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

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#### 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

(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

(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?

= (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

(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

(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)

#### Question 6. Multiply:

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

(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.

(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

(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.

(10x4 – 19x3 + 17x2 + 15x – 42) ÷ (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

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#### 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)

(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

(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

(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

(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.

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 :–