 3n= 6 x m x 3 x n
= 6 x 3 x m x n
= 18 mn
[v] -5x2, -2x= (-5) * (x2) * (-2) * (-x)
= (-5) * (-2) * (x2) * (-x)
= 10x3
Question 2: Complete the table below.
Solution:
| * | 7 | x | y | 2z | a | -5b | c |
| 7 | 49 | 7x | 7y | 14z | 7a | -35b | 7c |
| x | 7x | x2 | xy | 2zx | ax | -5bx | cx |
| 2y | 14y | 2xy | 2y2 | 4yz | 2ay | -10by | 2cy |
| -3a | -21a | -3ax | -3ay | -6az | -3a2 | 15ab | -3ac |
| b | 7b | bx | by | 2bz | ab | -5b2 | bc |
| y | 7y | xy | y2 | 2yz | ay | -5by | cy |
| 2x3 | 14x3 | 2x4 | 2x3y | 4x3z | 2x3a | -10x3b | 2x3c |
| a4 | 7a4 | a4x | a4y | 2a4z | a5 | -5a4b | a4c |
| z2 | 7z2 | xz2 | yz2 | 2z3 | az2 | -5bz2 | cz2 |
Question 3: Multiply the following monomial.
[i] xy, x2y, xy, x
[ii] m, n, mn, m3n, mn3
[iii] kl, lm, km, klm
Solution:
[i] xy, x2y, xy, x= (x) * (y) * (x) * (x) * (y) * (x) * (y) * (x)
= (x) * (x) * (x) * (x) * (x) * (y) * (y) * (y)
= x5y3
[ii] m, n, mn, m3n, mn3= m x n x m x n x m x m x m x n x m x n x n x n
= m x m x m x m x m x m x n x n x n x n x n x n
= m6n6
[iii] kl, lm, km, klm= k x l x l x m x k x m x k x l x m
= k x k x k x l x l x l x m x m x m
= k3l3m3
Question 4: Find the simple interest = [PTR] / [100], if principal P = 4x2, time (T) = 5x and rate of interest R = 5y.
Solution:
Simple interest = [PTR] / [100]
= [(4x2) * (5x) * (5y)] / [100]
= [(4 * (x) * (x) * 5 * (x) * 5 * (y)] / [100]
= [4 * 5 * 5 * (x) * (x) * (x) * (y)] / [100]
= [100x3y] / [100]
= x3y
RBSE Maths Chapter 9: Exercise 9.2 Textbook Important Questions and Solutions
Question 1: Multiply the binomials given below.
[i] (2x + 5) and (3x – 7)
[ii] (x – 8) and (3y + 5)
[iii] (1.5p – 0.5q) and (1.5p + 0.5q)
[iv] (a + 3b) and (x + 5)
[v] (2lm + 3l2) and (3lm – 5l2)
[vi] [(¾)a2 + 3b2] and [4a2 – (5/3)b2]
Solution:
[i] (2x + 5) and (3x – 7)= 2x (3x – 7) + 5 (3x – 7)
= (2x) * (3x) – (2x) * 7 + 5 * (3x) – (5 * 7)
= 2 * (x) * 3 * (x) – 2 * (x) * 7 + 5 * 3 * (x) – 35
= 2 * 3 * (x) * (x) – 2 * 7 * (x) + 15 * (x) – 35
= 6 * (x2) – 14 * (x) + 15x – 35
= 6x2 – 14x + 15x – 35
= 6x2 + x – 35
[ii] (x – 8) and (3y + 5)= x (3y + 5) – 8 (3y + 5)
= (x) * 3y + (x) * 5 – 8 * 3 * y – 8 * 5
= (x) * 3y + 5x – 8 * 3 * y – 40
= 3 * (x) * y + 5x – 24 * y – 40
= (3x) * y + 5x – 24y – 40
= 3xy + 5x – 24y – 40
[iii] (1.5p – 0.5q) and (1.5p + 0.5q)= 1.5p (1.5p + 0.5q) – 0.5q (1.5p + 0.5q)
= 1.5p x 1.5p + 1.5p x 0.5q – 0.5q x 1.5p – 0.5q x 0.5q
= 2.25p2 x + 0.75 x pq – 0.75 x qp – 0.25 q2
= 2.25p2 + 0.75pq – 0.75pq – 0.25q2
= 2.25p2 – 0.25q2
[iv] (a + 3b) and (x + 5)= a (x + 5) + 3b (x + 5)
= a x (x) + a x 5 + 3b x (x) + 3b x 5
= ax + 5a + 3bx + 3 x 5 x b
= ax + 5a+ 3bx + 15b
[v] (2lm + 3l2) and (3lm – 5l2)= 2lm (3lm – 5l2) + 3l2 (3lm – 5l2)
= [2lm x 3lm] – [2lm x 5l2] + 3l2 x 3lm – [3l2 x 5l2]
= 6l2m2 – 10l3m + 9l3m – 15l4
= 6l2m2 – l3m – 15l4
[vi] [(¾)a2 + 3b2] and [4a2 – (5/3)b2]![RBSE Class 8 Maths Solutions Chapter 9 Exercise 9.2 Question Number 2 : Subpart 6 [answer]](https://cdn1.byjus.com/wp-content/uploads/2020/04/rbse-class-8-maths-solutions-chapter-9-exercise-9-2-question-number-2-subpart-6-answer.png "RBSE Class 8 Maths Solutions Chapter 9")
Question 2: Find the product of the following.
[i] (3x + 8) (5 – 2x)
[ii] (x + 3y) (3x – y)
[iii] (a2 + b) (a + b2)
[iv] (p2 – q2) (2p + q)
Solution:
Question 3: Simplify the following.
[i] (x + 5) (x – 7) + 35
[ii] (a2 – 3) (b2 + 3) + 5
[iii] (t + s2) (t2 – s)
[iv] (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
[v] (a + b) (a2 – ab + b2)
[vi] (a + b + c) (a + b – c)
[vii] (a + b) (a – b) – a2 + b2
Solution:
RBSE Maths Chapter 9: Exercise 9.3 Textbook Important Questions and Solutions
Question 1: Find the products using a suitable identity.
[i] (x + 5) (x + 5)
[ii] (3x + 2) (3x + 2)
[iii] (5a – 7) (5a – 7)
[iv] (3p – [½]) (3p – [½])
[v] (1.2m – 0.3) (1.2m + 0.3)
[vi] (x2 + y2) (x2 – y2)
[vii] (7a – 9b) (7a – 9b)
Solution:
Question 2: Use the identity (x + a) (x + b) = x2 + (a + b)x + ab to find the following products.
[i] (x + 1) (x + 2)
[ii] (3x + 5) (3x + 1)
[iii] (4x – 5) (4x – 1)
Solution:
[i] (x + 1) (x + 2)Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 1; b = 2
= x2 + (1 + 2)x + 2 x 1
= x2 + 3x + 2
[ii] (3x + 5) (3x + 1)Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 5; b = 1
= [3x]2 + (5 + 1)* 3x + 5 x 1
= 9x2 + 6 * 3x + 5
= 9x2 + 18x + 5
[iii] (4x – 5) (4x – 1)Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = (-5); b = (-1)
= [4x]2 + ([-5] + [-1]) * 4x + (-5 * -1)
= 16x2 + (-6) * 4x + [5]
= 16x2 – 24x + 5
Question 3: Find the following squares by using identities.
[i] (b – 7)2
[ii] (xy + 3z)2
Solution:
Question 4: Simplify the following problems.
[i] (a2 – b2)2
[ii] (2n + 5)2 – (2n – 5)2
Solution:
Question 5: Show that
[i] (2a + 3b)2 – (2a – 3b)2 = 24ab
[ii] (4x + 5)2 – 80x = (4x – 5)2
Solution:
Question 6: Using the identities, evaluate the following.
[i] 992
[ii] 1032
Solution:
[i] 992= (100 – 1)2
(a – b)2 = a2 – 2ab + b2
a = 100; b = 1
= (100)2 – 2 * (100) * (1) + 12
= 10000 – 200 + 1
= 9801
[ii] 1032= (100 + 3)2
(a + b)2 = a2 + 2ab + b2
a = 100; b = 3
= (100)2 + 2 * (100) * (3) + 32
= 10000 + 600 + 9
= 10609
Question 7: Using a2 – b2 = (a + b) (a – b), find the values of the following.
[i] 102 – 992
[ii] (10.3)2 – (9.7)2
Solution:
[i] 102 – 992= (10 + 99) (10 – 99)
= (109) (-89)
= -9701
[ii] (10.3)2 – (9.7)2= (10.3 + 9.7) (10.3 – 9.7)
= (20) (0.6)
= 12
Question 8: Using a2 – b2 = (a + b) (a – b), find the values of the following.
[i] 103 x 102
[ii] 7.1 x 7.3
[iii] 102 x 99
[iv] 9.8 x 9.6
Solution:
[i] 103 x 102= (100 + 3) (100 + 2)
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 3; b = 2
= 1002 + (3 + 2) * 100 + [3 x 2]
= 10000 + 5 * 100 + 6
= 10000 + 500 + 6
= 10506
[ii] 7.1 x 7.3= (7 + 0.1) (7 + 0.3)
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 0.1; b = 0.3
= 72 + [0.1 + 0.3] * 7 + [0.1 x 0.3]
= 49 + [0.4] * 7 + 0.03
= 49 + 2.8 + 0.03
= 51.83
[iii] 102 x 99= (100 + 2) (100 – 1)
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 2; b = (-1)
= 1002 + (2 + [-1]) * 100 + (2) * (-1)
= 10000 + 100 – 2
= 10098
[iv] 9.8 x 9.6= (9 + 0.8) (9 + 0.6)
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab
a = 0.8; b = 0.6
= 92 + [0.8 + 0.6] * 9 + [0.8 x 0.6]
= 81 + [1.4] * 9 + 0.48
= 81 + 12.6 + 0.48
= 94.08
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