ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Check Your Progress

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Check Your Progress

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
Solution:

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

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

Question 3.
What must be added to 5x² – 3x + 1 to get 3x³ – 7x² + 8?
Solution:
Required expression
= (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) –\(\frac{4}{5}\)xy, \(\frac{5}{7}\)yz and –\(\frac{14}{9}\)zx
Solution:
Product of
(i) 3x²y and -4xy²
= 3x² × (-4xy²)
= -12x²⁺¹ y¹⁺²
= 12x³y³

Question 5.
Multiply:
(i) (3pq – 4p² + 5q² + 7) by -7pq
(ii) (\(\frac{3}{4}\)x²y – \(\frac{4}{5}\)xy + \(\frac{5}{6}\)xy²) by – 15xyz
Solution:
(i) (3pq – 4p² + 5q² + 7) by -7pq
= -7pq × 3pq – 7pq × (-4p²) + (-7pq) (5q²) – 7pq × 7
= -21p²q² + 28p³q – 35pq³ – 49pq

Question 6.
Multiply:
(i) (5x² + 4x – 2) by (3 – x – 4x²)
(ii) (7x² + 12xy – 9y²) by (3x² – 5xy + 3y²)
Solution:
(i) 5x² + 4x – 2) by (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) by (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.
Solution:
(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
We have, a = 1, 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
We have, x = 0, y = 1
= -2.7 × 0 – 4.1 × 0 × 1 – 5(1)² + 6.8 × 0 – 20 × 1
= 0 + 0 – 5 + 0 – 20 = -25

Question 8.
Carry out the following divisions:
(i) 66pq²r³ ÷ 11qr²
(ii) (x³ + 2x² + 3x) ÷ 2x
Solution:

Question 9.
Divide 10x⁴ – 19x³ + 17x² + 15x – 42 by 2x² – 3x + 5.
Solution:
(10x⁴ – 19x³ + 17x² + 15x – 42) ÷ (2x² – 3x + 5)

Quotient = 5x² – 2x – 7
Remainder = 4x – 7

Question 10.
Using identities, find the following products:
(i) (3x + 4y) (3x + 4y)
(ii) \(\left(\frac{5}{2} a-b\right)\left(\frac{5}{2} a-b\right)\)
(iii) (3.5m – 1.5n) (3.5m + 1.5n)
(iv) (7xy – 2)(7xy + 7)
Solution:
(i) (3x + 4y)(3x + 4y)
= (3x + 4y)²
= (3x)² + 2 × 3x × 4y + (4y)² {∵ (a + b)² = a² + 2ab + b²}
= 9x² + 24xy + 16y²

(iii) (3.5m – 1.5n)(3.5m + 1.5n)
= (3.5m)² – (1.5n)² {∵ (a – b)(a + b) = a² – b²}
= 12.25m² – 2.25n²

(iv) (7xy – 2)(7xy + 7)
= (7xy)² + (-2 + 7) × (7xy) + (-2) × 7
{∵(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
Solution:
(i) (105)² = (100 + 5)²
= (100)² + 2 × 100 × 5 + (5)²
{(a + b)² = a² + 2ab + b²}
= 10000 + 1000 + 25 = 11025

(ii) (97)² = (100 – 3)²
= (100)² – 2 × 100 × 3 + (3)²
{(a – b)² = a² – 2ab + b²}
= 10000 – 600 + 9
= 10009 – 600 = 9409

(iii) 201 × 199 = (200 + 1) (200 – 1)
= (200)² – (1)²
{(a + b) (a – b) = a² – b²}
= 40000 – 1 = 39999

(iv) 87² – 13²
= (87 + 13) (87- 13)
{a² – b² = (a + b) {a – b)}
= 100 × 74 = 7400

(v) 105 × 107 = (100 + 5) (100 + 7)
= (100)² + (5 + 7) × 100 + 5 × 7
{(x + a)(x – b)
= x² + (a + b)x + ab
= 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
Solution:
(i) (a + b)² = (a – b)² + 4ab
RHS = (a – b)² + 4ab = a²– 2ab + b² + 4ab
= a² + 2ab + b² = (a + b)² = L.H.S.

(ii) (2a + 3b)² + (2a – 3b)² = 8a² + 18b²
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 + \(\frac{1}{x}\) = 5, evaluate

Solution:
(i) x + \(\frac{1}{x}\) = 5
Squaring both sides,

Question 14.
If a + b = 5 and a² + b² = 13, find ab.
Solution:
a + b = 5 and a² + b² = 13
a + b = 5
Squaring both sides,
(a + b)² = (5)²
a² + b² + 2ab = 25
13 + 2ab = 25 ⇒ 2ab = 25 – 13 = 12
⇒ ab = \(\frac{12}{2}\) = 6
∴ ab = 6

ML Aggarwal Class 8 Solutions for ICSE Maths