# Frank Learning Maths Class 8 Solutions Chapter 5

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## Frank Learning Maths Class 8 Solutions Chapter 5 Algebraic Expressions

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 5 Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions. Here students will find solutions for Exercise 5.1, 5.2 and 5.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Algebraic Expressions Exercise 5.1 Solution

Question no – (1)

Solution :

(a) 6x + 4y – 3z

7x – 11y – 9z

14x + 8y – 6z

——————————

27x + y – 18z

or

= (6x + 4y – 3z) + (7x – 11y – 9z) + (14x + 8y – 6z)

= 27x + y – 18z

(b) 7p – 8q + 11r + 13

10p – 13q + 0 + 18

8p – 5q – 14r + 12

5p + 17q + 14r

——————————

30p – 9q + 11r + 43

(c) 30p – 9q + 11r + 43

8x2 + 7x + 12

17x2 – 15x – 21

– 9x2 + 11x + 4

4x2 – 18x + 19
——————————

20x2 – 15x + 14

(d) 6ax – 2by + 3cz

11ax – 16by – 15cz

– 9ax – 3by + 10cz
——————————

8ax – 21by – 2cz

(e) 15m2n – 17mn + 8mn2

13 m2n – 15mn – 9mn2

12m2n + 21mn – 10cz
————————————-

40m2n – 11mn – 15m2n

(f) 13x + 17y – 19z + 2

14x + 12y + 0 – 31

– 15x + 6z + 12

2x + y + 11z + 9
————————————-

14x + 30y – 2z – 8

Question no – (2)

Solution :

(a) 12x + 7y – 15z

17x + 8y – 15z

(-)    (-)      (+)

—————————-

– 5x – y + 0

(b) 21ax + 15by – 6cz

12ax + 6by + 11cz

(-)    (-)    (-)

—————————-
9ax + 9by – 17cz

(c) 16x3 + 14x2 – 9x + 15

17x– 19x2 + 13x – 21

(-)     (+)     (-)       (+)

—————————-

– x3 + 33x2 – 22x + 36

(d) 7p + 11q – 2r + 9

– 8p – 4q + 6r + 15

(+)    (+)    (-)    (-)
—————————-

15p + 15q – 8r – 6

Question no – (3)

Solution :

9a² – 17ab – 3b²a – 10

7a² + 18ab – 9b²a + 15

(-) (-) (+) (-)
————————————

2a² – 35ab + 6b²a – 25

(2a² – 35ab + 6b²a – 25) must be subtracted for get the given answer.

Question no – (4)

Solution :

We know that, Third side = perimeter – sum of two sides.

= (16x3 + 14x2 – 9x + 15) – (7x3 + 3x2 – 2x + 3 + 4x+ x2 – 5x + 7)

= (16x3 + 14x2 – 9x + 15) – (11x3 + 4x2 – 7x + 10)

= 5x3 + 10x2 – 2x + 5

Hence, the third side will be 5x3 + 10x2 – 2x + 5

Question no – (5)

Solution :

We know that, Perimeter of a rectangle is 2 (sum of adjacent sides)

= 2(-6p3 + 7p2q2 + 11pq + 7p2q2 + 4p2 + 9pq)

= 2(-6p3 + 14p2q2 + 20pq + 4p2)

= 8p2 + 28p2q2 + 40pq – 12p3

Therefore, its perimeter will be 8p2 + 28p2q2 + 40pq – 12p3

Question no – (6)

Solution :

(13 + 19b+ 12c + 14a – 21b + 11c) – (17a + 13b – 15c + 8a + 12b – 18c)

= 27a – 2b + 23c – 25a – 25b + 33c

= 2a – 27b + 56c …(Answer)

Question no – (7)

Solution :

As per the given question,

A = 5x + 11y – 75z

B = 12x + 13y + 19z

C = 7x – 6y + 21z

(a) A + B + C

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

= 24x – 8y + 25z

(b) A – B – C

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

= -14x + 30y – 55z

(c) A + B – C

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

= 10x + 4y – 17z

(d) 2A + 3B + C

= 10x + 22y – 30z + 36x – 39y + 57z + 7x – 6y + 217

= 53x – 23y + 48z

Question no – (8)

Solution :

Let other side = x

We know that, Perimeter = 2 (one side + other side)

16a + 8b – 6c = 2 (5a + 3b – 4c + x)

= 8a + 4b – 3c = 5a + 3b – 4c + x

= x = 3 + b + c

Hence, the other side will be 3 + b + c

Question no – (9)

Solution :

As we know that, Perimeter = Sum of 3 side

18p + 12q + 13r = 34 – 4r + 6p + 5q + 7r+ x

= 18p + 12q + 13r = 11p + 8q +3r + x

= x = 7p + 4q + 10r

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

Algebraic Expressions Exercise 5.2 Solution

Question no – (1)

Solution :

(a) (5a + 11b) (8a + 9b)

= (5a × 8a) (11b × 8a) + (5a × 9b) + (11b × 9b)

= 40a2 + 88ab + 45ab + 99b2

= 40a2 + 133ab + 99b2

(b) (11p – 9p) (4p + 17q)

= 44p2 – 36pq + 187pq – 153q2

= 44p2 – 18pq – 153q2

(c) (7x2 + 4y2) (3×2 + 8y2)

= 21x4 + 12x2y2 + 56x2y2 + 32y4

= 214 + 68x2y+ 32y4

(d) (13xy – 3z) (15xy – 8z)

= 195x2y2 – 45yz – 104xyz + 24z2

= 195x2y2 – 149xyz + 24z2

(e) (7x3y + 5zx) (11x3y  + 8xz)

= 77x3 + 55x4yz + 56x4yz + 40x2z2

= 77x3y + 1.11x4yz + 4ox2z2

(f) (2,5a + 0.3b) (1.5c + 0.7d)

= (2,5a + 0.3b/10) (1.5c + 0.7d/10)

= 3a5a + 45bc + 175ad + 21bd/10

= 37.5a + 4.5bc + 17.5ad + 2.1bd

(g) (2x/3 – 5y/4) (6x/5 + 7y/3)

= 3x2/5 – 3xy/2 + 14xy/9- 35y2/12

= 4x2/5 – (27xy – 28xy/18) – 35y2/12

= 4x2/5 + xy/18 – 35y2/12

(h) (2x3 – 3y3) (3x2 – 4y2)

= 6x5 – 9x2yu– 8x3y+ 12y5

Question no – (2)

Solution :

(a) (3a + 7b) (6a + 8b – 11c)

= 18a2 + 24ab – 33ac + 42ab + 56b2 – 77bc

= 18a2 + 6ab – 33ac – 77bc + 56b2

(b) (4x2 + 11) (3x2 + 17x – 16)

= 12x+ 68x3 =- 64x2 + 33x2 + 187x – 176

(c) (5p2q2 – 7) (7p2q – 13p + 6q)

= 35p4q3 – 65p2q2 + 30p2q3 – 49p2q + 91p – 42q

(d) (xyz) (6x3 + 7cx2 – 12x + 21)

= xyz (6x3 + 7cx2 – 12x + 21)

= 6x4yz + 7x3yz – 12x2yz + 21xyz

(e) (4m2n2) (3m2n + 7mn + 9mn2 – 15)

= 4m2n2 (3m2n + 7mn + 9mn2 – 15)

= 12m4n3 + 28m3n3 + 36m3n4 – 60m2n2

(f) (3x + 4y – 11) (5x – 3y – 8)

= 15x2 – 9xy – 24x + 20xy – 20y2 – 32y – 55x + 33y + 88

= 15x2 + 11xy – 79x + y – 12y2 + 88

(g) (2x2 + 4x + 15) (5x2 – 7x – 12)

= 10x4 + 10x3 – 24x2 + 20x3 – 28x2 – 48x + 75x2 – 105x – 180

= 10x– 14x3 + 51x2 – 153x – 180

(h) (5x3 – 3x2 + 6x) (4x2 – 7x + 9)

= 20x5 – 35x4 + 45x3 – 12x4 + 21x3 – 27x2 + 24x3 – 42x2 + 54x

= 20x4 – 47x4 + 90x3 – 69x2 + 54x

(i) (a + 1/a) (a3 + 1/a3)

= a4 + a2 + 1/a2 + 1a4

(j) (a – 1/a (a4 + 1/a4)

= a5 + 1/a3 – a3 – 1/a5

Question no – (3)

Solution :

(a) Given, a = 2

3a (a2 – 7) + 2 (a + 2)

= 6 (4 – 7) + 2 (2 + 2)

= = 6 (-3) + 2 × 4

= – 18 + 8

= -10

Therefore, the value will be -10.

(b) Given, a = 0

3a (a2 – 7) + 2 (a + 0)

= 0 + 2 (0 + 3)

= 6

Thus, the value will be 6

(c) Given, a = – 3

3a (a2 – 7) + 2 (a + -3)

= -9 (9 – 7) + (-3 + 3)

= -9 × 2 + 0

= -18

Therefore, the value will be -18

Question no – (4)

Solution :

(a) x = 2 and y = 3

2 × 4 (6 – 4) + 9 (2 + 2)

= 8 × 2 + 9 × 4

= 16 + 36

= 52

Thus, the value will be 52

(b) x = -1 and y = 5

2 ( -5 – 4) + 3 × 5 (- 1 + 2)

= 2 (-9) + 15 (1)

= – 18 + 15

= – 3

Therefore, the value will be -3

Question no – (5)

Solution :

(a) 4(2x + 5y + 8) +3 (5x + 7y – 9)

= 8x + 20y + 32 + 15x + 21y – 27

= 23x + 41y + 5…(Simplified)

(b) (p + q) (p – q) + (p + r) (r – p) – (r2 – q2)

= p2 –  q2 + r2 – p2 – r+ q2

= 0…(Simplified)

(c) (w + x) (y – z) + (w – x) (y +z) + 2 (wy + xz)

= wy + xy – wz – xz + wy – xy + wy – xz + 2 (wy + xz)

= 2 (wy – xz) + 2 (wy + xz)

= 2wy – 2xz + 2wy + 2xz

= 4wy…(Simplified)

(d) (a + b) (a2 – ab + b2)

= a3 + b3…(Simplified)

(e) (x – y) (x2 + xy + y2)

= x3 – y3…(Simplified)

(f) (a + b) (a2 + 2ab + b2)

= (a + b)3

Or,

= a3 + 3a2b + 3ab2 + b3…(Simplified)

(g) ab (cx + y) + bc (a – y) – 2ca (bx + y)

= abcx + aby + abcx – bay – 2abcx – 2acy

= aby – bcy – 2acy…(Simplified)

Question no – (6)

Solution :

 Length (l) Breadth (b) Height (h) Volume (=lbh) (a) 4xy -12xz -5zy 24x2y2z2 (b) 7a2c 2a2b2 2abc2 42a5b3c3 (c) -2mn -9lmn 6nl 108lm2n2 (d) -x4 -x3 -x2 -x9 (e) 3ax 13by 23cz 897abcxyz (f) 15x4 10×5 25 3750x9

Algebraic Expressions Exercise 5.2 Solution

Question no – (1)

Solution :

As per the given question,

a = 9,

b = 4

∴ L.H.S = (a – b)2

= (9 – 4)2

= 52

= 25

∴ R.H.S, = a2 – 2ab + b2

= 81 – 72 + 16

=  25

L.H.S = R.H.S …(Verified)

Question no – (2)

Solution :

According to the question,

x = 11

a = 4

b = -7

∴ L.H.S, = (x + a) (x + b)

= (11 + 4) (11 – 7)

= 15 × 4

= 60

∴ R.H.S, = x2 + (a + b) x + ab

= 121 + (-3) × 11 + 4 (- 7)

= 121 – 33 – 28

= 60

Hence, L.H.S = R.H.S …(Verified)

Question no – (3)

Solution :

(a) Given, (4x + 9y) (4x + 9y)

= (4x + 9y)2

= 16x2 + 72 xy + 18y2

Therefore, the product will be 16x2 + 72 xy + 18y2

(b) Given, (3a2 – 8b2) (3a2 – b2)

= (3a2 – 8b2)

= (3a2)2 – 2 × 3a2 × 8b2 + (8b2)2

= 9a4 – 48a2b2 + 64b4

Thus, the product will be, 9a4 – 48a2b2 + 64b4

(c) Given, (3x2/4m+ 4y2/7) (3x2/4 + 4y2/7)

= (3x2/4 + 4y2/7)2

= (3x2/4 + 4y2/7)2

= (3x2/4)2 + 2 × 3x2/4 × 4yz/7 + (4y2/7)2

Therefore, the product will be, (3x2/4)2 + 2 × 3x2/4 × 4yz/7 + (4y2/7)2

(d) Given, (13p – 15q) (13p + 15q)

= (13p)– (15q)2

= 169p2 – 225q2

(e) Given, (0.5x + 0.9y) (0.5x – 0.9y)

= (0.5x/10)2 – (0.9y/10)2

= 0.5z/10 – 81y2/100

= 25x2 – 81y2

Therefore, the product will be 25x2 – 81y2

(f) Given, (-3p + 7q) (- 3p + 7q)

= (7q – 3p)2

= 49q2 – 42pq + 9p2

Thus, the product will be 49q2 – 42pq + 9p2

(g) Given, (x2y2 + 2) 9x2y2 + 2)

= (x2y2 + 2)2

= x2y2 +2)

= x4y4 + 4x2y2 + 4

Therefore, the product will be x4y4 + 4x2y2 + 4

(h) Given, (11x + 9) (11x + 15)

= (11z + 9) (11x + 9 + 6)

= (11x + 9)2 + 6 (11x + 9)

= 121x2 + 198x + 81 + 66x + 54

= 121x2 + 264x + 135

Thus, the product will be 121x2 + 264x + 135

(i) Given, 9m+ 7) (9m2 – 5)

= (9m2 + 5 + 2) (9m2 – 5)

= (9m2 + 5) (9m2 – 5) + 2 (9m2 – 5)

= (9m2)2 (5)2 + 18m2 + 10

= 81 m4 – 25 + 18m2 – 10

= 81m4 + 18m2 – 35

Hence, the product will be 81m4 + 18m2 – 35

(j) Given, (7a + 13b) (7a – 18b)

= (7a + 13b) (7 – 13b – 5b)

= (7a + 13b) (7 – 13b) – 5b (7a + 13b)

= (7a)2 – (13b)2 – 35ab2 – 354b + 45b2

= 49a2 – 169b2 – 35ab + 65b2

= 49a2 – 25ab – 104b2

Therefore, the product will be 49a2 – 25ab – 104b2

(k) Given, (x2y2 + 17) (x2y2 – 21)

= (x2y2 + 17) (x2y2 – 17 – 4)

= x4y4 – 289 – 4x2y2 – 68

= x4y4 – 4x2y2 – 357

Therefore, the product will be x4y4 – 4x2y2 – 357

Question no – (4)

Solution :

(a) (a2b2 + c2d2)2

= a4b4 + 2a2b2c2d+ c4d4

(b) (3/2m – 2/3n)2

= (3/2m – 2/3n)2

= 9m2/4 – 2mn – 4n2/9

(c) (0.6a – 0.7b)2

= 0.36a2– 0 84ab + 0.49b2

(d) (4x2 + 3y2)2

= 16x4 + 24x2y2 + 9y4

Question no – (5)

Solution :

(a) (a2 + b2)2 – (a2 – b2)2

= 4a2b2…(Simplified)

(b) (3.5a – 4.5b)– (3.5a + 4.5b)2

= 4 × 3.5a/10 × (- 4.5b/10)

= – 63ab…(Simplified)

(d) (12a + 17b)2 – (12a – 17b)2

= 4×12a × 17b

= 816ab…(Simplified)

(e) 1.98 × 1.98 – 1.02 × 1.02/0.96

= (1.98)2 – 1.02)2/0.96

= (1.98 + 1.02) (198 – 1.02)/096

= 3 × 96/0.96

= 3…(Simplified)

(f) (x + y) (x – y) (x2 + y2)

= (x2 – y2) (x2 + y2)

= x4 – y4…(Simplified)

(g) (r + s)2 – (r + s) (r – s)

= (r + s) – (r + s – r + s)

= (r + s) × 2s

= 2 (sr + s2)…(Simplified)

Question no – (6)

Solution :

(a) Given, (5x + 11)2 – (5x – 11)2 = 220x

= 4 × 5x × 11

= 220x …(Proved)

(b) Given, (4ax + 3y)2 – (4ax – 3y)2 = 48axy

= 4 × 4ax × 3y

= 48axy …(Proved)

(c) Given, (5/6 p – 6/5q) + 2pq = 25/36 = 25/36 p2 + 36/25 q2

= (5/6 p)2 + (6/5 q)2

= (5/6p – 6/5 q) + 2. 5/6.6/5 q

= (5/6 p – 6/5q)2 + 2pq (Proved)

(d) Given, (x + y) (x – y) + (y + z) (y – z) + (z + x) (z – x) = 0

= x2 – y2 + y2 – x2 + x2 – x2

= 0 …(Proved)

Question no – (7)

Solution :

(a) (93)2

= (90 + 3)2

= 8100 + 2 × 90 × 3 + 9

= 8100 + 540 + 9

= 8649

(b) (103)2

= (100 + 3)2

= 10000 + 600 + 9

= 10609

(c) (54)2

= (50 + 4)2

= 2500 + 400 + 6

= 2916

(d) (997)2

= (1000 – 3)2

= 1000000 – 6000 + 9

= 994,009

(e) (10.1)2

= (10 + 1)2

= 100 + 2 + 01

= 102.01

(f) (8.6)2

= 1/100 [(86)2]

= 1/100 [(90 – 4)2]

= 1/100 [8100 – 120 + 16]

= 7396/100 = 73.96

(g) 103 × 97

= (100 + 3) (100 – 3)

= (100)2 – (3)2

= 1000000 – 9

= 999,991

(h) 53 × 47

= (50 + 3) (50 – 3)

= (50)2 – (3)2

= 2500 – 9

= 2491

(i) (12.5)2 – (7.5)2

= (12.5 + 7.5) (12.5 – 7.5)

= (20.0) × (5)

= 100

(j) (10.2)2 – (9.8)2

= (10.2 + 9.8) (10.2 – 9.8)

= 20 + 4

= 20.4

(k) 93 × 109

= (100 – 7) (100 + 7 + 2)

= (100 – 7) (100 + 7) + 2 × 93

= (100)2 – (7)2 + 186

= 10000 – 49 + 186

= 10,137

(l) 54 × 59

= (50 + 4) × (50 + 4 + 5)

= (50 + 4)2+5 × 54

= 2500 + 400 + 16 + 270

= 3186

(m) 12.5 × 10.5

= 12.5/10 × 12.5/10

= 1/100 [(10 + 5 + 20) × (100 + 5)]

= 1/100 [(100 + 5)2 + 20 × 105)]

= 1/100 910000 + 1000 + 25 + 2100)

= 1/100 × 12125

(n) 997 × 1007

= (100 – 3) (100 + 3 + 4)

= (100)2 (3)+ 4 × 997

= 100000 – 9 3988

= 1003979

(o) 82 × 85

= (80 + 2) (80 + 2 + 3)

= (80 + 2)2 × 3 × 82

= 6400 + 320 + 4 + 246

= 6970

Question no – (8)

Solution :

In the question we get,

(x + y) = 13

xy = 22

x2 + y= (x + y)2 – 2xy

= (13)2 – 2 × 22

= 169 – 44

= 125

Therefore, the value of x2 + y is2 125

Question no – (9)

Solution :

As per the given question,

(x – y) = 5

xy = 36

∴ x+ y2 = (x – y)2 + 2xy

= 25 + 72

= 97

Therefore, the value of x+ ywill be 97

Question no – (10)

Solution :

Given in the question,

(x2 + y2) = 74

xy = 35

x2 + y2 = (x + y)2 – 2xy

= 74 = (x + y)2 – 2 × 35

= (x + y) = 74 + 70

= x + y = √144

= 12

Now,  x2 + y2 = (x – y)2 – 2xy

= 7x (x – y)2  + 2 × 35

=  (x – y)2 = 74 – 70

=  x – y = √4

= 2

Question no – (11)

Solution :

(a) Given, x2 + 1/x2

= x2 + 1/x2

= (x + 1/x)2 2.x 1/x

= 52 – 2

= 25 – 2

= 23

(b) Given, x + 1/x = 5,

x4 + 1/x4

= (x2 + 1/x2)2 – 2.x4 1/x4

= (23)2 – 2

= 529 – 2

= 527

Question no – (12)

Solution :

According to the given question,

x2 + 1/x2 = 3

(a) x – 1/x

= x2 × 1/x2 = 3

= x2 + 1/x2 = (x – 1/x)2 + 2.x 1/4

= 3 = (x – 1/x)2 + 2

= (x – 1/x)2 = 1

= x – 1/x = 1

(b) x + 1/x

= x2 + 1/x2 = (x + 1/x)2 – 2.x.1/x

= 3 = (x + 1/x)2 = 2

= x + 1/x = √5

Previous Chapter Solution :

Updated: June 5, 2023 — 7:24 am