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

Question 1.

Add the following expressions:

(i) -5x^{2}y + 3xy^{2} – 7xy + 8, 12x^{2}y – 5xy^{2} + 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) 10x^{2} – 8y^{2} + 5y – 3 from 8x^{2} – 5xy + 2y^{2 }+ 5x – 3y

Solution:

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

(ii) 10x^{2} – 8y^{2} + 5y – 3 from 8x^{2} – 5xy + 2y^{2} + 5x – 3y

Question 3.

What must be added to 5x^{2} – 3x + 1 to get 3x^{3} – 7x^{2} + 8?

Solution:

Required expression

= (3x^{3} – 7x^{2} + 8) – (5x^{2} – 3x + 1)

= 3x^{3} – 7x^{2} + 8 – 5x^{2} + 3x – 1

= 3x^{3} – 12x^{2} + 3x + 7

Question 4.

Find the product of

(i) 3x^{2}y and -4xy^{2}

(ii) –\(\frac{4}{5}\)xy, \(\frac{5}{7}\)yz and –\(\frac{14}{9}\)zx

Solution:

Product of

(i) 3x^{2}y and -4xy^{2}

= 3x^{2} × (-4xy^{2})

= -12x^{2+1} y^{1+2}

= 12x^{3}y^{3}

Question 5.

Multiply:

(i) (3pq – 4p^{2} + 5q^{2} + 7) by -7pq

(ii) (\(\frac{3}{4}\)x^{2}y – \(\frac{4}{5}\)xy + \(\frac{5}{6}\)xy^{2}) by – 15xyz

Solution:

(i) (3pq – 4p^{2} + 5q^{2} + 7) by -7pq

= -7pq × 3pq – 7pq × (-4p^{2}) + (-7pq) (5q^{2}) – 7pq × 7

= -21p^{2}q^{2} + 28p^{3}q – 35pq^{3} – 49pq

Question 6.

Multiply:

(i) (5x^{2} + 4x – 2) by (3 – x – 4x^{2})

(ii) (7x^{2} + 12xy – 9y^{2}) by (3x^{2} – 5xy + 3y^{2})

Solution:

(i) 5x^{2} + 4x – 2) by (3 – x – 4x^{2})

= 5x^{2}(3 – x – 4x^{2}) + 4x(3 – x – 4x^{2}) – 2(3x – x – 4x^{2})

= 15x^{2} – 5x^{3} – 20x^{4} + 12x – 4x^{2} – 16x^{3} – 6x + 2x + 8x^{2}

= -20x^{4} – 21x^{3} + 19x^{2} + 14x – 6

(ii) (7x^{2} + 12xy – 9y^{2}) by (3x^{2} – 5xy + 3y^{2})

= 7x^{2}(3x^{2} – 5xy + 3y^{2}) + 12xy(3x^{2} – 5xy + 3y^{2}) – 9y^{2}(3x^{2} – 5xy + 3y^{2})

= 21x^{4} – 35x^{3}y + 21x^{2}y^{2} + 36x^{3}y – 60x^{2}y^{2} + 36xy^{3} – 27x^{2}y^{2} + 45xy^{3} – 27y^{4}

= 21x^{4} + x^{3}y + 81xy^{3} – 66x^{2}y^{2} – 27y^{4}

Question 7.

Simplify the following expressions and evaluate them as directed:

(i) (3ab – 2a^{2} + 5b^{2}) x (2b^{2} – 5ab + 3a^{2}) + 8a^{3}b – 7b^{4} for a = 1, b = -1

(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x^{2} – 10y for x = 0, y = 1.

Solution:

(i) (3ab – 2a^{2} + 5b^{2}) × (2b^{2} – 5ab + 3a^{2}) + 8a^{3}b – 7b^{4}

= 3ab(2b^{2} – 5ab + 3a^{2}) – 2a^{2}(2b^{2} – 5ab + 3a^{2}) + 5b^{2}(2b^{2} – 5ab + 3a^{2}) + 8a^{3}b – 7b^{4}

= 6ab^{32} – 15a^{2}b^{2} + 9a^{3}b – 4a^{2}b^{2} + 10a^{3}b – 6a^{4} + 10b^{4} – 25ab^{3} + 15a^{2}b^{2} + 8a^{3}b – 7b^{4}

= 27a^{3}b – 4a^{2}b^{2} – 19ab^{3} – 6a^{4} + 3b^{4}

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) – 7.8x^{2} – 10y

1.7x(2y + 3x + 4) – 2.5y(2y + 3x + 4) – 7.8x^{2} – 10y

= 3.4xy + 5.1x^{2} + 6.8x – 5y^{2} – 7.5xy – 10y – 7.8x^{2} – 10y

= -2.7x^{2} – 4.1xy – 5y^{2} + 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) 66pq^{2}r^{3} ÷ 11qr^{2}

(ii) (x^{3} + 2x^{2} + 3x) ÷ 2x

Solution:

Question 9.

Divide 10x^{4} – 19x^{3} + 17x^{2} + 15x – 42 by 2x^{2} – 3x + 5.

Solution:

(10x^{4} – 19x^{3} + 17x^{2} + 15x – 42) ÷ (2x^{2} – 3x + 5)

Quotient = 5x^{2} – 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} = a^{2} + 2ab + b^{2}}

= 9x^{2} + 24xy + 16y^{2}

(iii) (3.5m – 1.5n)(3.5m + 1.5n)

= (3.5m)^{2} – (1.5n)^{2} {∵ (a – b)(a + b) = a^{2} – b^{2}}

= 12.25m^{2} – 2.25n^{2}

(iv) (7xy – 2)(7xy + 7)

= (7xy)^{2 }+ (-2 + 7) × (7xy) + (-2) × 7

{∵(x + a)(x + b) = x^{2} + (a + b)x + ab}

= 49x^{2}y^{2} + 35xy – 14

Question 11.

Using suitable identities, evaluate the following:

(i) 105^{2}

(ii) 97^{2}

(iii) 201 × 199

(iv) 87^{2} – 13^{2}

(v) 105 × 107

Solution:

(i) (105)^{2} = (100 + 5)^{2}

= (100)^{2} + 2 × 100 × 5 + (5)^{2}

{(a + b)^{2} = a^{2} + 2ab + b^{2}}

= 10000 + 1000 + 25 = 11025

(ii) (97)^{2} = (100 – 3)^{2}

= (100)^{2} – 2 × 100 × 3 + (3)^{2}

{(a – b)^{2} = a^{2} – 2ab + b^{2}}

= 10000 – 600 + 9

= 10009 – 600 = 9409

(iii) 201 × 199 = (200 + 1) (200 – 1)

= (200)^{2} – (1)^{2}

{(a + b) (a – b) = a^{2} – b^{2}}

= 40000 – 1 = 39999

(iv) 87^{2} – 13^{2}

= (87 + 13) (87- 13)

{a^{2} – b^{2} = (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)

= x^{2} + (a + b)x + ab

= x^{2} + (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} = 8a^{2} + 18b^{2}

Solution:

(i) (a + b)^{2} = (a – b)^{2} + 4ab

RHS = (a – b)^{2} + 4ab = a^{2}– 2ab + b^{2} + 4ab

= a^{2} + 2ab + b^{2} = (a + b)^{2} = L.H.S.

(ii) (2a + 3b)^{2} + (2a – 3b)^{2} = 8a^{2} + 18b^{2}

LHS = (2a + 3b)^{2} + (1a – 3b)^{2}

= (2a)^{2} + 2 × 2a × 3b + (3b)^{2} + (2a)^{2} – 2 × 2a × 3b + (3b)^{2}

= 4a^{2} + 12ab + 9b^{2} + 4a^{2} – 12ab + 9b^{2}

= 8a^{2} + 18b^{2} = 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 a^{2} + b^{2} = 13, find ab.

Solution:

a + b = 5 and a^{2} + b^{2} = 13

a + b = 5

Squaring both sides,

(a + b)^{2} = (5)^{2}

a^{2} + b^{2} + 2ab = 25

13 + 2ab = 25 ⇒ 2ab = 25 – 13 = 12

⇒ ab = \(\frac{12}{2}\) = 6

∴ ab = 6