ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities

Class 8: Maths Chapter 10 solutions. Complete Class 8 Maths Chapter 10 Notes.

ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities

ML Aggarwal 8th Maths Chapter 10, Class 8 Maths Chapter 10 solutions

Exercise 10.1

1. Identify the terms, their numerical as well as literal coefficients in each of the following expressions:
(i) 12x²yz – 4xy²
(ii) 8 + mn + nl – lm

(iii) x²/3 + y/6 – xy²

(iv) -4p + 2.3q + 1.7r
Solution:

2. Identify monomials, binomials, and trinomials from the following algebraic expressions :
(i) 5p × q × r²
(ii) 3x² + y ÷ 2z
(iii) -3 + 7x²
(iv) (5a² – 3b² + c)/2
(v) 7x⁵ – 3x/y
(vi) 5p ÷ 3q – 3p² × q²
Solution:

(i) 5p × q × r² = 5pqr²

As this algebraic expression has only one term, its therefore a monomial.

(ii) 3x² + y ÷ 2z = 3x²/2z + y/2z

As this algebraic expression has two terms, its therefore a binomial.

(iii) -3 + 7x²

As this algebraic expression has two terms, its therefore a binomial.

(iv)

As this algebraic expression has three terms, its therefore a trinomial.

(v) 7x⁵ – 3x/y

As this algebraic expression has two terms, its therefore a binomial.

(vi) 5p ÷ 3q – 3p² × q² = 5p/3q – 3p²q²

As this algebraic expression has two terms, its therefore a binomial.

3. Identify which of the following expressions are polynomials. If so, write their degrees.
(i) 2/5x⁴ – √3x² + 5x – 1
(ii) 7x³ – 3/x² + √5
(iii) 4a³b² – 3ab⁴ + 5ab + 2/3
(iv) 2x²y – 3/xy + 5y³ + √3
Solution:

(i) It is a polynomial and the degree of this expression is 4.

(ii) It is not a polynomial.

(iii) It is a polynomial and the degree of this expression is 5.

(iv) It is not a polynomial.

4. Add the following expressions:
(i) ab – bv, bv – ca, ca – ab
(ii) 5p²q² + 4pq + 7, 3 + 9pq – 2p²q
(iii) l² + m² + n², lm + mn, mn + nl, nl + lm
(iv) 4x³ – 7x² + 9, 3x² – 5x + 4, 7x³ – 11x + 1, 6x² – 13x
Solution:

(i) ab – bc, bc – ca, ca – ab

On adding the expressions, we have

⇒ ab – bc + bc – ca + ca – ab = 0

(ii) 5p²q² + 4pq + 7,3 + 9pq – 2p²q²

On adding the expressions, we have

= 5p²q² + 4pq + 7 + 3 + 9pq – 2p²q²

= 5p²q² – 2p²q² + 4pq + 9pq + 7 + 3

= 3p²q² + 13pq + 10

(iii) l² + m² + n², lm + mn, mn + nl, nl + lm

On adding the expressions, we have

= l² + m² + n² + lm + mn + mn + nl + nl + lm

= l² + m² + n² + 2lm + 2mn + 2nl

(iv) 4x³ – 7x² + 9, 3x² – 5x + 4, 7x³ – 11x + 1, 6x² – 13x

On adding the expressions, we have

= 4x³ – 7x² + 9 + 3x² – 5x + 4 + 7x³ – 11² + 1 + 6x² – 13x

= 4x² + 7x³ – 7x² + 3x² + 6x² – 5x – 11x – 13x + 9 + 4 + 1

= 11x³ – 2x² – 29x + 14

5. Subtract:
(i) 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5
(ii) 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz
(iii) 4p²q – 3pq + 5pq² – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:

(i) Subtracting 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5, we have

= (14a – 5ab + 7b – 5) – (8a + 3ab – 2b + 7)

= 14a – 5ab + 7b – 5 – 8a – 3ab + 2b – 7

= 6a – 8ab + 9ab – 12

(ii) Subtracting 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz, we have

= (12xy – 3yz – 4zx + 5xyz) – (8xy + 4yz + 5zx)

= 12xy – 3yz – 4zx + 5xyz – 8xy – 4yz – 5zx

= 4xy – 7yz – 9zx + 5xyz

(iii) Subtracting 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q, we have

= (18 – 3p – 11q + 5pq – 2pq² + 5p²q) – (4p²q – 3pq + 5pq² – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq² + 5p²q – 7p²q + 3pq – 5pq² + 8p – 7q + 10

= 28 + 5p – 78q + 8pq – 7pq² + p²q

6. Subtract the sum of 3x² + 5xy + 7y² + 3 and 2x² – 4xy – 3y² + 7 from 9x² – 8xy + 11y²
Solution:

First, adding 3x² + 5xy + 7y² + 3 and 2x² – 4xy – 3y² + 7, we have

= 3x² + 5xy + 7y² + 3 + 2x² – 4xy – 3y² + 7

= 5x² + xy + 4y² + 10

Now,

Subtracting 5x² + xy + 4y² + 10 from 9x² – 8xy + 11y²

= (9x² – 8xy + 11y²) – (5x² + xy + 4y² + 10)

= 9x² – 8xy + 11y² – 5x² – xy – 4y² – 10

= 4x² – 9xy + 7y² – 10

7. What must be subtracted from 3a² – 5ab – 2b² – 3 to get 5a² – 7ab – 3b² + 3a?
Solution:

From the question, its understood that we have to subtract 5a² – 7ab – 3b² + 3a from 3a² – 5ab – 2b² – 3

= 3a² – 5ab – 2b² – 3 – (5a² – 7ab – 3b² + 3a)

= 3a² – 5ab – 2b² – 3 – 5a² + 7ab + 3b² – 3a

= -2a² + 2ab + b² – 3a – 3

8. The perimeter of a triangle is 7p² – 5p + 11 and two of its sides are p² + 2p – 1 and 3p² – 6p + 3. Find the third side of the triangle.
Solution:

Given,

Perimeter of a triangle = 7p² – 5p + 11

And, two of its sides are p² + 2p – 1 and 3p² – 6p + 3

We know that,

Perimeter of a triangle = Sum of three sides of triangle

⇒ 7p² – 5p + 11 = (p² + 2p – 1) + (3p² – 6p + 3) + (Third side of triangle)

7p² – 5p + 11 = (4p² – 4p + 2) + (Third side of triangle)

⇒ Third side of triangle = (7p² – 5p + 11) – (4p² – 4p + 2)

= (7p² – 4p²) + (- 5p + 4p) + (11 – 2)

= 3p² – p + 9

Thus, the third side of the triangle is 3p² – p + 9.

Exercise 10.2

1. Find the product of:
(i) 4x³ and -3xy
(ii) 2xyz and 0
(iii) –(2/3)p²q, (3/4)pq² and 5pqr
(iv) -7ab, -3a³ and –(2/7)ab²
(v) –½x² – (3/5)xy, (2/3)yz and (5/7)xyz
Solution:

Product of:

(i) 4x³ and -3xy = 4x³ × (-3xy) = -12x³⁺¹ y = -12x⁴y

(ii) 2xyz and 0 = 2xyz × 0 = 0

2. Multiply:
(i) (3x – 5y + 7z) by – 3xyz
(ii) (2p² – 3pq + 5q² + 5) by – 2pq
(iii) (2/3a²b – 4/5ab² + 2/7ab + 3) by 35ab
(iv) (4x² – 10xy + 7y² – 8x + 4y + 3) by 3xy
Solution:

(i) – 3xyz × (3x – 5y + 7z)

= (- 3xyz) × 3x + (- 3xyz) × (- 5y) + (- 3xyz) × (7z)

= – 9x²yz + 15xyz² – 21xyz²

(ii) -2pq × (2p² – 3pq + 5q² + 5)

= (-2pq) × 2p² + (-2pq) × (-3pq) + (- 2pq) × (5q²) + (-2pq) × 5

= -4p³q + 6p²q² – 10pq³ – 10pq

(iii)
by 35ab

= (2/3)a²b × 35ab – (4/5)ab² × 35ab + (2/7)ab × 35ab + 3 × 35ab

= (70/3)a³b² – 28a²b³ + 10a²b² + 105ab

(iv) (4x² – 10xy + 7y² – 8x + 4y + 3) by 3xy

= 4x² × 3xy – 10xy × 3xy + 7y² × 3xy – 8x × 3xy + 4y × 3xy + 3 × 3xy

= 12x³y – 30x²y² + 21xy³ – 24x²y + 12xy² + 9xy

3. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(i) (p²q, pq²)
(ii) (5xy, 7xy²)
Solution:

(i) Given, sides of a rectangle are p²q and pq²

Hence,

Area = p²q × pq² = p²⁺¹ × q²⁺¹ = p³q³

(ii) Given, sides are 5xy and 7xy²

Hence,

Area = 5xy × 7xy² = 35x¹⁺¹ × y¹⁺² = 35x²y³

4. Find the volume of rectangular boxes with the following length, breadth and height respectively:
(i) 5ab, 3a²b, 7a⁴b²
(ii) 2pq, 4q², 8rp
Solution:

Given are the length, breadth and height of a rectangular box:

(i) 5ab, 3a²b, 7a⁴b²

∴ Volume = Length × breadth × height

= 5ab × 3a²b × 7a⁴b²

= 5 × 3 × 7 × a¹⁺²⁺⁴ × b¹⁺¹⁺²

= 105a⁷b⁴

(ii) 2pq, 4q², 8rp

∴ Volume = Length × breadth × height

= 2pq × 4q² × 8rp

= 2 × 4 × 8 × p¹⁺¹ × q¹⁺² × r

= 64p²q³r

5. Simplify the following expressions and evaluate them as directed:
(i) x²(3 – 2x + x²) for x = 1; x = -1; x = 2/3 and x = –1/2
(ii) 5xy(3x + 4y – 7) – 3y(xy – x² + 9) – 8 for x = 2, y = -1
Solution:

(i) x²(3 – 2x + x²)

For x = 1; x = -1; x = 2/3 and x = –1/2

x²(3 – 2x + x²) = 3x² – 2x³ + x⁴

(a) For x = 1

3x² – 2x³ + x⁴ = 3(1)² – 2(1 )³ + (1)⁴

= 3 × 1 – 2 × 1 + l

= 3 – 2 + 1 = 2

(b) For x = -1

3x² – 2x³ + x⁴ = 3(-1)² – 2(-1)³ + (-1)⁴

= 3 × 1 – 2 × (-1) + 1

= 3 + 2 + 1 = 6

(c) For x = 2/3

3x² – 2x³ + x⁴ = 3(2/3)² – 2(2/3)³ + (2/3)⁴

= 3 × (4/9) – 2 × (8/27) + (16/81)

= (4/3) – (16/27) + (16/81)

= (108 – 48 + 16)/81

= (124 – 48)/81

= 76/81

(d) For x = -1/2

3x² – 2x³ + x⁴ = 3(-1/2)² – 2(-1/2)³ + (-1/2)⁴

= 3 × (1/4) – 2 × (-1/8) + (1/16)

= (3/4) + ¼ + (1/16)

= (12 + 4 + 1)/16

= 17/16

(ii) 5xy(3x + 4y – 7) – 3y(xy – x² + 9) – 8

= 15x²y + 20xy² – 35xy – 3xy² + 3 x²y – 21y – 8

= 18x²y + 17xy² – 35xy – 27y – 8

When x = 2, y = -1, we have

= 18(2)² × (-1) + 17(2) (-1)² – 35(2) (-1) – 27(-1) – 8

= 18 × 4 × (-1) + 17 × 2 × 1 – 35 × 2 × (-1) – 27 × (-1) – 8

= -74 + 34 + 70 + 27 – 8

= 131 – 80 = 51

6. Add the following:
(i) 4p(2 – p²) and 8p³ – 3p
(ii) 7xy(8x + 2y – 3) and 4xy²(3y – 7x + 8)
Solution:

Adding,

(i) 4p(2 – p²) and 8p³ – 3p

= 8p – 4p³ + 8p³ – 3p

= 5p + 4p³

= 4p³ + 5p

(ii) 7xy(8x + 2y – 3) and 4xy²(3y – 7x + 8)

= 56x²y + 14xy² – 21xy + 12xy³ – 28x²y² + 32xy²

= 12xy³ – 28x²y² + 56x²y +46xy² – 21xy

7. Subtract:
(i) 6x(x – y + z)- 3y(x + y – z) from 2z(-x + y + z)
(ii) 7xy(x² -2xy + 3y²) – 8x(x²y – 4xy + 7xy²) from 3y(4x²y – 5xy + 8xy²)
Solution:

Subtracting,

(i) 6x(x – y + z) – 3y(x + y – z) from 2z(-x + y + z)

⇒ 6x² – 6xy + 6xz – 3xy – 3y² + 3yz from – 2xz + 2yz + 2z²

= (-2xz + 2yz + 2z²) – (6x² – 6xy + 6xz – 3xy – 3y² + 3yz)

= – 2xz + 2yz + 2z² – 6x² + 6xy – 6xz + 3xy + 3y² – 3yz

= 9xy – yz – 8zx – 6x² + 3y² + 2z²

(ii) 7xy(x² – 2xy + 3y²) – 8x(x²y – 4xy + 7xy²) from 3y(4x²y – 5xy + 8xy²)

⇒ 7x³y – 14x²y² + 21xy³ – 8x³y + 32x²y – 56x²y² from 12x²y² – 15xy² + 24xy³

= (12x²y² – 15xy² + 24xy³) – (7x³y – 14x²y² + 21xy³ – 8x³y + 32x²y – 56x²y²

= 12x²y² – 15xy² + 24xy³ – 7x³y + 14x²y² – 12xy³ + 8x³y – 32x²y + 56x²y²

= 82x²y² + 3xy³ + x³y – 15xy² – 32x²y

Exercise 10.3

1. Multiply:
(i) (5x – 2) by (3x + 4)
(ii) (ax + b) by (cx + d)
(iii) (4p – 7) by (2 – 3p)
(iv) (2x² + 3) by (3x – 5)
(v) (1.5a – 2.5b) by (1.5a + 2.56)
(vi) 
Solution:

(i) (5x – 2) by (3x + 4)

= (5x – 2) × (3x + 4)

= 5x (3x + 4) – 2 (3x + 4)

= 15x² + 20x – 6x – 8

= 15x²+ 14x – 8

(ii) (ax + b) by (cx + d)

= (ax + b) × (cx + d)

= ax (cx + d) + b (cx + d)

= acx² + adx + bcx + bd

(iii) (4p – 7) by (2 – 3p)

= (4p – 7) × (2 – 3p)

= 4p(2 – 3p) -7(2 – 3p)

= 8p – 12p² – 14 + 21p

= 29p – 12p² – 14

(iv) (2x² + 3) by (3x – 5)

= (2x² + 3) (3x – 5)

= 2x²(3x – 5) + 3(3x – 5)

= 6x³ – 10x² + 9x – 15

(v) (1.5a – 2.5b) by (1.5a + 2.5b)

= (1.5a – 2.5b) (1.5a + 2.5b)

= 1.5a(1.5 + 2.5b) – 2.5b(1.5a + 2.5b)

= 2.25a² + 3.75ab – 3.75a6 – 6.25b²

= 2.25a² – 6.25b²

2. Multiply:
(i) (x – 2y + 3) by (x + 2y)
(ii) (3 – 5x + 2x²) by (4x – 5)

Solution:

(i) (x – 2y + 3) by (x + 2y)

= (x – 2y + 3) × (x + 2y)

= x (x + 2y) – 2y(x + 2y) + 3 (x + 2y)

= x² + 2xy – 2xy – 4y² + 3x + 6y

= x² – 4y² + 3x + 6y

(ii) (3 – 5x + 2x²) by (4x – 5)

= (4x – 5) (3 – 5x + 2x²)

= 4x(3 – 5x + 2x²) – 5(3 – 5x + 2x²)

= 12x – 20x² + 8x³ – 15 + 25x – 10x²

= 8x³ – 30x² + 37x – 15

3. Multiply:
(i) (3x² – 2x – 1) by (2x² + x – 5)
(ii) (2 – 3y – 5y²) by (2y – 1 + 3y²)
Solution:

(i) (3x² – 2x – 1) by (2x² + x – 5)

= (3x² – 2x – 1) (2x² + x – 5)

= 3x²(2x² + x – 5) – 2x(2x² + x – 5) -1(2x² + x – 5)

= 6x⁴ + 3x³ – 15x² – 4x³ – 2x² + 10x – 2x² – x + 5

= 6x⁴ – x³ – 19x² + 9x + 5

(ii) (2 – 3y – 5y²) by (2y- 1 + 3y²)

= (2 – 3y – 5y²) × (2y- 1 + 3y²)

= 2(2y – 1 + 3y² ) – 3y (2y – 1 + 3y²) -5y²(2y – 1 + 3y²)

= 4y – 2 + 6y² – 6y² + 3y – 9y³ – 10y³ + 5y² – 15y⁴

= -15y⁴ – 19y³ + 5y² + 7y – 2

4. Simplify:
(i) (x² + 3) (x – 3) + 9
(ii) (x + 3) (x – 3) (x + 4) (x – 4)
(iii) (x + 5) (x + 6) (x + 7)
(iv) (p + q – 2r) (2p – q + r) – 4qr
(v) (p + q) (r + s) + (p – q)(r – s) – 2(pr + qs)
(vi) (x + y + z) (x – y + z) + (x + y – z) (-x + y + z) – 4zx
Solution:

(i) (x² + 3) (x – 3) + 9

= x² (x – 3) + 3(x – 3) + 9

= x² – 3x² + 3x – 9 + 9

= x³ – 3x² + 3x

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

= {(x + 3) (x – 3)} × {(x + 4) (x – 4)}

= {x (x – 3) + 3 (x – 3)} {x (x – 4) + 4 (x – 4)}

= (x² – 3x + 3x – 9) {x² – 4x + 4x – 16}

= (x² – 9) (x² – 16)

= x² (x² – 16) – 9 (x² – 16)

= x⁴ – 16x² – 9x² + 144

= x⁴ – 25x² + 144

(iii) (x + 5) (x + 6) (x + 7)

= {(x + 5) × (x + 6)} (x + 7)

= (x² + 6x + 5x + 30) (x + 7)

= (x² + 11x + 30) (x + 7)

= x(x²+ 11x + 30) + 7(x²+ 11x + 30)

= x³ + 11x² + 30x + 7x² + 77x + 210

= x³ + 18x² + 107x + 210

(iv) (p + q – 2r)(2p – q + r) – 4qr

= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr

= 2p² – pq + pr + 2pq – q² + qr – 4pr + 2qr – 2r² – 4qr

= 2p² – q² – 2r² + pq – 3pr – 2qr

(v) (p + q)(r + s) + (p – q) (r – s) – 2(pr + qs)

= (pr + ps + qr + qs) + (pr – ps – qr + qs) – 2pr – 2qs

= 0

(vi) (x + y + z)(x – y + z) + (x + y – z)(-x + y + z) – 4zx

= x² – xy + xz + xy – y² + yz + xz – yz + z² – x² + xy + xz

– xy + x² + yx + xz – yz – z² – 4zx

= 0

5. If two adjacent sides of a rectangle are 5x² + 25xy + 4y² and 2x² – 2xy + 3y², find its area.
Solution:

Given,

The adjacent sides of a rectangle are 5x² + 25xy + 4y² and 2x² – 2xy + 3y²

So,

Area of rectangle = Product of two adjacent sides

= (5x² + 25xy + 4y²) (2x² – 2xy + 3y²)

= 10x⁴– 10x³y+ 15x²y² + 50x³y – 50x²y² + 75xy³ + 8x²y² – 8xy³ + 12y⁴

= 10x⁴ + 40x³y – 27x²y² + 67xy³ + 12y⁴

Thus,

The area of the rectangle is 10x⁴ + 40x³y – 27x²y² + 67xy³ + 12y⁴.

Exercise 10.4

1. Divide:
(i) – 39pq²r⁵ by – 24p³q³r
(ii) –3/4 a²b³ by 6/7 a³b²
Solution:

(i) – 39pq²r⁵ (÷) – 24p³q³r

= – 39pq²r⁵/ – 24p³q³r

2. Divide:
(i) 9x⁴ – 8x³ – 12x + 3 by 3x
(ii) 14p²q³ – 32p³q² + 15pq² – 22p + 18q by – 2p²q.

Solution:

3. Divide:
(i) 6x² + 13x + 5 by 2x + 1
(ii) 1 + y³ by 1 + y
(iii) 5 + x – 2x² by x + 1
(iv) x³ – 6x² + 12x – 8 by x – 2
Solution:

(i) 6x² + 13x + 5 ÷ 2x + 1

∴ Quotient = 3x + 5 and remainder = 0

(ii) 1 + y³ ÷ 1 + y

∴ Quotient = y² – y + 1 and remainder = 0

(iii) On arranging the terms of dividend in descending order of powers of x and then dividing, we get

– 2x² + x + 5 ÷ x + 1

∴ Quotient = – 2x + 3 and remainder = 2

(iv) x³ – 6x² + 12x – 8 ÷ x – 2

∴ Quotient = x² – 4x + 4 and remainder = 0

4. Divide:
(i) 6x³ + x² – 26x – 25 by 3x – 7
(ii) m³ – 6m² + 7 by m – 1
Solution:

(i) 6x³ + x² – 26x – 25 ÷ 3x – 7

∴ Quotient = 2x² + 5x + 3 and remainder = – 4

(ii) m³ – 6m² + 7 ÷ m – 1

∴ Quotient = m² – 5m – 5 and remainder = 2.

5. Divide:
(i) a³ + 2a² + 2a + 1 by a² + a + 1
(ii) 12x³ – 17x² + 26x – 18 by 3x² – 2x + 5
Solution:

(i) a³ + 2a² + 2a + 1 ÷ a² + a + 1

∴ Quotient = a + 1 and remainder = 0.

(ii) 12x³ – 17x² + 26x – 18 ÷ 3x² – 2x + 5


∴ Quotient = 4x – 3 and remainder = -3

6. If the area of a rectangle is 8x² – 45y² + 18xy and one of its sides is 4x + 15y, find the length of adjacent side.
Solution:

Given,

Area of rectangle = 8x² – 45y² + 18xy

And, one side = 4x + 15y

∴ Second (adjacent) side = Area of rectangle/ One side

= 8x² – 45y² + 18xy ÷ 4x + 15y

Thus, length of the adjacent side is 2x – 3y.

Exercise 10.5

1. Using suitable identities, find the following products:
(i) (3x + 5) (3x + 5)
(ii) (9y – 5) (9y – 5)
(iii) (4x + 11y) (4x – 11y)
(iv) (3m/2 + 2n/3) (3m/2 – 2n/3)
(v) (2/a + 5/b) (2a + 5/b)
(vi) (p²/2 + 2/q²) (p²/2 – 2/q²)
Solution:

(i) (3x + 5) (3x + 5)

= (3x + 5)²

= (3x)² + 2 × 3x × 5 + (5)² [Using, (a + b)² = a² + 2ab + b²]

= 9x² + 30x + 25

(ii) (9y – 5) (9y – 5)

= (9y – 5)²

= (9y)² – 2 × 9y × 5 + (5)² [Using, (a – b)² = a² – 2ab + b²]

= 81y² – 90y + 25

(iii) (4x + 11y)(4x – 11y)

= (4x)² – (11y)²

= 16x² – 121y² [Using, (a + b)(a – b) = a² – b²]

(iv) (3m/2 + 2n/3) (3m/2 – 2n/3)

= (3m/2)² – (2n/3)²

= 9m²/4 – 4n²/9 [Using, (a + b)(a – b) = a² – b²]

(v) (2/a + 5/b) (2a + 5/b)

= (2/a + 5/b)²

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

= 4/a² + 20a/b + 25/b²

(vi) (p²/2 + 2/q²) (p²/2 – 2/q²)

= (p²/2)² – (2/q²)² [Using, (a + b)(a – b) = a² – b²]

= p⁴/4 – 4/q⁴

2. Using the identities, evaluate the following:
(i) 81²
(ii) 97²
(iii) 105²
(iv) 997²
(v) 6.1²
(vi) 496 × 504
(vii) 20.5 × 19.5
(viii) 9.62
Solution:

(i) (81)² = (80 + 1)²

= (80)² + 2 × 80 × 1 + (1)²  [Using, (a + b)² = a² + 2ab + b²]

= 6400 + 160+ 1

= 6561

(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

(ii) (105)² = (100 + 5)²

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

= 10000+ 1000 + 25

= 11025

(iv) (997)² = (1000 – 3)²

= (1000)² – 2 × 1000 × 3 + (3)²  [Using, (a – b)² = a² – 2ab + b²]

= 1000000 – 6000 + 9

= 1000009 – 6000

= 994009

(v) (6.1)² = (6 + 0.1)²

= (6)² + 2 × 6 × 0.1 +(0.1)²  [Using, (a + b)² = a² + 2ab + b²]

= 36 + 1.2 + 0.01

= 37.21

(vi) 496 × 504

= (500 – 4) (500 + 4) [Using, (a + b) (a – b) = a² – b²]

= (500)² – (4)²

= 250000 – 16

= 249984

(vii) 20.5 × 19.5

= (20 + 0.5) (20 – 0.5) [Using, (a + b) (a – b) = a² – b²]

= (20)² – (0.5)²

= 400 – 0.25

= 399.75

(viii) (9.6)² = (10 – 0.4)²

= (10)² – 2 × 10 × 0.4 + (0.4)²  [Using, (a – b)² = a² – 2ab + b²]

= 100 – 8.0 + 0.16

= 92.16

3. Find the following squares, using the identities:
(i) (pq + 5r)² (ii) (5a/2 – 3b/5)²

(iii) (√2a + √3b)² (iv) (2x/3y – 3y/2x)²
Solution:

(i) (pq + 5r)²

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

= p²q² + 10pqr + 25r²

(ii) (5a/2 – 3b/5)²

= (5a/2)² – 2 × (5a/2) × (-3b/5) + (3b/5)² [Using, (a – b)² = a² – 2ab + b²]

= 25a²/4 – 3ab + 9b²/25

(iii) (√2a + √3b)²

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

= 2a² + 2√6ab + 3b²

(iv) (2x/3y – 3y/2x)²

4. Using the identity, (x + a) (x + b) = x² + (a + b)x + ab, find the following products:
(i) (x + 7) (x + 3)
(ii) (3x + 4) (3x – 5)
(iii) (p² + 2q) (p² – 3q)
(iv) (abc + 3) (abc – 5)
Solution:

(i) (x + 7) (x + 3)

= (x)² + (7 + 3)x + 7 × 3

= x² + 10x + 21

(ii) (3x + 4) (3x – 5)

= (3x)² + (4 – 5) (3x) + 4 × (-5)

= 9x² – 3x – 20

(iii) (P² + 2q)(p² – 3q)

= (p²)² + (2q – 3q)p² + 2q × (-3q)

= p⁴ – p²q – 6pq

(iv) (abc + 3) (abc – 5)

= (abc)² + (3 – 5)abc + 3 × (-5)

= a²b²c² – 2abc – 15

5. Using the identity, (x + a) (x + b) = x² + (a + b)x + ab, evaluate the following:
(i) 203 × 204
(ii) 8.2 × 8.7
(iii) 107 × 93
Solution:

(i) 203 × 204

= (200 + 3) (200 + 4)

= (200)² + (3 + 4) × 200 + 3 × 4

= 40000 + 1400 + 12

= 41412

(ii) 8.2 × 8.7

= (8 + 0.2) (8 + 0.7)

= (8)² + (0.2 + 0.7) × 8 + 0.2 × 0.7

= 64 + 8 × (0.9) + 0.14

= 64 + 7.2 + 0.14

= 71.34

(iii) 107 × 93

= (100 + 7) (100 – 7)

= (100)² + (7 – 7) × 100 + 7 × (-7)

= 10000 + 0 – 49

= 9951

6. Using the identity a² – b² = (a + b) (a – b), find
(i) 53² – 47²
(ii) (2.05)² – (0.95)²
(iii) (14.3)² – (5.7)²
Solution:

(i) 53² – 47²

= (50 + 3) (50 – 3)

= (50)² – (3)²

= 2500 – 9

= 2491

(ii) (2.05)² – (0.95)²

= (2.05 + 0.95) (2.05 – 0.95)

= 3 × 1.10

= 3.3

(iii) (14.3)² – (5.7)²

= (14.3 + 5.7) (14.3 – 5.7)

= 20 × 8.6

= 172

7. Simplify the following:
(i) (2x + 5y)² + (2x – 5y)²
(ii) (7a/2 – 5b/2)² – (5a/2 – 7b/2)²
(iii) (p² – q²r)² + 2p²q²r
Solution:

(i) (2x + 5y)² + (2x – 5y)² [Using, (a ± b)² = a² ± 2ab + b²]

= (2x)² + 2 × 2x × 5y + (5y)² + (2x)² – 2 × 2x × 5y + (5y)²

= 4x² + 20xy + 25y² + 4x² – 20xy + 25y²

= 8x² + 50y²

(ii) (7a/2 – 5b/2)² – (5a/2 – 7b/2)²

(iii) (p² – q²r)² + 2p²q²r [Using, (a – b)² = a² – 2ab + b²]

= (p²)² – 2 × p² × q²r + (q²r)² + 2p²q²r

= p⁴ – 2p²q + q⁴r² + 2p²q²r

= p⁴ + q⁴r²

8. Show that:
(i) (4x + 7y)² – (4x – 7y)² = 112xy
(ii) (3p/7 – 7q/6)² + pq = 9p²/49 + 49q²/36
(iii) (p – q)(p + q) + (q – r)(q + r) + (r – p) (r + p) = 0
Solution:

(i) Taking LHS, we have

LHS = (4x + 7y)² – (4x – 7y)² [Using, (a ± b)² = a² ± 2ab + b²]

= [(4x)² + 2 × 4x × 7y + (7y)²] – [(4x)² – 2 × 4x + 7y + (7y)²]

= (16x² + 56xy + 49y²) – (16x² – 56xy + 49y²)

= l6x² + 56xy + 49y² – 16x² + 56xy – 49y²

= 112xy = RHS

(ii) Taking LHS, we have

(iii) Taking LHS, we have

LHS = (p – q) (p + q) + (q – r) (q + r) + (r – p)(r + p)

= p² – q² + q² – r² + r² – p² [Using, (a + b) (a – b) = a² – b²]

= 0 = RHS

9. If x + 1/x = 2, evaluate:
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
Solution:

(i) We have, x + 1/x = 2

On squaring on both sides, we get

(x + 1/x)² = 2²

x² + 2 × x × 1/x + 1/x² = 4

x² + 2 + 1/x² = 4

x² + 1/x² = 4 – 2

Thus,

x² + 1/x² = 2

(ii) Again squaring, we get

(x² + 1/x²)² = 2²

x⁴ + 2 × x² × 1/x² + 1/x⁴ = 4

x⁴ + 2 + 1/x⁴ = 4

x⁴ + 1/x⁴ = 4 – 2

Thus,

x⁴ + 1/x⁴ = 2

10. If x – 1/x = 7, evaluate:
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
Solution:

We have, x – 1/x = 7

On squaring on both sides, we get

(x – 1/x)² = 7²

x² – 2 × x² × 1/x² + 1/x² = 49

x² – 2 + 1/x² = 49

x² + 1/x² = 49 + 2

Thus,

x² + 1/x² = 51

(ii) Again squaring, we get

(x² + 1/x²)² = 51²

x⁴ + 1/x⁴ + 2 × x² × 1/x² = 2601

x⁴ + 1/x⁴ + 2 = 2601

x⁴ + 1/x⁴ = 2601 – 2

Thus,

x⁴ + 1/x⁴ = 2599

11. If x² + 1/x² = 23, evaluate:
(i) x + 1/x (ii) x – 1/x
Solution:

We have, x² + 1/x² = 23

(i) (x + 1/x)² = x² + 1/x² + 2

= 23 + 2

= 25

Taking square root on both sides, we get

(x + 1/x) = ±5

Thus, x + 1/x = 5 or -5

(ii) (x – 1/x)² = x² + 1/x² – 2

= 23 – 2

= 21

Taking square root on both sides, we get

(x + 1/x) = ±√21

Thus, x + 1/x = √21 or -√21

12. If a + b = 9 and ab = 10, find the value of a² + b².
Solution:

Given,

a + b = 9 and ab = 10

Now, squaring a + b = 9 on both sides, we have

(a + b)² = (9)

a² + b² + 2ab = 81

a² + b² + 2 × 10 = 81

a² + b² + 20 = 81

a² + b² = 81 – 20 = 61

∴ a² + b² = 61

13. If a – b = 6 and a² + b² = 42, find the value of
Solution:

Given

a – b = 6 and a² + b² = 42

a – b = 6

Now, squaring a – b = 6 on both sides, we have

(a – b)² = (6)²

a² + b² – 2ab = 36

42 – 2ab = 36

2ab = 42 – 36 = 6

ab = 6/2 = 3

∴ ab = 3

14. If a² + b² = 41 and ab = 4, find the values of
(i) a + b
(ii) a – b
Solution:

Given, a² + b² = 41 and ab = 4

(i) (a + b)² = a² + b² + 2ab

= 41 + 2 × 4

= 41 + 8

= 49

∴ a + b = ±7

(ii) (a – b)² = a² + b² – 2ab

= 41 – 2 × 4

= 41 – 8

= 33

∴ a – b = ±√33

Check Your Progress

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:

(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

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

= (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

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

From the question, the required expression is

= (3x³ – 7x² + 8) – (5x² – 3x + 1)

= 3x³ – 7x² + 8 – 5x² + 3x – 1

= 3x³ – 12x² + 3x + 7

4. Find the product of
(i) 3x²y and -4xy²
(ii) –(4/5)xy, (5/7)yz and –(14/9)zx
Solution:

Product of:

(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

5. Multiply:
(i) (3pq – 4p² + 5q² + 7) by -7pq
(ii) (3/4x²y – 4/5xy + 5/6xy²) by – 15xyz
Solution:

(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)

6. Multiply:
(i) (5x² + 4x – 2) by (3 – x – 4x²)
(ii) (7x² + 12xy – 9y²) by (3x² – 5xy + 3y²)
Solution:

(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

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

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

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

(i) 66pq²r³/ 11qr²

= 6pq^2-1r^3-2

= 6pqr

(ii) (x³ + 2x² + 3x)/ 2x

= x³/2x + 2x²/2x + 3x/2x

= ½ x² + x + 3/2

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

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

Performing long division, we have

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

10. Using identities, find the following products:
(i) (3x + 4y) (3x + 4y)
(ii) (5a/2 – b) (5a/2 – b)
(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  [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 + (-2 + 7) × (7xy) + (-2) × 7 [Using, (x + a)(x + b) = x² + (a + b)x + ab]

= 49x²y² + 35xy – 14

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 [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

12. Prove that following:
(i) (a + b)² – (a – b)² + 4ab
(ii) (2a + 3b)² + (2a – 3b)² = 8a² + 18b²
Solution:

(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

13. If x + 1/x = 5, evaluate
(i) x² + 1/x² (ii) x⁴ + 1/x⁴
Solution:

(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

Hence, 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

Hence,

x⁴ + 1/x⁴ = 527

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

Given,

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

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ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities

Download PDF: ML Aggarwal Solutions for Class 8 Maths Chapter 10- Algebraic Expressions and Identities PDF

Chapterwise ML Aggarwal Solutions for Class 8 Maths :

• Chapter 1- Rational Numbers
• Chapter 2- Exponents and Powers
• Chapter 3- Squares and Square Roots
• Chapter 4- Cubes and Cube Roots
• Chapter 5- Playing with Numbers
• Chapter 6- Operation On Sets Venn Diagram
• Chapter 7- Percentage
• Chapter 8- Simple and Compound Interest
• Chapter 9- Direct and Inverse Variation
• Chapter 10- Algebraic Expressions and Identities
• Chapter 11- Factorisation
• Chapter 12- Linear Equations and Inequalities in One Variable
• Chapter 13- Understanding Quadrilaterals
• Chapter 14- Constructions of Quadrilaterals
• Chapter 15- Circle
• Chapter 16- Symmetry Reflection and Rotation
• Chapter 17- Visualising Solid Shapes
• Chapter 18- Mensuration
• Chapter 19- Data Handling

About ML Aggarwal

M. L. Aggarwal, is an Indian mechanical engineer, educator. His achievements include research in solutions of industrial problems related to fatigue design. Recipient Best Paper award, Manipal Institute of Technology, 2004. Member of TSTE.