# Frank ICSE Class 8 Solutions Chapter 10

## Frank ICSE Mathematics Class 8 Solutions Chapter 10 Algebraic Expressions

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 10, Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions, Exercise 10.1, 10.2, 10.3 and 10.4 Also our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 10 solutions. Here in this post all the solutions are based on latest Syllabus.

Algebraic Expressions Exercise 10.1 Solution :

Question no – (1)

Solution :

(a) 4x + 8

Terms = 4x

Coefficients = 4

Constants = 8

(b) 7a2 – a + 16

Terms = 7a2

Coefficients = 7

Constants = 16

(c) 17ab + 11bc – 3cd

Terms = 17ab 11bc – 3cd

Coefficients = 17 1 – 3

Constants = 0

(d) 0.75pq + 0.25p – 0.32

Terms = 0.75pq 0.25p

Coefficients = 0.75 0.25

Constants = – 0.32

(e) m/2 – n/2 + 5

Terms = m/2 – n/3

Coefficients = 1/2 – 1/3

Constants = 0

(f) 9x2 + 7y2 + 12z2

Terms = 9x2 7y2 12z2

Coefficients = 9 7 12

Constants = 0

Question no – (2)

Solution :

(a) Binomial

(b) Monomial

(c) Trinomial

(d) Binomial

(e) Monomial

(f) Binomial

(g) Trinomial

(h) Trinomial

Question no – (3)

Solution :

= 14xy, – 6yx; 7x2y, 21yx2; 12abc – 17bac; 17xy2, 4xy2; – n2m2, 8m2n2; 18pqr, 15pqr; 18p2qr, 7rqp2

Question no – (4)

Solution :

(a) 6x + 4y – 3z

7x – 11y – 9z

14x + 8y – 6z
—————————————
27x + y – 182

(b) 7p – 8q + 11r + 13

10p – 13p + 18

8p – 5q – 14r + 12

5p + 17q + 14r
—————————————
30p – 9q + 11r + 43

(c) 8x2 + 7x  12

17x2 – 15x – 21

– 9x2 + 11x + 4

4x2 – 18x + 19
—————————————
20x2 – 18x + 14

(d) 6ax – 2by + 3cz

– 11ax + 16by – 15cz

– 9ax – 3by + 10cz
—————————————
– 14ax + 11by – 2cz

(e) 15m2n – 17mn + 8mn2

13m2n – 15mn – 9mn2

12m2n + 21mn – 14mn2
—————————————
40m2n – 1mn – 15mn2

(f) 13x + 17y – 19z + 2

14x + 12y – 31

– 15x + 6z + 12

2x + y + 11z + 9
—————————————
14x + 30y – 2z – 8

Question no – (5)

Solution :

(a) 17x + 8y – 15z

+ 12x + 7y – 15z
—————————————
5x + y + 0

(b) 12ax + 6by + 111cz

+ 21ax + 15by – 6cz
—————————————
-9ax – 9by + 17cz

(c) 16x3 + 14x2 – 9x + 15

+ 17x3 – 19x2 – 21
—————————————–
x3 + 33×2 – 22x + 36

(d) 7p + 11q – 2r + 9

– 8p – 4q + 6r + 15
—————————————
15p + 15q + 8r – 6

Question no – (6)

Solution :

= [(13 + 19b + 12c) + (14a + 21b + 11c)]

= [(17a + 13b – 15c) + (8a + 12b – 18c)]

= [(13a + 14a) + 19b) + (12c + 11c)]

= [17a + 8a) + (13b + 126) + (- 15c – 18c)]

= [27a – 2b + 23c] – [25a + 25b – 33c]

= (27a – 25a) + (- 2b + 256) + (23c – 33c)

= 2a + 23b – 10c

Question no – (7)

Solution :

(a) A + B + C

= (5x + 11y + 152) + (12x – 13y + 152) + (12x – 13y + 19z) + (7x – 6y  21z)

= (5x + 12 + 7x) + 911y – 13y – 6y) + (15z + 19z + 21z)

= 24x – 7y + 25z

(b) A – B – C

= (5x + 11y – 15z) – (12x – 13y + 192) – (7x – 6y + 21z)

= (5x – 12x – 7x) + (11y + 13y + 6y) + (15z – 19z – 21z)

= – 14x + 30y – 55z

(c) A + B – C

= (5x + 11y – 15z) + (12x – 13y + 19z) – (7x – 6y + 21z)

= (5x – 12x – 7x) + (11y – 13y + 6y) + (- 15z + 19z – 121z)

= – 14x + 4y – 17z

(d) 2A + 3B + C

= 2(5x + 11y – 15z) + 3 (12x – 13y + 19z) + (7x – 6y + 21z)

= (10x + 22y – 20z) + (36x – 39y + 57z) + (7x – 6y + 21z)

= (10x + 36x + 7x) + (22y – 39y – 6y) + (- 30z + 57z + 21z)

= 910x + 36x + 7x) + (22y – 39y – 6y) + (- 30z + 57z + 21z)

= 53x – 23y + 48z

Question no – (8)

Solution :

As we know that,

Other side = 1/2 (Perimeter) – (given side)

Now,

= 1/2 (16a + 8b – 6c) – (5a + 3b – 4c)

= (8a + 4b – 3c) – (5a + 3b – 4c)

= (8a – 5a) + (4b – 3b) – (3c + 4c)

= 3a + b – 7c

Therefore, the other side will be 3a + b – 7c

Question no – (9)

Solution :

As we know that,

Third side = [Perimeter – (sum of 2 other side)

Now,

= (18p + 12q + 13r) – (5p + 3q – 4r) + (6p + 5q + 7r)

= (18p + 12q + 13r) – [11 + 8q + 3r]

= (18p – 11p) + (12q – 8q) + (13r – 3r)

= 7p + 4q + 10r

Therefore, the third side will be 7p + 4q + 10r

Algebraic Expressions Exercise 10.2 Solution :

Question no – (1)

Solution :

(a) (m + 3)n

= m. m + 3.m

= m2 + 3m

(b) 5(3x – 2)

= 5.x – 5.2

= 5x – 10

(c) 8(2x – 3y)

= 8.2x – 8.3y

= 16x – 24y

(d) 3x (x + y)

= 3x.x + 3x.y

= 3x2 + 3xy

(e) 2x (12x – 11y)

= 2x (12x – 1y)

= 2x.12x – 2x.11y

= 24x2 – 22xy

(f) x(3 – 2x + 5y)

= x.3 – x.2 + x.5y

= 3x – 2x2 + 5xy

(g) 15x (x + 5) + 7

= 15x.x + `5x.5 + 7

= 15×2 + 75n + 7

(h) 12x (x + 3y + 8z)

= 12x.x + 12x.3y + 12x.8z

= 12×2 + 36y + 96xz

(i) 10a(a – 5b + 3c)

= 10a.a – 10a.5b + 10a.3c

= 10a2 – 50ab + 30ac

(j) 6x (x + 3y + 5z)

= 6x.x + 6x.3y+ 6x.5z

= 6x2 + 18xy + 30xz

(k) 4x (x2 + 3x + 9)

= 4x.x2 + 4x.3x + 4x.9

= 4x3 + 12×2 + 36x

Question no – (2)

Solution :

(a) 3 + 2(x + 1)

= 3 + 2x + 2

= 2x + 3 + 2

= 2x + 5

(b) 7(x + 1) + 4(x + 3)

= 7x + 7 + 4x + 12

= 7x + 4x + 7 + 12

= 11x + 19

(c) 3(x – 8) + 5(2x – 3)

= 3x – 24 + 10x – 15

= 3x + 10x – 24 – 15

= 13x – 39

(d) 4(x + 11) + 13(2 + x)

= 4x + 44 + 26 + 13x

= 17x + 70

(e) 12(x – 6) + (x + 7)y

= 12x – 12 + xy + 7y

(f) x(x + 3) + x(3x + 1)

= x2 + 3x + 3x2 + x

= (x2 + 3x2) + (3x + x)

= 4x2 + 4x

(g) 10x(x + 5) – 2

= 10x2 + 50x – 2

(h) x(x + 8y) + 5x(x + y)

=  x2 + 8xy + 5x2 + 5xy

= 5x2 + x2 + 8xy + 5xy

= 6x2 + 13xy

(i) 4(a + 10b) + 3(a + b)

= 4a + 10b + 3a + 3b

= 4a + 3a + 10b + 3b

= 7a + 13b

Question no – (3)

Solution :

(a) (5a + 11b) (8a + 9b)

= (5a. 8a) + (5a, 9b) + (11b.8a) + (11b.9b)

= 40a2 + 45ab + 88ab + 99b2

= 40a2 + 133ab + 99b2

(b) (11p – 9p) (4p + 17p)

= (110.4p) – (9q .4p) + (11p.17q – (9q .17q)

= 44p2 – 36pq + 187pq – 153q2

= 44p2 + 151pq – 153q2

(c) (7x2 + 4y2) (3x2 + 8y2)

= (7x2. 3x2) + (7x2. 8y2) + (4y2.3x2) + (4y2.8y2)

= 21x4 + 56x2y2 + 12x2y2 + 32y4

= 21x4 + 68x2y2 + 32x4

(d) (13xy – 3z) (15xy – 8z)

= (13xy.15xy) – (3z.15xy) – (13xy.8z) + (3z, 8z)

= 195x2y2 – 45xyz – 104xyz + 24x2

= 195x2y2 – 149xyz + 25z2

Question no – (4)

Solution :

(a) (6a + 8b – 11) (3a + 7b)

6a + 8b – 11

3a + 7b
————————————————-
18a2 + 24ab – 33a

+ 421b + 56b2 – 77b
————————————————-
18a2 + 66ab – 33a + 56b2 – 77b

The product will be 18a2 + 66ab – 33a + 56b2 – 77b

(b) (3x2 + 7x – 6) (4x + 11)

3x2 + 7x – 6

4x + 11
————————————————-
12x3  28x2 – 24x

33x2 + 77x – 66
————————————————-
12x3 + 6 1 x2 + 53x – 66

(c) (7p2 – 13pq + 6q2)

7p2 – 13pq + 6q2

5p – 79
————————————————-
35p3 – 65p2q + 30pq2

– 49p2q + 91pq2 – 42q3
————————————————-
35p3 – 114p2q + 121p2q – 42q3

(d) (3m2 + 7mn + 9n2) (4m – 5n)

= 3m2 + 7mn +9n2

4m – 5n
————————————————-
13m3 + 28m2n + 36mn2

– 15m2n – 35mn2 – 45n2
————————————————-
12m3+ 13m2n +mn2 – 45x3

(f) (3m2 + 7mn + 9n2) (4m – 5n)

3m2 + 7mn +9n2

4m – 5n
————————————————-
13m3 + 28m2n + 36mn2

– 15m2n – 35mn2 – 45n2
————————————————-
12m3+ 13m2n +mn2 – 45x3

Algebraic Expressions Exercise 10.3 Solution :

Question no – (1)

Solution :

(a) 16x2 + 72xy + 81y2

(b) 9a2 + 48ab + 64b2

(c) 9x2 + 24xy + 16y2

(d) 4p2 + 60pq +225q2

(e) 25x2 + 81y2 + 90xy

(f) 9p2 + 42pq + 49q2

(g) x2y2 + 4xy + 4

(h) 121x2 + 110x + 25

(i) 64m2 + 80m + 4

(j) 49a2/9 + 2ab + 9b2/49

(k) 25a2b2 + 70ab + 49

(l) 3 + 2√2

Question no – (2)

Solution :

(a) a2 – 6a + 9

(b) 4x2 – 20x + 25

(c) 25a2 -60ab + 36b2

(d) a2b2 – 2abcd+ c2d2

(e) 9m2/4 – 2mn + 4n2/9

(f) 36a2b2 – 84abc + 49c2

(g) 16x2 – 24xy + 9y2

(h) 9x2 + 1/9x2 – 2

Question no – (3)

Solution :

(a) 24ab

(b) 80ab

(c) 180xy

(d) 420ab

Question no – (4)

Solution :

(a) a2 – 4

(b) x2 – 49

(c) x2 – 9

(d) 64x2 – 25y2

(e) 16xy – y2

(f) p2 – 4q2

(g) 9x2 – 16

(h) 49x2 – 16/y2

Question no – (5)

Solution :

(i) – (a) (104)2

= (100 + 4)2

= (100)2 + 2.100.4 + (4)2

= 10000 + 800 + 16

= 10816

(i) – (b) (103)2

= (100 + 3)2

= (100)2 + 2.100.3 + (3)2

= 10000 + 600 + 9

= 10609

(i) – (c) (54)2

= (50 + 4)2

= (50)2 + 2.504 + (4)2

= 2500 + 400 + 16

= 2916

(i) – (d) (52)2

= (50 + 2)2

= (50)2 + 2.50.2 + (2)2

= 2500 + 200 + 4

= 2704

(i) – (e) (10.1)2

= (10 + 01)2

= (10)2 + 2.10 (0.1) + (0.1)2

= 100 + 2 + 001

= 102021

(i) – (f) (10.2)2

= (10 + 0.2)2

= (10)2 + 2.10. (0.2)2

= 100 + 4 + 0.04

= 104.04

(ii) – (a) (97)2

= (100 – 3)2

= (100)2 – 2.100.(3) + (3)2

= 10000 – 600 + 9

= 9409

(ii) – (b) (98)2

= (100 – 2)2

= (100)2 + (2)2

= 10000 – 400 + 4

= 9604

(ii) – (c) (49)2

= (50 – 1)2

= (50)2 – 2.50.1 + (1)2

= 2500 – 100 + 1

= 2401

(ii) – (d) (48)2

= (50 – 2)2

= (50)2 – 2.50.2 + (2)2

= 2500 – 200 + 4

= 2304

(ii) – (e) (9.9)2

= (10 – 0.1)2

= (10)2 – 2.(10) (0.1) + (0.1)2

= 100 – 2 + 0.01

= 98.01

(ii) – (f) (9.8)2

= (10 – 0.2)2

= (10)2 – 2 (10) (0.2)2 + (0.2)

= 100 – 4 + 0.04

= 96.04

(iii) – (a) 59 × 61

= (60 – 1) (60 + 1)

= (60)2 – (1)2

= 3600 – 1

= 3599

(iii) – (b) 102 × 98

= (100 + 2) (100 – 2)

= (100)2 – (2)2

= 10000 – 4

= 9996

(iii) – (c) 48 × 52

= (50 – 2) (50 + 2)

= (50)2 – (2)2

= 2500 – 4

= 2496

(iii) – (d) 103 × 97

= (100 + 3) (100 – 3)

= (100)2 – (3)2

= 10000 – 9

= 9991

(iii) – (e) 49 × 51

= (50 – 1) (50 + 1)

= (50)2 – (1)2

= 2500 – 1

= 2499

(iii) – (f) 612 – 602

= (61 + 60)

= (61 + – 60)

= (121) (10

= 121

(iii) – (g) 732 – 722

= (73 + 72) (73 – 72)

= 145 × 1

= 45

(iii) – (h) 762 – 242

= (76 + 24)

= (76 – 24)

= 100 × 48

= 4800

(iii) – (i) 572 – 432

= (57 + 43) (57 – 43)

= 100 × 14

= 1400

Question no – (6)

Solution :

In the question we get,

(x + y) = 13

xy = 22

Now, x2 + y2

= (x + y)2 – 2xy

= (13)2 – 2.22

= 169 – 44

= 125

Therefor, The value of x2 + ywill be 125

Question no – (7)

Solution :

In the question we get,

(x – y) = 5

xy = 36

Now, (x – y)2 + 2xy

= (5)2 + 2.36

= 25 + 72

= 97

Question no – (8)

Solution :

From the question we get,

(x2 + y2) = 74

xy = 35

Now, (a) x + y

= (x + y)2 = x2 + y2 + 2xy

= or, (x + y)2 = 74 + 2.35

or, (x + y)2 = 74 + 70

or, (x + y)2 = 144

or, (x + y) = √144

or, (x + y) = 12

Therefore, The value of x + y will be 12

(b) x – y

= (x – y)2 = x2 + y2 – 2xy

or, (x – y)2 = 74 – 2.35

or,(x – y)2 = 74 – 70

or, (x – y)2 = 4

or, (x – y) = √4

or, (x – y) = 2

Therefore, The value of x – y will be 2

Question no – (9)

Solution :

In the question we get,

(2x + 3y) = 14

xy = 8

Now, (2x)2 + 93y)2

= (2x + 3y)2 – 2 (2x) (3y)

= (14)2 – 12xy = (14)2 – 12(8)

= 196 – 96

= 100

Therefor, The value of  4x2 + 9y2  will be 100.

Question no – (10)

Solution :

In the question we get,

(x + y) = 15

xy = 54

Now, (x – y) = √(x +y)2 – 4xy

= √(15)2 – 4.54

= √225 – 216

= √9 = 3

= (x2 – y2)

= (x + y) (x – y)

= 15 × 3

= 45

Question no – (11)

Solution :

In the question we get,

x + 1/x = 5

(a) x2 + 1/x2

= (x + 1/x)2 – 2.x 1/x

= (52)2 – 2 = 25 – 2

= 23

Therefor, The value of x2 + 1/x2 will be 23

(b) x4 + 1/x4

= (x2 + 1/x) – 2.x2 – 1/x

= (23)2 – 2

= 529 – 2

= 527

Therefore, the value of x4 + 1/x4 will be 527

Question no – (12)

Solution :

In the question we get,

x + 1/x = √3

(a) x2 + 1/x2

= (x + 1/x)2 – 2x 1/x

= (√3)2 – 2

= 3 – 2

= 1
Therefore, the value of x2 + 1/x2 will be 1

(b) x4 + 1/x4

= (x2 + 1/x) – 2.x2 – 1/x2

= (1)2 – 2

= 1 – 2

= – 1

Hence, the value of x4 + 1/x4 will be – 1

Question no – (13)

Solution :

(a)  If x2 + 1/x2 = 62

Now, x2 + 1/x = 62

or, (x + 1/x)2 – 2. x. 1/x = 62

or, (x + 1/x)2 – 2 = 62

or, (x + 1/x)2 = 64

or, x + 1/x

= √64

= 8

Thus, the value of x2 + 1/x  will be 8

(b)  x2 + 1/x2 = 79

= x2 + 1/x2 = 79

= (x + 1/x)2 – 2.x. 1/x = 79

or, (x + 1/x)2 – 2 = 79

or, (x + 1/x) = 81

(x + 1/x)

= √81

= 9

Hence, the value of x2 + 1/x2 will be 9

Question no – (14)

Solution :

(a) x2 + 1/x2 = 102

Now, x2 + 1/x2 = 102

or, (x – 1/x)2 + 2.x = 1/x = 102

or, (x – 1/x)2 = 100

or (x – 1/x)

= √10

Therefore, The value of x2 + 1/x2 will be √10

(b) x2 + 1/m2 = 38

Now, x2 + 1/m2 = 38

or, (x – 1x)2 + 2.x /x = 38

or, (x – 1/x) = 36

or, (x – 1/x)

= √36

= 6

Therefor, The value of x2 + 1/m2 will be 6

Question no – (15)

Solution :

In the question we get,

9x2 + y2 = 397

xy = 38

Now, = (3x + y)

= √9x2 + y2 + 2xy

= √397 + 2.38

= √397 + 76

= √473

Therefore, the value of 9x2 + y2 will be √473

Algebraic Expressions Exercise 10.4 Solution :

Question no – (1)

Solution :

(a) 48x3y3z3 by 4x2y2z2

= (4x2y2) (12xyz)/(4x2y2z2)

= 12xyz2

(b) – 81a4b5c7 by a3b2c2

= (27a3b2c2) (- 3ab3c5)/(27a3b2c2)

= – 3ab3c

(c) 72a3b3c3 by 24a2bc2

= (24a2bc2) (3ab2c)/(24a2bc2)

= 3ab2c

(d) 28x2y2x2 by – 4xyz2

= (- 4xyz2) (- 7xy)/(- 4myz2)

= – 7xy

(e) 55a8b8 by 11a5b5

= (11a5b5) (5a3b3)/(11a5b5)

= 5a3b3

(f) 24a3b3 – 18a2b2 + 12ab by – 6ab

= (- 6ab) (- 4a2b2 + 3ab – 2)/(- 6ab)

= – 4a2b2 + 3ab – 2

Question no – (2)

Solution :

(a) y2 + 5y – 36 by y + 9

(b) x2 + 14x + 48 by x + 6

(c) 5×2 + 31x – 28 by 5x – 4

(d) 3×2 + 10x + 3 by 3x + 1

(e) x2 – 8x – 33 by x + 3

(f) 6×2 – 31x + 40 by 2x – 5

(g) x2 + 2x – 35 by x + 7

(h) 15×2 + x – 6 by 3x + 2

(i) 4a2 – 24a2 + 64a – 128 by a – 4

Question no – (3)

Solution :

(a) 3×2 + 5×2 – 8x – 28 by x – 2

(b) 4×3 + 6×2 – 8x – 5 by 2x + 1

(c) 2×3 + 5×2 – 11 – 14 by 2x + 7

(d) 6×3 + 5×2 – 3x – 2 by 3x – 2

(e) 6×3 + 19×2 + 11x – 6 by 3x – 1

(f) 2×3 – 7×2 – 8x – 3 by 2x – 3

Revision Exercise Questions Solution :

Question no – (1)

Solution :

= [(14a + 9b – 8c) + (12b + 13c – 15a)] – [(8a – 17b + 4c) + (11a – 16b + 17c)]

= [(14a + 15a) + (b + 12b) + (8c + 13c)] – [(8a + 11a) + (- 17b – 16b) + (4c + 17c)]

= [- a + 21b + 5c] – [19a – 33b + 21c]

= (- a – 19a) + (12b + 33b) + (5c + 21c)

= – 2a + 45b – 16c

Question no – (2)

Solution :

8y3 + 7y2 + 3y

3y – 4
——————————————-
24y4 + 21y3 + 9y2

– 32y3 – 28y2 – 12y
——————————————-
24y4 – 11y3 – 19y2 – 12y

Question no – (3)

Solution :

(a) (503)2

= (503)2 = (500 + 3)2

= (500)2 + 2.500.3 + (3)2

= 250000 + 3000 + 9

= 253009

(b) (8.7)– (1.3)2

= (87 + 1.3) (87 – 1.3)

= 10 × 7.4

= 74

Question no – (4)

Solution :

(a) (2x + 3y/4)2

= (2x)2 + 2.2x. 3y/4 + (3y/4)2

= 4x2 + 3xy + 9y2/16

(b) (4x2 + 5y2) (4x2 – 5y2)

= (4x2)2 – 5y)2

= 16x4 – 25y4

(c) (x + 2/5) (x – 2/5)

= x2 – (2/5)2

= x2 – 4/25

(d) (xy – 2z)2

= (xy)2 – 2.x.y (+ 2z) + (2z)2

= x2y2 – 4xy2 +4z2

(e) (a + bc) (a – bc) (a2 + b2c2)

= (a2 – b2c2) (a2 + b2c2)

= (a2)2 – (b2c2)

= a4 – b4c4

(f) (2a/3b – 3b/2a)2

= (2a/3b)2 – 2 2a/3b 3b/2a

= 4a2/9b2 – 2 + 9b2/4a2

(g) (- 3a – 4b)2

= [(- 3a + 4b)]2 – (3a + 4b)

= (3a)2 + 3a.4b + (4b)2

= 9a2 + 24ab + 16b2

(h) (a – 7) (a + 7)

= a2 – (7)2

= a2 – 49

Question no – (5)

Solution :

= 104 × 96

= (100 + 4) (100 – 4)

= (100)2 – (4)2

= 10000 – 16

= 9984

Therefore, The value of 104 × 96 will be 9984.

Next Chapter Solution :

Updated: June 24, 2023 — 11:35 am