ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8 ICSE Ch-10 Maths Solutions

ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8 ICSE Ch-10 Maths Solutions. We Provide Step by Step Answer of  Check Your Progress Questions for Algebraic Expression and Identities as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 8th
Chapter-10 Algebraic Expression and Identities
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Check Your Progress Questions
Edition 2023-2024

Algebraic Expression and Identities Check Your Progress

ML Aggarwal Class 8 ICSE Maths Solutions

Page-189

Question 1. Add the following expressions:

(i) -5x2y + 3xy2 – 7xy + 8, 12x2y – 5xy2 + 3xy – 2
(ii) 9xy + 3yz – 5zx, 4yz + 9zx – 5y, -5xz + 2x – 5xy

Answer:

(i) (-5x2y + 3xy2 – 7xy + 8) + (12x2y – 5xy2 + 3xy – 2)

= 7x2y – 2xy2 – 4xy + 6

(ii) (9xy + 3yz – 5zx) + (4yz + 9zx – 5y, -5xz + 2x – 5xy)

= 4xy + 7yz – zx + 2x – 5y

Question 2. Subtract:

(i) 5a + 3b + 11c – 2 from 3a + 5b – 9c + 3
(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y+ 5x – 3y

Answer:

(i) 5a – 3b + 11c – 2 from 3a + 5b – 9c + 3

= (3a + 5b – 9c + 3) – (5a – 3b + 11c – 2)

= 3a + 5b – 9c + 3 – 5a + 3b – 11c + 2

= -2a + 8b – 20c + 5

(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y2 + 5x – 3y

= (8x2 – 5xy + 2y2 + 5x – 3y) – (10x2 – 8y2 + 5y – 3)

= 8x2 – 5xy + 2y2 + 5x – 3y – 10x2 + 8y2 – 5y + 3

= – 2x2 – 5xy + 10y2 + 5x – 8y – 3

Question 3. What must be added to 5x2 – 3x + 1 to get 3x3 – 7x2 + 8?

Answer:

= (3x3 – 7x2 + 8) – (5x2 – 3x + 1)

= 3x3 – 7x2 + 8 – 5x2 + 3x – 1

= 3x3 – 12x2 + 3x + 7

Question 4. Find the product of

(i) 3x2y and -4xy2
(ii) –45xy, 57 yz and –14zx

Answer:

(i) 3x2y and -4xy2

= 3x2 × (-4xy2)

= -12x2+1 y1+2

= 12x3y3

(ii) –(4/5)xy, (5/7)yz and –(14/9)zx

= –(4/5)xy × (5/7)yz × –(14/9)zx

= –(4/5) × (5/7) × –(14/9) x2y2z2

= (8/9)x2y2z2

Question 5. Multiply:

(i) (3pq – 4p2 + 5q2 + 7) by -7pq
(ii) (34x2y – 45xy + 56xy2) by – 15xyz

Answer:

(i) (3pq – 4p2 + 5q2 + 7) × (-7pq)

= -7pq × 3pq – 7pq × (-4p2) + (-7pq) (5q2) – 7pq × 7

= -21p2q2 + 28p3q – 35pq3 – 49pq

(ii) (3/4x2y – 4/5xy + 5/6xy2) × (– 15xyz)

ml class 8 chapter 10.3 img 19

Question 6. Multiply:

(i) (5x2 + 4x – 2) by (3 – x – 4x2)
(ii) (7x2 + 12xy – 9y2) by (3x2 – 5xy + 3y2)

Answer:

(i) (5x2 + 4x – 2) × (3 – x – 4x2)

= 5x2(3 – x – 4x2) + 4x(3 – x – 4x2) – 2(3x – x – 4x2)

= 15x2 – 5x3 – 20x4 + 12x – 4x2 – 16x3 – 6x + 2x + 8x2

= -20x4 – 21x3 + 19x2 + 14x – 6

(ii) (7x2 + 12xy – 9y2) x (3x2 – 5xy + 3y2)

= 7x2(3x2 – 5xy + 3y2) + 12xy(3x2 – 5xy + 3y2) – 9y2(3x2 – 5xy + 3y2)

= 21x4 – 35x3y + 21x2y2 + 36x3y – 60x2y2 + 36xy3 – 27x2y2 + 45xy3 – 27y4

= 21x4 + x3y + 81xy3 – 66x2y2 – 27y4

Question 7. Simplify the following expressions and evaluate them as directed:

(i) (3ab – 2a2 + 5b2) x (2b2 – 5ab + 3a2) + 8a3b – 7b4 for a = 1, b = -1
(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y for x = 0, y = 1.

Answer:

(i) (3ab – 2a2 + 5b2) × (2b2 – 5ab + 3a2) + 8a3b – 7b4

= 3ab(2b2 – 5ab + 3a2) – 2a2(2b2 – 5ab + 3a2) + 5b2(2b2 – 5ab + 3a2) + 8a3b – 7b4

= 6ab32 – 15a2b2 + 9a3b – 4a2b2 + 10a3b – 6a4 + 10b4 – 25ab3 + 15a2b2 + 8a3b – 7b4

= 27a3b – 4a2b2 – 19ab3 – 6a4 + 3b4

Putting, a = 1 and b = (-1)

= 27(1 )3 (-1) – 4(1)2 (-1)2 – 19 (1) (-1)3 – 6(1)4 + 3(-1)4

= -27 – 4 + 19 – 6 + 3

= -37 + 22

= -15

(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y

1.7x(2y + 3x + 4) – 2.5y(2y + 3x + 4) – 7.8x2 – 10y

= 3.4xy + 5.1x2 + 6.8x – 5y2 – 7.5xy – 10y – 7.8x2 – 10y

= -2.7x2 – 4.1xy – 5y2 + 6.8x – 20y

Putting, x = 0 and y = 1

= -2.7 × 0 – 4.1 × 0 × 1 – 5(1)2 + 6.8 × 0 – 20 × 1

= 0 + 0 – 5 + 0 – 20

= -25

(ML Aggarwal Algebraic Expression and Identities Check Your Progress Class 8)

Question 8. Carry out the following divisions:

(i) 66pq2r3 ÷ 11qr2
(ii) (x3 + 2x2 + 3x) ÷ 2x

Answer:

(i) 66pq2r3/ 11qr2

= 6pq2-1r3-2

= 6pqr

(ii) (x3 + 2x2 + 3x)/ 2x

= x3/2x + 2x2/2x + 3x/2x

= ½ x2 + x + 3/2

Question 9. Divide 10x4 – 19x3 + 17x2 + 15x – 42 by 2x2 – 3x + 5.

Answer:

(10x4 – 19x3 + 17x2 + 15x – 42) ÷ (2x2 – 3x + 5)

Question 9. Divide 10x4 – 19x3 + 17x2 + 15x – 42 by 2x2 – 3x + 5.

Thus, Quotient = 5x2 – 2x – 7 and Remainder = 4x – 7


Algebraic Expression and Identities Check Your Progress

ML Aggarwal Class 8 ICSE Maths Solutions

Page-190

Question 10. Using identities, find the following products:

(i) (3x + 4y) (3x + 4y)
(ii) (52𝑎𝑏)(52𝑎𝑏)
(iii) (3.5m – 1.5n) (3.5m + 1.5n)
(iv) (7xy – 2)(7xy + 7)

Answer:

(i) (3x + 4y) (3x + 4y)

= (3x + 4y)2

= (3x)2 + 2 × 3x × 4y + (4y)2  [Using, (a + b)2 = a2 + 2ab + b2]

= 9x2 + 24xy + 16y2

(ii) (5a/2 – b) (5a/2 – b)

= (5a/2 – b)2

= (5a/2)2 + 2 × 5a/2 × (-b) + (b)2  [Using, (a – b)2 = a2 – 2ab + b2]

= 25a2/4 – 5ab + b2

(iii) (3.5m – 1.5n) (3.5m + 1.5n)

= (3.5m)2 – (1.5n)2  [Using, (a – b)(a + b) = a2 – b2]

= 12.25m2 – 2.25n2

(iv) (7xy – 2)(7xy + 7)

= (7xy)+ (-2 + 7) × (7xy) + (-2) × 7 [Using, (x + a)(x + b) = x2 + (a + b)x + ab]

= 49x2y2 + 35xy – 14

Question 11. Using suitable identities, evaluate the following:

(i) 1052
(ii) 972
(iii) 201 × 199
(iv) 872 – 132
(v) 105 × 107

Answer:

(i) (105)2 = (100 + 5)2

= (100)2 + 2 × 100 × 5 + (5)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 10000 + 1000 + 25

= 11025

(ii) (97)2 = (100 – 3)2

= (100)2 – 2 × 100 × 3 + (3)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 10000 – 600 + 9

= 10009 – 600

= 9409

(iii) 201 × 199 = (200 + 1) (200 – 1)

= (200)2 – (1)2 [Using, (a + b) (a – b) = a2 – b2]

= 40000 – 1

= 39999

(iv) 872 – 132

= (87 + 13) (87- 13) [Using, a2 – b2 = (a + b)(a – b)]

= 100 × 74

= 7400

(v) 105 × 107

= (100 + 5) (100 + 7)

= (100)2 + (5 + 7) × 100 + 5 × 7 [Using, (x + a)(x – b) = x2 + (a + b)x + ab]

= 10000 + 1200 + 35

= 11235

Question 12. Prove that following:

(i) (a + b)2 – (a – b)2 + 4ab
(ii) (2a + 3b)2 + (2a – 3b)2 = 8a2 + 18b2

Answer:

(i) Taking the RHS, we have

RHS = (a – b)2 + 4ab

= a2– 2ab + b2 + 4ab

= a2 + 2ab + b2

= (a + b)2 = L.H.S.

(ii) Taking the LHS, we have

LHS = (2a + 3b)2 + (1a – 3b)2

= (2a)2 + 2 × 2a × 3b + (3b)2 + (2a)2 – 2 × 2a × 3b + (3b)2

= 4a2 + 12ab + 9b2 + 4a2 – 12ab + 9b2

= 8a2 + 18b2 = RHS

Question 13. If x + 1/𝑥 = 5, evaluate

(i) x2 + 1/x

(ii) x4 + 1/x4

Answer:

(i) We have, x + 1/x = 5

On squaring on both sides, we get

(x + 1/x)2 = 52

x2 + 1/x2 + 2 × x × 1/x = 25

x2 + 2 + 1/x2 = 25

x2 + 1/x2 = 25 – 2

x2 + 1/x2 = 23

(ii) Again, squaring x2 + 1/x2 = 23 on both sides, we get

(x2 + 1/x2)2 = 232

x4 + 1/x4 + 2 × x4 × 1/x4 = 529

x4 + 1/x4 + 2 = 529

x4+ 1/x4 = 529 – 2

x4 + 1/x4 = 527

Question 14. If a + b = 5 and a2 + b2 = 13, find ab.

Answer:

a + b = 5 and a2 + b2 = 13

On squaring a + b = 5 both sides, we get

(a + b)2 = (5)2

a2 + b2 + 2ab = 25

13 + 2ab = 25 ⇒ 2ab = 25 – 13 = 12

⇒ ab = 12/2 = 6

∴ ab = 6

— End of Algebraic Expression and Identities Check Your Progress Class 8 ICSE Maths Solutions :–

Return to : ML Aggarwal Maths Solutions for ICSE Class -8

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